Solutions to old Exam 2 problems
Hi students!
I am putting this version of my review for the Final exam
review (place and time TBA) here on the website. DO NOT
PRINT!!; it is very long!! Enjoy!!
Your course chair, Bill
PS. There are probably errors in some of the solutions
presented here and for a few problems you need to complete
them or simplify the answers; some questions are left to you
the student. Also you might need to add more detailed
explanations or justifications on the actual similar problems on
your exam. I will keep updating these solutions with better
corrected/improved versions. The first 6 slides are from Exam
2 practice problems but the material falls on our Final exam.
After our exam, I will place the solutions to it right after this
slide.
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Solution:
Note that the region R defined by
√
{(x, y ) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} is equal to the region
{(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2 }.
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Solution:
Note that the region R defined by
√
{(x, y ) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} is equal to the region
{(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2 }.
Thus,
Z 1Z 1 p
Z Z p
x 3 + 1 dA =
R
0
√
x 3 + 1 dx dy
y
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Solution:
Note that the region R defined by
√
{(x, y ) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} is equal to the region
{(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2 }.
Thus,
Z 1Z 1 p
Z Z p
x 3 + 1 dA =
R
Z
0
1
Z
0
x2
p
x 3 + 1 dy dx
0
√
x 3 + 1 dx dy =
y
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Solution:
Note that the region R defined by
√
{(x, y ) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} is equal to the region
{(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2 }.
Thus,
Z 1Z 1 p
Z Z p
x 3 + 1 dA =
R
Z
0
1
Z
0
x2
0
Z
p
3
x + 1 dy dx =
0
1
hp
√
x 3 + 1 dx dy =
y
x3 + 1 y
ix 2
0
dx
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Solution:
Note that the region R defined by
√
{(x, y ) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} is equal to the region
{(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2 }.
Thus,
Z 1Z 1 p
Z Z p
x 3 + 1 dA =
R
Z
0
1
Z
0
x2
0
Z
p
3
x + 1 dy dx =
0
1
hp
x3
√
x 3 + 1 dx dy =
y
+ 1y
ix 2
0
Z
dx =
0
1
p
x 3 + 1 x 2 dx
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Solution:
Note that the region R defined by
√
{(x, y ) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} is equal to the region
{(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2 }.
Thus,
Z 1Z 1 p
Z Z p
x 3 + 1 dA =
R
Z
0
1
x2
Z
0
Z
p
3
x + 1 dy dx =
0
=
1
hp
0
1
3
Z
0
1
1
(x 3 + 1) 2 · 3x 2 dx
x3
√
x 3 + 1 dx dy =
y
+ 1y
ix 2
0
Z
dx =
0
1
p
x 3 + 1 x 2 dx
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Solution:
Note that the region R defined by
√
{(x, y ) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} is equal to the region
{(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2 }.
Thus,
Z 1Z 1 p
Z Z p
x 3 + 1 dA =
R
Z
0
1
x2
Z
Z
p
3
x + 1 dy dx =
0
=
0
0
1
3
Z
0
1
1
hp
x3
√
x 3 + 1 dx dy =
y
+ 1y
ix 2
Z
dx =
0
0
1
1
3 1 2
(x 3 + 1) 2 · 3x 2 dx = ( (x 3 + 1) 2 )
3 3
0
1
p
x 3 + 1 x 2 dx
Problem 23 - Exam 2 Fall 2006
Evaluate the integral
Z 1Z 1 p
0
√
x 3 + 1 dx dy
y
by reversing the order of integration.
Solution:
Note that the region R defined by
√
{(x, y ) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} is equal to the region
{(x, y ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x 2 }.
Thus,
Z 1Z 1 p
Z Z p
x 3 + 1 dA =
R
Z
0
1
x2
Z
Z
p
3
x + 1 dy dx =
0
=
0
0
1
3
Z
0
1
1
hp
x3
√
x 3 + 1 dx dy =
y
+ 1y
ix 2
0
Z
dx =
1
p
x 3 + 1 x 2 dx
0
1
1
3 1 2
2 3
(x 3 + 1) 2 · 3x 2 dx = ( (x 3 + 1) 2 ) = (2 2 − 1).
3 3
9
0
Problem 32(3) - From Exam 2
Find the iterated integral,
Z 1Z
0
x
2−x
(x 2 − y )dy dx.
Problem 32(3) - From Exam 2
Find the iterated integral,
Z 1Z
0
Solution:
Z
0
1 Z 2−x
x
2−x
(x 2 − y )dy dx.
x
(x 2 − y ) dy dx
Problem 32(3) - From Exam 2
Find the iterated integral,
Z 1Z
0
Solution:
Z
1 Z 2−x
2
2−x
(x 2 − y )dy dx.
x
Z
(x − y ) dy dx =
0
x
0
1
y2
x y−
2
2
2−x
dx
x
Problem 32(3) - From Exam 2
Find the iterated integral,
Z 1Z
0
Solution:
Z
1 Z 2−x
2
2−x
(x 2 − y )dy dx.
x
Z
1
(x − y ) dy dx =
0
Z
=
0
x
1
x 2 (2−x)−
0
(2 − x)2
x2
−(x 3 − ) dx
2
2
y2
x y−
2
2
2−x
dx
x
Problem 32(3) - From Exam 2
Find the iterated integral,
Z 1Z
0
Solution:
Z
2−x
(x 2 − y )dy dx.
x
2−x
y2
x y−
dx
(x − y ) dy dx =
2 x
0
0
x
Z 1
Z 1
(2 − x)2
x2
=
x 2 (2−x)−
−(x 3 − ) dx = 2
−x 3 +x 2 +x −1 dx
2
2
0
0
1 Z 2−x
2
Z
1
2
Problem 32(3) - From Exam 2
Find the iterated integral,
Z 1Z
0
Solution:
Z
2−x
(x 2 − y )dy dx.
x
2−x
y2
x y−
dx
(x − y ) dy dx =
2 x
0
0
x
Z 1
Z 1
(2 − x)2
x2
=
x 2 (2−x)−
−(x 3 − ) dx = 2
−x 3 +x 2 +x −1 dx
2
2
0
0
1
1
2
= − x 4 + x 3 + x 2 − 2x 2
3
0
1 Z 2−x
2
Z
1
2
Problem 32(3) - From Exam 2
Find the iterated integral,
Z 1Z
0
Solution:
Z
2−x
(x 2 − y )dy dx.
x
2−x
y2
x y−
dx
(x − y ) dy dx =
2 x
0
0
x
Z 1
Z 1
(2 − x)2
x2
=
x 2 (2−x)−
−(x 3 − ) dx = 2
−x 3 +x 2 +x −1 dx
2
2
0
0
1
1
2
1 2
= − x 4 + x 3 + x 2 − 2x = − + + 1 − 2
2
3
2 3
0
1 Z 2−x
2
Z
1
2
Problem 32(3) - From Exam 2
Find the iterated integral,
Z 1Z
0
Solution:
Z
2−x
(x 2 − y )dy dx.
x
2−x
y2
x y−
dx
(x − y ) dy dx =
2 x
0
0
x
Z 1
Z 1
(2 − x)2
x2
=
x 2 (2−x)−
−(x 3 − ) dx = 2
−x 3 +x 2 +x −1 dx
2
2
0
0
1
1
2
1 2
5
= − x 4 + x 3 + x 2 − 2x = − + + 1 − 2 = − .
2
3
2 3
6
0
1 Z 2−x
2
Z
1
2
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z 1Z 1
x 3 sin(y 3 ) dy dx
0
x2
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
0
x2
0
1
√
Z
0
y
x 3 sin(y 3 ) dx dy
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
x2
0
0
Z
=
0
1
√y
x4
3
sin(y )
dy
4
0
1
√
Z
0
y
x 3 sin(y 3 ) dx dy
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
x2
0
0
Z
=
0
1
1
√
Z
y
x 3 sin(y 3 ) dx dy
0
√y
Z 1 2
x4
y
3
sin(y )
sin(y 3 ) dy .
dy =
4
4
0
0
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
x2
0
0
Z
=
0
1
1
√
Z
y
x 3 sin(y 3 ) dx dy
0
√y
Z 1 2
x4
y
3
sin(y )
sin(y 3 ) dy .
dy =
4
4
0
0
Let u = y 3 and then du = 3y 2 dy .
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
x2
0
0
Z
=
0
1
1
√
Z
y
x 3 sin(y 3 ) dx dy
0
√y
Z 1 2
x4
y
3
sin(y )
sin(y 3 ) dy .
dy =
4
4
0
0
Let u = y 3 and then du = 3y 2 dy . Making this substitution,
Z 2
Z
y
1
sin(y 3 ) dy =
sin (y 3 )3y 2 dy
4
12
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
x2
0
0
Z
=
0
1
1
√
Z
y
x 3 sin(y 3 ) dx dy
0
√y
Z 1 2
x4
y
3
sin(y )
sin(y 3 ) dy .
dy =
4
4
0
0
Let u = y 3 and then du = 3y 2 dy . Making this substitution,
Z 2
Z
y
1
1
sin(y 3 ) dy =
sin (y 3 )3y 2 dy = − cos(y 3 ).
4
12
12
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
x2
0
0
Z
=
0
1
1
√
Z
y
x 3 sin(y 3 ) dx dy
0
√y
Z 1 2
x4
y
3
sin(y )
sin(y 3 ) dy .
dy =
4
4
0
0
Let u = y 3 and then du = 3y 2 dy . Making this substitution,
Z 2
Z
y
1
1
sin(y 3 ) dy =
sin (y 3 )3y 2 dy = − cos(y 3 ).
4
12
12
Hence,
Z 1 2
y
sin(y 3 ) dy
4
0
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
x2
0
0
Z
=
0
1
1
√
Z
y
x 3 sin(y 3 ) dx dy
0
√y
Z 1 2
x4
y
3
sin(y )
sin(y 3 ) dy .
dy =
4
4
0
0
Let u = y 3 and then du = 3y 2 dy . Making this substitution,
Z 2
Z
y
1
1
sin(y 3 ) dy =
sin (y 3 )3y 2 dy = − cos(y 3 ).
4
12
12
Hence,
1
Z 1 2
y
1
sin(y 3 ) dy = − cos(y 3 )
12
0 4
0
Problem 32(4) - From Exam 2
Find the iterated integral,
Z 1Z 1
0
x 3 sin(y 3 )dy dx.
x2
(Hint: Reverse the order of integration)
Solution:
Reverse the order of integration:
Z
Z 1Z 1
3
3
x sin(y ) dy dx =
x2
0
0
Z
=
0
1
1
√
Z
y
x 3 sin(y 3 ) dx dy
0
√y
Z 1 2
x4
y
3
sin(y )
sin(y 3 ) dy .
dy =
4
4
0
0
Let u = y 3 and then du = 3y 2 dy . Making this substitution,
Z 2
Z
y
1
1
sin(y 3 ) dy =
sin (y 3 )3y 2 dy = − cos(y 3 ).
4
12
12
Hence,
1
Z 1 2
y
1
1
sin(y 3 ) dy = − cos(y 3 ) =
(1 − cos(1)).
4
12
12
0
0
Problem 33(2) - From Exam 2
Evaluate the following double integral.
Z Z
2
e y dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Problem 33(2) - From Exam 2
Evaluate the following double integral.
Z Z
2
e y dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
e
R
y2
Z
1Z y
dA =
0
0
2
e y dx dy
Problem 33(2) - From Exam 2
Evaluate the following double integral.
Z Z
2
e y dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
e
R
y2
Z
1Z y
dA =
y2
Z
e dx dy =
0
0
0
1h
2
ey x
iy
0
dy
Problem 33(2) - From Exam 2
Evaluate the following double integral.
Z Z
2
e y dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
e
y2
1Z y
Z
dA =
R
Z
0
1
Z
e dx dy =
0
=
y2
2
0
e y y dy
0
1h
2
ey x
iy
0
dy
Problem 33(2) - From Exam 2
Evaluate the following double integral.
Z Z
2
e y dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
e
y2
1Z y
Z
dA =
R
e dx dy =
0
Z
=
0
1
Z
y2
2
0
e y y dy =
1h
0
1
2
Z
0
1
2
2
ey x
e y (2y ) dy
iy
0
dy
Problem 33(2) - From Exam 2
Evaluate the following double integral.
Z Z
2
e y dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
e
y2
1Z y
Z
dA =
R
e dx dy =
0
Z
=
Z
y2
1
2
0
e y y dy =
0
1 y 2 1
= e 2
0
1h
0
1
2
Z
0
1
2
2
ey x
e y (2y ) dy
iy
0
dy
Problem 33(2) - From Exam 2
Evaluate the following double integral.
Z Z
2
e y dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
e
y2
1Z y
Z
dA =
R
e dx dy =
0
Z
=
0
1
Z
y2
2
0
e y y dy =
1h
0
1
2
Z
1
0
1 y 2 1
1
= e = (e − e 0 )
2
2
0
2
2
ey x
e y (2y ) dy
iy
0
dy
Problem 33(2) - From Exam 2
Evaluate the following double integral.
Z Z
2
e y dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
e
y2
1Z y
Z
dA =
R
e dx dy =
0
Z
=
0
1
Z
y2
2
0
e y y dy =
1h
2
ey x
0
1
2
Z
1
2
e y (2y ) dy
0
1 y 2 1
1
1
= e = (e − e 0 ) = (e − 1).
2
2
2
0
iy
0
dy
Problem 33(3) - From Exam 2
Evaluate the following double integral.
Z Z p
x y 2 − x 2 dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Problem 33(3) - From Exam 2
Evaluate the following double integral.
Z Z p
x y 2 − x 2 dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
R
Z
p
2
2
x y − x dA =
0
1Z y
0
p
y 2 − x 2 x dx dy
Problem 33(3) - From Exam 2
Evaluate the following double integral.
Z Z p
x y 2 − x 2 dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
Z
p
2
2
x y − x dA =
0
R
=−
1
2
Z
0
1Z y
0
1Z y
1
(y 2 − x 2 ) 2 (−2x) dx dy
0
p
y 2 − x 2 x dx dy
Problem 33(3) - From Exam 2
Evaluate the following double integral.
Z Z p
x y 2 − x 2 dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
Z
p
2
2
x y − x dA =
0
R
1
=−
2
Z
0
1Z y
0
1Z y
p
y 2 − x 2 x dx dy
0
1
(y − x ) (−2x) dx dy = −
2
2
2
1
2
Z
0
1
y
2 2
2 32 (y − x ) dy
3
0
Problem 33(3) - From Exam 2
Evaluate the following double integral.
Z Z p
x y 2 − x 2 dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
Z
p
2
2
x y − x dA =
0
R
1
=−
2
Z
0
1Z y
0
1Z y
p
y 2 − x 2 x dx dy
0
1
(y − x ) (−2x) dx dy = −
2
2
=
1
2
2
1
3
Z
0
1
y 3 dy
Z
0
1
y
2 2
2 32 (y − x ) dy
3
0
Problem 33(3) - From Exam 2
Evaluate the following double integral.
Z Z p
x y 2 − x 2 dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
Z
p
2
2
x y − x dA =
0
R
1
=−
2
Z
0
1Z y
0
1Z y
p
y 2 − x 2 x dx dy
0
1
(y − x ) (−2x) dx dy = −
2
Z
1 1 3
1 1 4 1
y dy = · y =
3
3 4
2
1
2
2
0
0
Z
0
1
y
2 2
2 32 (y − x ) dy
3
0
Problem 33(3) - From Exam 2
Evaluate the following double integral.
Z Z p
x y 2 − x 2 dA, R = {(x, y ) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y }.
R
Solution:
Z Z
Z
p
2
2
x y − x dA =
0
R
1
=−
2
Z
0
1Z y
0
1Z y
p
y 2 − x 2 x dx dy
0
y
Z
1 12 2
2 32 (y − x ) (−2x) dx dy = −
(y − x ) dy
2 0 3
0
1
Z 1
1
1 1 1
y 3 dy = · y 4 = .
=
3
3 4
12
2
1
2
2
0
0
Problem 34(2) - From Exam 2
Find the volume V of the solid under the surface z = 2x + y 2 and
above the region bounded by curves x − y 2 = 0 and x − y 3 = 0.
Problem 34(2) - From Exam 2
Find the volume V of the solid under the surface z = 2x + y 2 and
above the region bounded by curves x − y 2 = 0 and x − y 3 = 0.
Solution:
Z
1 Z y2
V=
0
y3
(2x + y 2 ) dx dy
Problem 34(2) - From Exam 2
Find the volume V of the solid under the surface z = 2x + y 2 and
above the region bounded by curves x − y 2 = 0 and x − y 3 = 0.
Solution:
Z
1 Z y2
2
Z
(2x + y ) dx dy =
V=
0
y3
0
1h
x 2 + y 2x
iy 2
y3
dy
Problem 34(2) - From Exam 2
Find the volume V of the solid under the surface z = 2x + y 2 and
above the region bounded by curves x − y 2 = 0 and x − y 3 = 0.
Solution:
Z
1 Z y2
0
Z
0
Z
(2x + y ) dx dy =
V=
=
2
y3
1
y 4 + y 4 − (y 6 + y 2 y 3 ) dy
0
1h
x 2 + y 2x
iy 2
y3
dy
Problem 34(2) - From Exam 2
Find the volume V of the solid under the surface z = 2x + y 2 and
above the region bounded by curves x − y 2 = 0 and x − y 3 = 0.
Solution:
Z
1 Z y2
0
Z
0
1h
Z
(2x + y ) dx dy =
V=
=
2
y3
1
y 4 + y 4 − (y 6 + y 2 y 3 ) dy =
0
Z
0
x 2 + y 2x
iy 2
y3
dy
1
−y 6 − y 5 + 2y 4 dy
Problem 34(2) - From Exam 2
Find the volume V of the solid under the surface z = 2x + y 2 and
above the region bounded by curves x − y 2 = 0 and x − y 3 = 0.
Solution:
Z
1 Z y2
0
Z
1h
Z
(2x + y ) dx dy =
V=
=
2
y3
1
y 4 + y 4 − (y 6 + y 2 y 3 ) dy =
0
0
Z
0
1
y7
y 6 2 5 =− −
+ y 7
6
5 0
x 2 + y 2x
iy 2
y3
dy
1
−y 6 − y 5 + 2y 4 dy
Problem 34(2) - From Exam 2
Find the volume V of the solid under the surface z = 2x + y 2 and
above the region bounded by curves x − y 2 = 0 and x − y 3 = 0.
Solution:
Z
1 Z y2
0
Z
0
1h
Z
(2x + y ) dx dy =
V=
=
2
y3
1
y 4 + y 4 − (y 6 + y 2 y 3 ) dy =
x 2 + y 2x
iy 2
0
Z
y3
dy
1
−y 6 − y 5 + 2y 4 dy
0
1
y7
y 6 2 5 1 1 2
=− −
+ y =− − + .
7
6
5 0
7 6 5
Problem 1(a) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the parametric equations for the line L passing through the
points A and C .
Problem 1(a) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the parametric equations for the line L passing through the
points A and C .
Solution:
A vector parallel to the line L is:
−→
v = AC = h1 − 1, 2 − 0, 3 − 0, i = h0, 2, 3i.
Problem 1(a) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the parametric equations for the line L passing through the
points A and C .
Solution:
A vector parallel to the line L is:
−→
v = AC = h1 − 1, 2 − 0, 3 − 0, i = h0, 2, 3i.
A point on the line is A = (1, 0, 0).
Problem 1(a) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the parametric equations for the line L passing through the
points A and C .
Solution:
A vector parallel to the line L is:
−→
v = AC = h1 − 1, 2 − 0, 3 − 0, i = h0, 2, 3i.
A point on the line is A = (1, 0, 0).
Therefore parametric equations for the line L are:
x =1
y = 2t
z = 3t.
Problem 1(b) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find an equation of the plane in R3 which contains the points
A, B, C .
Problem 1(b) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find an equation of the plane in R3 which contains the points
A, B, C .
Solution:
Since a plane is determined by its normal vector n and a point on
it, say the point A, it suffices to find n.
Problem 1(b) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find an equation of the plane in R3 which contains the points
A, B, C .
Solution:
Since a plane is determined by its normal vector n and a point on
it, say the point A, it suffices to find n. Note that:
i j k −→
−→
n = AB × AC = 1 1 0 = h3, −3, 2i.
0 2 3 Problem 1(b) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find an equation of the plane in R3 which contains the points
A, B, C .
Solution:
Since a plane is determined by its normal vector n and a point on
it, say the point A, it suffices to find n. Note that:
i j k −→
−→
n = AB × AC = 1 1 0 = h3, −3, 2i.
0 2 3 So the equation of the plane is:
h3, −3, 2i · hx − 1, y − 0, z − 0i
Problem 1(b) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find an equation of the plane in R3 which contains the points
A, B, C .
Solution:
Since a plane is determined by its normal vector n and a point on
it, say the point A, it suffices to find n. Note that:
i j k −→
−→
n = AB × AC = 1 1 0 = h3, −3, 2i.
0 2 3 So the equation of the plane is:
h3, −3, 2i · hx − 1, y − 0, z − 0i = 3(x − 1) − 3y + 2z = 0.
Problem 1(c) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the area of the triangle T with vertices A, B and C .
Problem 1(c) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the area of the triangle T with vertices A, B and C .
Solution:
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Problem 1(c) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the area of the triangle T with vertices A, B and C .
Solution:
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Then the area of the triangle ∆ with these vertices can be found
−→
−→
by taking the area of the parallelogram spanned by AB and AC
and dividing by 2.
Problem 1(c) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the area of the triangle T with vertices A, B and C .
Solution:
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Then the area of the triangle ∆ with these vertices can be found
−→
−→
by taking the area of the parallelogram spanned by AB and AC
and dividing by 2. Thus:
−→
−→
|AB × AC |
Area(∆) =
2
Problem 1(c) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the area of the triangle T with vertices A, B and C .
Solution:
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Then the area of the triangle ∆ with these vertices can be found
−→
−→
by taking the area of the parallelogram spanned by AB and AC
and dividing by 2. Thus:
−→
−→
i j k |AB × AC |
1 Area(∆) =
= 1 1 0 2
2 0 2 3 Problem 1(c) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the area of the triangle T with vertices A, B and C .
Solution:
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Then the area of the triangle ∆ with these vertices can be found
−→
−→
by taking the area of the parallelogram spanned by AB and AC
and dividing by 2. Thus:
−→
−→
i j k |AB × AC |
1 Area(∆) =
= 1 1 0 2
2 0 2 3 1
= |h3, −3, 2i|
2
Problem 1(c) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the area of the triangle T with vertices A, B and C .
Solution:
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Then the area of the triangle ∆ with these vertices can be found
−→
−→
by taking the area of the parallelogram spanned by AB and AC
and dividing by 2. Thus:
−→
−→
i j k |AB × AC |
1 Area(∆) =
= 1 1 0 2
2 0 2 3 1
1√
= |h3, −3, 2i| =
9+9+4
2
2
Problem 1(c) - Fall 2008
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Find the area of the triangle T with vertices A, B and C .
Solution:
Consider the points A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3).
Then the area of the triangle ∆ with these vertices can be found
−→
−→
by taking the area of the parallelogram spanned by AB and AC
and dividing by 2. Thus:
−→
−→
i j k |AB × AC |
1 Area(∆) =
= 1 1 0 2
2 0 2 3 1
1√
1√
= |h3, −3, 2i| =
9+9+4=
22.
2
2
2
Problem 2 - Fall 2008
Find the volume under the graph of f (x, y ) = x + 2xy and over
the bounded region in the first quadrant {(x, y ) | x ≥ 0, y ≥ 0}
bounded by the curve y = 1 − x 2 and the x and y -axes.
Problem 2 - Fall 2008
Find the volume under the graph of f (x, y ) = x + 2xy and over
the bounded region in the first quadrant {(x, y ) | x ≥ 0, y ≥ 0}
bounded by the curve y = 1 − x 2 and the x and y -axes.
Solution:
Z 1Z
1−x 2
(x + 2xy ) dy dx
0
0
Problem 2 - Fall 2008
Find the volume under the graph of f (x, y ) = x + 2xy and over
the bounded region in the first quadrant {(x, y ) | x ≥ 0, y ≥ 0}
bounded by the curve y = 1 − x 2 and the x and y -axes.
Solution:
Z 1Z
1−x 2
Z
(x + 2xy ) dy dx =
0
0
0
1
1−x 2
dx
(xy + xy 2 )
0
Problem 2 - Fall 2008
Find the volume under the graph of f (x, y ) = x + 2xy and over
the bounded region in the first quadrant {(x, y ) | x ≥ 0, y ≥ 0}
bounded by the curve y = 1 − x 2 and the x and y -axes.
Solution:
Z 1Z
1−x 2
Z
(x + 2xy ) dy dx =
0
Z
=
0
0
1
x(1 − x 2 ) + x(1 − x 2 )2 dx
0
1
1−x 2
dx
(xy + xy 2 )
0
Problem 2 - Fall 2008
Find the volume under the graph of f (x, y ) = x + 2xy and over
the bounded region in the first quadrant {(x, y ) | x ≥ 0, y ≥ 0}
bounded by the curve y = 1 − x 2 and the x and y -axes.
Solution:
Z 1Z
1−x 2
Z
1
(x + 2xy ) dy dx =
0
Z
=
0
0
0
1
x(1 − x 2 ) + x(1 − x 2 )2 dx =
1−x 2
dx
(xy + xy 2 )
0
Z
0
1
x 5 − 3x 3 + 2x dx
Problem 2 - Fall 2008
Find the volume under the graph of f (x, y ) = x + 2xy and over
the bounded region in the first quadrant {(x, y ) | x ≥ 0, y ≥ 0}
bounded by the curve y = 1 − x 2 and the x and y -axes.
Solution:
Z 1Z
1−x 2
Z
1
(x + 2xy ) dy dx =
0
Z
=
0
0
1
x(1 − x 2 ) + x(1 − x 2 )2 dx =
0
1−x 2
dx
(xy + xy 2 )
0
Z
0
1
x 6 3x 4
2 =(
−
+ x )
6
4
0
1
x 5 − 3x 3 + 2x dx
Problem 2 - Fall 2008
Find the volume under the graph of f (x, y ) = x + 2xy and over
the bounded region in the first quadrant {(x, y ) | x ≥ 0, y ≥ 0}
bounded by the curve y = 1 − x 2 and the x and y -axes.
Solution:
Z 1Z
1−x 2
Z
1
(x + 2xy ) dy dx =
0
Z
=
0
0
0
1
x(1 − x 2 ) + x(1 − x 2 )2 dx =
1−x 2
dx
(xy + xy 2 )
0
Z
1
x 5 − 3x 3 + 2x dx
0
1
x 6 3x 4
1 3
2 =(
−
+ x ) = − + 1
6
4
6 4
0
Problem 2 - Fall 2008
Find the volume under the graph of f (x, y ) = x + 2xy and over
the bounded region in the first quadrant {(x, y ) | x ≥ 0, y ≥ 0}
bounded by the curve y = 1 − x 2 and the x and y -axes.
Solution:
Z 1Z
1−x 2
Z
1
(x + 2xy ) dy dx =
0
Z
=
0
0
0
1
x(1 − x 2 ) + x(1 − x 2 )2 dx =
1−x 2
dx
(xy + xy 2 )
0
Z
1
x 5 − 3x 3 + 2x dx
0
1
x 6 3x 4
1 3
5
2 =(
−
+ x ) = − + 1 = .
6
4
6 4
12
0
Problem 3 - Fall 2008
Let
Z
1
Z
2
I=
0
sin(y 2 ) dy dx.
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Problem 3 - Fall 2008
Let
Z
1
Z
2
I=
0
sin(y 2 ) dy dx.
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Solution:
1
See the blackboard for a sketch.
Problem 3 - Fall 2008
Let
1
Z
Z
2
sin(y 2 ) dy dx.
I=
0
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Solution:
1
See the blackboard for a sketch.
2
Z
1
Z
I=
0
0
1
2y
sin(y 2 ) dx dy
Problem 3 - Fall 2008
Let
1
Z
Z
2
sin(y 2 ) dy dx.
I=
0
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Solution:
1
See the blackboard for a sketch.
2
1
Z
Z
I=
0
3
By Fubini’s Theorem,
Z 1 Z 21 y
Z
sin(y 2 ) dx dy =
0
0
0
1
2y
sin(y 2 ) dx dy
0
1
21 y
sin(y )x dy
2
0
Problem 3 - Fall 2008
Let
1
Z
Z
2
sin(y 2 ) dy dx.
I=
0
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Solution:
1
See the blackboard for a sketch.
2
1
Z
Z
I=
0
3
By Fubini’s Theorem,
Z 1 Z 21 y
Z
sin(y 2 ) dx dy =
0
0
0
1
2y
sin(y 2 ) dx dy
0
1
21 y
sin(y )x dy
2
0
Problem 3 - Fall 2008
Let
1
Z
Z
2
sin(y 2 ) dy dx.
I=
0
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Solution:
1
See the blackboard for a sketch.
2
1
Z
Z
I=
0
3
By Fubini’s Theorem,
Z 1 Z 21 y
Z
sin(y 2 ) dx dy =
0
0
0
1
2y
sin(y 2 ) dx dy
0
1
21 y
Z 1
1
sin(y )x dy =
(sin(y 2 ) · y ) dy
2
0
0
2
Problem 3 - Fall 2008
Let
1
Z
Z
2
sin(y 2 ) dy dx.
I=
0
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Solution:
1
See the blackboard for a sketch.
2
1
Z
Z
I=
0
3
By Fubini’s Theorem,
Z 1 Z 21 y
Z
sin(y 2 ) dx dy =
0
0
0
=
1
4
Z
0
1
2y
sin(y 2 ) dx dy
0
1
21 y
Z 1
1
sin(y )x dy =
(sin(y 2 ) · y ) dy
2
0
0
1
sin(y 2 ) · 2y dy
2
Problem 3 - Fall 2008
Let
1
Z
Z
2
sin(y 2 ) dy dx.
I=
0
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Solution:
1
See the blackboard for a sketch.
2
Z
1
Z
I=
0
3
By Fubini’s Theorem,
Z 1 Z 21 y
Z
sin(y 2 ) dx dy =
0
0
1
2y
sin(y 2 ) dx dy
0
21 y
Z 1
1
sin(y )x dy =
(sin(y 2 ) · y ) dy
2
0
0
0
Z 1
2 1
cos(y ) 1
=
sin(y 2 ) · 2y dy =
4 0
4 0
1
2
Problem 3 - Fall 2008
Let
1
Z
Z
2
sin(y 2 ) dy dx.
I=
0
2x
1
Sketch the region of integration.
2
Write the integral I with the order of integration reversed.
3
Evaluate the integral I. Show your work.
Solution:
1
See the blackboard for a sketch.
2
Z
1
Z
I=
0
3
By Fubini’s Theorem,
Z 1 Z 21 y
Z
sin(y 2 ) dx dy =
0
0
1
2y
sin(y 2 ) dx dy
0
21 y
Z 1
1
sin(y )x dy =
(sin(y 2 ) · y ) dy
2
0
0
0
Z 1
2 1
cos(y ) 1
1
= (cos 1 − 1).
=
sin(y 2 ) · 2y dy =
4
4
4 0
0
1
2
Problem 4(a, b, c) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z.
1 What is the gradient ∇F(x, y , z) of F at the point (1, 2, −1)?
2 Calculate the directional derivative of F at the point (1, 2, −1)
in the direction h1, 1, 1i?
3 What is the maximal rate of change of F at the point
(1, 2, −1)?
Problem 4(a, b, c) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z.
1 What is the gradient ∇F(x, y , z) of F at the point (1, 2, −1)?
2 Calculate the directional derivative of F at the point (1, 2, −1)
in the direction h1, 1, 1i?
3 What is the maximal rate of change of F at the point
(1, 2, −1)?
Solution:
∇f = h2x + y 2 , 2xy , 1i.
Problem 4(a, b, c) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z.
1 What is the gradient ∇F(x, y , z) of F at the point (1, 2, −1)?
2 Calculate the directional derivative of F at the point (1, 2, −1)
in the direction h1, 1, 1i?
3 What is the maximal rate of change of F at the point
(1, 2, −1)?
Solution:
∇f = h2x + y 2 , 2xy , 1i.
So,
∇f (1, 2, −1) = h6, 4, 1i.
Problem 4(a, b, c) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z.
1 What is the gradient ∇F(x, y , z) of F at the point (1, 2, −1)?
2 Calculate the directional derivative of F at the point (1, 2, −1)
in the direction h1, 1, 1i?
3 What is the maximal rate of change of F at the point
(1, 2, −1)?
Solution:
∇f = h2x + y 2 , 2xy , 1i.
So,
∇f (1, 2, −1) = h6, 4, 1i.
The unit vector u in the direction h1, 1, 1i is u =
h1,1,1i
√ .
3
Problem 4(a, b, c) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z.
1 What is the gradient ∇F(x, y , z) of F at the point (1, 2, −1)?
2 Calculate the directional derivative of F at the point (1, 2, −1)
in the direction h1, 1, 1i?
3 What is the maximal rate of change of F at the point
(1, 2, −1)?
Solution:
∇f = h2x + y 2 , 2xy , 1i.
So,
∇f (1, 2, −1) = h6, 4, 1i.
The unit vector u in the direction h1, 1, 1i is u =
Du f (1, 2, −1) = ∇f (1, 2, −1) · u
h1,1,1i
√ .
3
Problem 4(a, b, c) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z.
1 What is the gradient ∇F(x, y , z) of F at the point (1, 2, −1)?
2 Calculate the directional derivative of F at the point (1, 2, −1)
in the direction h1, 1, 1i?
3 What is the maximal rate of change of F at the point
(1, 2, −1)?
Solution:
∇f = h2x + y 2 , 2xy , 1i.
So,
∇f (1, 2, −1) = h6, 4, 1i.
The unit vector u in the direction h1, 1, 1i is u =
Du f (1, 2, −1) = ∇f (1, 2, −1) · u = h6, 4, 1i ·
h1,1,1i
√ .
3
√1 h1, 1, 1i
3
Problem 4(a, b, c) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z.
1 What is the gradient ∇F(x, y , z) of F at the point (1, 2, −1)?
2 Calculate the directional derivative of F at the point (1, 2, −1)
in the direction h1, 1, 1i?
3 What is the maximal rate of change of F at the point
(1, 2, −1)?
Solution:
∇f = h2x + y 2 , 2xy , 1i.
So,
∇f (1, 2, −1) = h6, 4, 1i.
The unit vector u in the direction h1, 1, 1i is u =
Du f (1, 2, −1) = ∇f (1, 2, −1) · u = h6, 4, 1i ·
h1,1,1i
√ .
3
√1 h1, 1, 1i
3
=
11
√
.
3
The maximum rate of change is the length of the gradient:
Problem 4(a, b, c) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z.
1 What is the gradient ∇F(x, y , z) of F at the point (1, 2, −1)?
2 Calculate the directional derivative of F at the point (1, 2, −1)
in the direction h1, 1, 1i?
3 What is the maximal rate of change of F at the point
(1, 2, −1)?
Solution:
∇f = h2x + y 2 , 2xy , 1i.
So,
∇f (1, 2, −1) = h6, 4, 1i.
The unit vector u in the direction h1, 1, 1i is u =
Du f (1, 2, −1) = ∇f (1, 2, −1) · u = h6, 4, 1i ·
h1,1,1i
√ .
3
√1 h1, 1, 1i
3
=
11
√
.
3
The maximum rate of change is the length of the gradient:
√
MRC(f ) = |∇f (1, 2, −1)| = |h6, 4, 1i| = 53.
.
Problem 4(d) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z. Find the equation
of the tangent plane to the level surface F(x, y , z) = 4 at the point
(1, 2, −1).
Problem 4(d) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z. Find the equation
of the tangent plane to the level surface F(x, y , z) = 4 at the point
(1, 2, −1).
Solution:
Recall that the gradient of F(x, y , z) = x 2 + xy 2 + z is normal
n to the surface.
Problem 4(d) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z. Find the equation
of the tangent plane to the level surface F(x, y , z) = 4 at the point
(1, 2, −1).
Solution:
Recall that the gradient of F(x, y , z) = x 2 + xy 2 + z is normal
n to the surface.
Calculating, we obtain:
∇F(x, y , z) = h2x, 2xy , 1i
n = ∇F(1, 2, −1) = h6, 4, 1i.
Problem 4(d) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z. Find the equation
of the tangent plane to the level surface F(x, y , z) = 4 at the point
(1, 2, −1).
Solution:
Recall that the gradient of F(x, y , z) = x 2 + xy 2 + z is normal
n to the surface.
Calculating, we obtain:
∇F(x, y , z) = h2x, 2xy , 1i
n = ∇F(1, 2, −1) = h6, 4, 1i.
The equation of the tangent plane is:
Problem 4(d) - Fall 2008
Consider the function F(x, y , z) = x 2 + xy 2 + z. Find the equation
of the tangent plane to the level surface F(x, y , z) = 4 at the point
(1, 2, −1).
Solution:
Recall that the gradient of F(x, y , z) = x 2 + xy 2 + z is normal
n to the surface.
Calculating, we obtain:
∇F(x, y , z) = h2x, 2xy , 1i
n = ∇F(1, 2, −1) = h6, 4, 1i.
The equation of the tangent plane is:
h6, 4, 1i·hx −1, y −2, z +1i = 6(x −1)+4(y −2)+(z +1) = 0.
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 1 − r 2 .
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 1 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 1
V=
(1 − r 2 )r dr dθ
0
0
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 1 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 1
Z 2π Z 1
2
V=
(1 − r )r dr dθ =
(r − r 3 ) dr dθ
0
0
0
0
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 1 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 1
Z 2π Z 1
2
V=
(1 − r )r dr dθ =
(r − r 3 ) dr dθ
0
0
Z
=
0
2π
0
1 2 1 4
r − r
2
4
1
dθ
0
0
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 1 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 1
Z 2π Z 1
2
V=
(1 − r )r dr dθ =
(r − r 3 ) dr dθ
0
0
Z
=
0
2π
0
1 2 1 4
r − r
2
4
1
0
Z
dθ =
0
0
2π
1 1
( − ) dθ
2 4
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 1 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 1
Z 2π Z 1
2
V=
(1 − r )r dr dθ =
(r − r 3 ) dr dθ
0
0
Z
=
0
2π
0
1
1 2 1 4
r − r
dθ =
2
4
0
Z
1 2π
=
dθ
4 0
0
Z
0
2π
1 1
( − ) dθ
2 4
Problem 5 - Fall 2008
Find the volume V of the solid under the surface z = 1 − x 2 − y 2
and above the xy -plane.
Solution:
The domain of integration for the function z = 1 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 1 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 1
Z 2π Z 1
2
V=
(1 − r )r dr dθ =
(r − r 3 ) dr dθ
0
0
Z
=
0
2π
0
1
Z
0
2π
1 2 1 4
1 1
r − r
dθ =
( − ) dθ
2
4
2 4
0
0
Z 2π
1
π
=
dθ = .
4 0
2
Problem 6(a) - Fall 2008
Determine whether the following vector fields are conservative or
not. Find a potential function for those which are indeed
conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Problem 6(a) - Fall 2008
Determine whether the following vector fields are conservative or
not. Find a potential function for those which are indeed
conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function F(x, y ).
Problem 6(a) - Fall 2008
Determine whether the following vector fields are conservative or
not. Find a potential function for those which are indeed
conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function F(x, y ).
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where
P(x, y ) = x 2 + e x + xy and Q(x, y ) = xy − sin(y ).
Problem 6(a) - Fall 2008
Determine whether the following vector fields are conservative or
not. Find a potential function for those which are indeed
conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function F(x, y ).
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where
P(x, y ) = x 2 + e x + xy and Q(x, y ) = xy − sin(y ).
Since Py (x, y ) = x 6= y = Qx (x, y ), the vector field F(x, y ) is
not conservative.
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2
∂y
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2 =⇒ f(x, y ) = x 3 y + xy 2 + y 3 + g (x),
∂y
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2 =⇒ f(x, y ) = x 3 y + xy 2 + y 3 + g (x),
∂y
where g (x) is some function of x.
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2 =⇒ f(x, y ) = x 3 y + xy 2 + y 3 + g (x),
∂y
where g (x) is some function of x.
Since
∂f
∂x
= 3x 2 y + e x + y 2 ,
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2 =⇒ f(x, y ) = x 3 y + xy 2 + y 3 + g (x),
∂y
where g (x) is some function of x.
Since
∂f
∂x
= 3x 2 y + e x + y 2 ,
∂(x 3 y + xy 2 + y 3 + g (x))
∂f
=
= 3x 2 y + y 2 + g 0 (x)
∂x
∂x
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2 =⇒ f(x, y ) = x 3 y + xy 2 + y 3 + g (x),
∂y
where g (x) is some function of x.
Since
∂f
∂x
= 3x 2 y + e x + y 2 ,
∂(x 3 y + xy 2 + y 3 + g (x))
∂f
=
= 3x 2 y + y 2 + g 0 (x) = 3x 2 y + e x + y 2
∂x
∂x
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2 =⇒ f(x, y ) = x 3 y + xy 2 + y 3 + g (x),
∂y
where g (x) is some function of x.
Since
∂f
∂x
= 3x 2 y + e x + y 2 ,
∂(x 3 y + xy 2 + y 3 + g (x))
∂f
=
= 3x 2 y + y 2 + g 0 (x) = 3x 2 y + e x + y 2
∂x
∂x
=⇒ g 0 (x) = e x
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2 =⇒ f(x, y ) = x 3 y + xy 2 + y 3 + g (x),
∂y
where g (x) is some function of x.
Since
∂f
∂x
= 3x 2 y + e x + y 2 ,
∂(x 3 y + xy 2 + y 3 + g (x))
∂f
=
= 3x 2 y + y 2 + g 0 (x) = 3x 2 y + e x + y 2
∂x
∂x
=⇒ g 0 (x) = e x =⇒ g (x) = e x + constant.
Problem 6(b) - Fall 2008
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
1
F(x, y ) = hx 2 + e x + xy , xy − sin(y )i.
2
G(x, y ) = h3x 2 y + e x + y 2 , x 3 + 2xy + 3y 2 i.
Solution:
On this slide we only consider the function G(x, y ).
∂
∂
Since ∂y
(3x 2 y + e x + y 2 ) = 3x 2 + 2y = ∂x
(x 3 + 2xy + 3y 2 ), there
exists a potential function f(x, y ), where ∇f = G.
Note that:
∂f
= x 3 + 2xy + 3y 2 =⇒ f(x, y ) = x 3 y + xy 2 + y 3 + g (x),
∂y
where g (x) is some function of x.
Since
∂f
∂x
= 3x 2 y + e x + y 2 ,
∂(x 3 y + xy 2 + y 3 + g (x))
∂f
=
= 3x 2 y + y 2 + g 0 (x) = 3x 2 y + e x + y 2
∂x
∂x
=⇒ g 0 (x) = e x =⇒ g (x) = e x + constant.
Hence, f(x, y ) = x 3 y + xy 2 + y 3 + e x is a potential function.
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
So, C01 (t) = h1, 0, 0i and C2 (t) = h0, 2, 1i.
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
So, C01R(t) = h1, 0, 0i and C2 (t) = h0, 2, 1i.
Thus, C1 yz dx + xy dy + xy dz =
R1
0 [(0 · 0 · 1) + (t · 0 · 0) + (t · 0 · 0)] dt = 0.
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
So, C01R(t) = h1, 0, 0i and C2 (t) = h0, 2, 1i.
Thus, C1 yz dx + xy dy + xy dz =
R1
· 0 · 1) + (t · 0 · 0) + (t · 0 · 0)] dt = 0.
0 [(0 R
Also, C2 yz dx + xz dy + xy dz
R1
= 0 [(2t · t · 0) + (1 · t · 2) + (1 · 2t · 1)] dt
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
So, C01R(t) = h1, 0, 0i and C2 (t) = h0, 2, 1i.
Thus, C1 yz dx + xy dy + xy dz =
R1
· 0 · 1) + (t · 0 · 0) + (t · 0 · 0)] dt = 0.
0 [(0 R
Also, C2 yz dx + xz dy + xy dz
R1
R1
= 0 [(2t · t · 0) + (1 · t · 2) + (1 · 2t · 1)] dt = 0 4t dt
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
So, C01R(t) = h1, 0, 0i and C2 (t) = h0, 2, 1i.
Thus, C1 yz dx + xy dy + xy dz =
R1
· 0 · 1) + (t · 0 · 0) + (t · 0 · 0)] dt = 0.
0 [(0 R
Also, C2 yz dx + xz dy + xy dz
R1
R1
= 0 [(2t · t · 0) + (1 · t · 2) + (1 · 2t · 1)] dt = 0 4t dt =
1
4t 2 2 0
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
So, C01R(t) = h1, 0, 0i and C2 (t) = h0, 2, 1i.
Thus, C1 yz dx + xy dy + xy dz =
R1
· 0 · 1) + (t · 0 · 0) + (t · 0 · 0)] dt = 0.
0 [(0 R
Also, C2 yz dx + xz dy + xy dz
R1
R1
= 0 [(2t · t · 0) + (1 · t · 2) + (1 · 2t · 1)] dt = 0 4t dt =
=
4
2
1
4t 2 2 0
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
So, C01R(t) = h1, 0, 0i and C2 (t) = h0, 2, 1i.
Thus, C1 yz dx + xy dy + xy dz =
R1
· 0 · 1) + (t · 0 · 0) + (t · 0 · 0)] dt = 0.
0 [(0 R
Also, C2 yz dx + xz dy + xy dz
R1
R1
= 0 [(2t · t · 0) + (1 · t · 2) + (1 · 2t · 1)] dt = 0 4t dt =
=
4
2
= 2.
1
4t 2 2 0
Problem 7 - Fall 2008
Evaluate the line integral
Z
yz dx + xz dy + xy dz,
C
where C is the curve starting at (0, 0, 0), traveling along a line
segment to (1, 0, 0) and then traveling along a second line segment
to (1, 2, 1).
Solution:
The parameterizations C1 (t), C2 (t) are:
0≤t≤1
C1 (t) = ht, 0, 0i
C2 (t) = h1, 2t, ti
0≤t≤1
So, C01R(t) = h1, 0, 0i and C2 (t) = h0, 2, 1i.
Thus, C1 yz dx + xy dy + xy dz =
R1
· 0 · 1) + (t · 0 · 0) + (t · 0 · 0)] dt = 0.
0 [(0 R
Also, C2 yz dx + xz dy + xy dz
R1
R1
= 0 [(2t · t · 0) + (1 · t · 2) + (1 · 2t · 1)] dt = 0 4t dt =
=
4
2
= 2. So, the entire integral equals 0 + 2 = 2.
1
4t 2 2 0
Problem 8(a) - Fall 2008
Use Green’s Theorem to show that if D ⊂ R2 is the bounded
region with boundary a positively oriented simple closed curve C,
then the area of D can be calculated by the formula:
I
1
Area(D) =
−y dx + x dy
2 C
Problem 8(a) - Fall 2008
Use Green’s Theorem to show that if D ⊂ R2 is the bounded
region with boundary a positively oriented simple closed curve C,
then the area of D can be calculated by the formula:
I
1
Area(D) =
−y dx + x dy
2 C
Solution:
Recall Green’s Theorem:
Z Z
I
∂Q
∂P
P dx + Q dy =
(
−
) dA.
∂x
∂y
C
D
Problem 8(a) - Fall 2008
Use Green’s Theorem to show that if D ⊂ R2 is the bounded
region with boundary a positively oriented simple closed curve C,
then the area of D can be calculated by the formula:
I
1
Area(D) =
−y dx + x dy
2 C
Solution:
Recall Green’s Theorem:
Z Z
I
∂Q
∂P
P dx + Q dy =
(
−
) dA.
∂x
∂y
C
D
Hence,
1
2
I
Z Z
−y dx + x dy =
C
(
D
∂x
∂−y
−
) dA
∂x
∂y
Problem 8(a) - Fall 2008
Use Green’s Theorem to show that if D ⊂ R2 is the bounded
region with boundary a positively oriented simple closed curve C,
then the area of D can be calculated by the formula:
I
1
Area(D) =
−y dx + x dy
2 C
Solution:
Recall Green’s Theorem:
Z Z
I
∂Q
∂P
P dx + Q dy =
(
−
) dA.
∂x
∂y
C
D
Hence,
1
2
=
I
Z Z
−y dx + x dy =
C
1
2
(
D
Z Z
(1 + 1) dA
D
∂x
∂−y
−
) dA
∂x
∂y
Problem 8(a) - Fall 2008
Use Green’s Theorem to show that if D ⊂ R2 is the bounded
region with boundary a positively oriented simple closed curve C,
then the area of D can be calculated by the formula:
I
1
Area(D) =
−y dx + x dy
2 C
Solution:
Recall Green’s Theorem:
Z Z
I
∂Q
∂P
P dx + Q dy =
(
−
) dA.
∂x
∂y
C
D
Hence,
1
2
=
I
Z Z
−y dx + x dy =
C
1
2
(
D
∂x
∂−y
−
) dA
∂x
∂y
Z Z
Z Z
(1 + 1) dA =
D
dA
D
Problem 8(a) - Fall 2008
Use Green’s Theorem to show that if D ⊂ R2 is the bounded
region with boundary a positively oriented simple closed curve C,
then the area of D can be calculated by the formula:
I
1
Area(D) =
−y dx + x dy
2 C
Solution:
Recall Green’s Theorem:
Z Z
I
∂Q
∂P
P dx + Q dy =
(
−
) dA.
∂x
∂y
C
D
Hence,
1
2
=
I
Z Z
−y dx + x dy =
C
1
2
(
D
∂x
∂−y
−
) dA
∂x
∂y
Z Z
Z Z
(1 + 1) dA =
D
dA = Area(D)
D
Problem 8(b) - Fall 2008
Consider the ellipse 4x 2 + y 2 = 1. Use the above area formula to
calculate the area of the region D ⊂ R2 with boundary this ellipse.
(Hint: This ellipse can be parametrized by r(t) = h 12 cos(t), sin(t)i
for 0 ≤ t ≤ 2π.)
Problem 8(b) - Fall 2008
Consider the ellipse 4x 2 + y 2 = 1. Use the above area formula to
calculate the area of the region D ⊂ R2 with boundary this ellipse.
(Hint: This ellipse can be parametrized by r(t) = h 12 cos(t), sin(t)i
for 0 ≤ t ≤ 2π.)
Solution:
The ellipse has parametric equations x =
y = sin t, where 0 ≤ t ≤ 2π.
1
2
cos t and
Problem 8(b) - Fall 2008
Consider the ellipse 4x 2 + y 2 = 1. Use the above area formula to
calculate the area of the region D ⊂ R2 with boundary this ellipse.
(Hint: This ellipse can be parametrized by r(t) = h 12 cos(t), sin(t)i
for 0 ≤ t ≤ 2π.)
Solution:
The ellipse has parametric equations x =
y = sin t, where 0 ≤ t ≤ 2π.
1
2
cos t and
Using the formula in the previous theorem, we have:
I
1
Area(D) =
x dy − y dx
2 C
Problem 8(b) - Fall 2008
Consider the ellipse 4x 2 + y 2 = 1. Use the above area formula to
calculate the area of the region D ⊂ R2 with boundary this ellipse.
(Hint: This ellipse can be parametrized by r(t) = h 12 cos(t), sin(t)i
for 0 ≤ t ≤ 2π.)
Solution:
The ellipse has parametric equations x =
y = sin t, where 0 ≤ t ≤ 2π.
1
2
cos t and
Using the formula in the previous theorem, we have:
I
1
Area(D) =
x dy − y dx
2 C
Z
1 2π 1
1
=
( cos t)(cos t) dt − (sin t)(− sin t) dt
2 0 2
2
Problem 8(b) - Fall 2008
Consider the ellipse 4x 2 + y 2 = 1. Use the above area formula to
calculate the area of the region D ⊂ R2 with boundary this ellipse.
(Hint: This ellipse can be parametrized by r(t) = h 12 cos(t), sin(t)i
for 0 ≤ t ≤ 2π.)
Solution:
The ellipse has parametric equations x =
y = sin t, where 0 ≤ t ≤ 2π.
1
2
cos t and
Using the formula in the previous theorem, we have:
I
1
Area(D) =
x dy − y dx
2 C
Z
1 2π 1
1
=
( cos t)(cos t) dt − (sin t)(− sin t) dt
2 0 2
2
Z 2π
1
1
=
(cos2 t + sin2 t) dt
2 0 2
Problem 8(b) - Fall 2008
Consider the ellipse 4x 2 + y 2 = 1. Use the above area formula to
calculate the area of the region D ⊂ R2 with boundary this ellipse.
(Hint: This ellipse can be parametrized by r(t) = h 12 cos(t), sin(t)i
for 0 ≤ t ≤ 2π.)
Solution:
The ellipse has parametric equations x =
y = sin t, where 0 ≤ t ≤ 2π.
1
2
cos t and
Using the formula in the previous theorem, we have:
I
1
Area(D) =
x dy − y dx
2 C
Z
1 2π 1
1
=
( cos t)(cos t) dt − (sin t)(− sin t) dt
2 0 2
2
Z 2π
1
1
2π
=
(cos2 t + sin2 t) dt =
2 0 2
4
Problem 8(b) - Fall 2008
Consider the ellipse 4x 2 + y 2 = 1. Use the above area formula to
calculate the area of the region D ⊂ R2 with boundary this ellipse.
(Hint: This ellipse can be parametrized by r(t) = h 12 cos(t), sin(t)i
for 0 ≤ t ≤ 2π.)
Solution:
The ellipse has parametric equations x =
y = sin t, where 0 ≤ t ≤ 2π.
1
2
cos t and
Using the formula in the previous theorem, we have:
I
1
Area(D) =
x dy − y dx
2 C
Z
1 2π 1
1
=
( cos t)(cos t) dt − (sin t)(− sin t) dt
2 0 2
2
Z 2π
1
1
2π
π
= .
=
(cos2 t + sin2 t) dt =
2 0 2
4
2
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Solution:
The velocity v(t) is equal to r0 (t):
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Solution:
The velocity v(t) is equal to r0 (t):
1
v(t) = r0 (t) = h2t, 2t, i
2
1
v(4) = h8, 8, i.
2
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Solution:
The velocity v(t) is equal to r0 (t):
1
v(t) = r0 (t) = h2t, 2t, i
2
1
v(4) = h8, 8, i.
2
The acceleration a(t) is equal to v0 (t):
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Solution:
The velocity v(t) is equal to r0 (t):
1
v(t) = r0 (t) = h2t, 2t, i
2
1
v(4) = h8, 8, i.
2
The acceleration a(t) is equal to v0 (t):
a(t) = v0 (t) = h2, 2, 0i
a(4) = h2, 2, 0i.
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Solution:
The velocity v(t) is equal to r0 (t):
1
v(t) = r0 (t) = h2t, 2t, i
2
1
v(4) = h8, 8, i.
2
The acceleration a(t) is equal to v0 (t):
a(t) = v0 (t) = h2, 2, 0i
a(4) = h2, 2, 0i.
The speed s(4) is equal to |v(4)|:
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Solution:
The velocity v(t) is equal to r0 (t):
1
v(t) = r0 (t) = h2t, 2t, i
2
1
v(4) = h8, 8, i.
2
The acceleration a(t) is equal to v0 (t):
a(t) = v0 (t) = h2, 2, 0i
a(4) = h2, 2, 0i.
The speed s(4) is equal to |v(4)|:
s(4) = |v(4)|
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Solution:
The velocity v(t) is equal to r0 (t):
1
v(t) = r0 (t) = h2t, 2t, i
2
1
v(4) = h8, 8, i.
2
The acceleration a(t) is equal to v0 (t):
a(t) = v0 (t) = h2, 2, 0i
a(4) = h2, 2, 0i.
The speed s(4) is equal to |v(4)|:
r
1
s(4) = |v(4)| = 64 + 64 +
4
Problem 9(a) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(a) find the velocity, speed, and acceleration of a particle whose
position function is r(t) at time t = 4.
Solution:
The velocity v(t) is equal to r0 (t):
1
v(t) = r0 (t) = h2t, 2t, i
2
1
v(4) = h8, 8, i.
2
The acceleration a(t) is equal to v0 (t):
a(t) = v0 (t) = h2, 2, 0i
a(4) = h2, 2, 0i.
The speed s(4) is equal to |v(4)|:
r
r
1
1
s(4) = |v(4)| = 64 + 64 + = 128 + .
4
4
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( )
2
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1
2
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1 = (2t + 1)(t − 1)
2
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1 = (2t + 1)(t − 1) = 0
2
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1 = (2t + 1)(t − 1) = 0
2
1
=⇒ t = 1 or t = − .
2
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1 = (2t + 1)(t − 1) = 0
2
1
=⇒ t = 1 or t = − .
2
Next evaluate the points on r(t) at these times to obtain the
2 points of intersection:
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1 = (2t + 1)(t − 1) = 0
2
1
=⇒ t = 1 or t = − .
2
Next evaluate the points on r(t) at these times to obtain the
2 points of intersection:
1
r(1) = h12 − 1, 12 , i
2
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1 = (2t + 1)(t − 1) = 0
2
1
=⇒ t = 1 or t = − .
2
Next evaluate the points on r(t) at these times to obtain the
2 points of intersection:
1
1
r(1) = h12 − 1, 12 , i = h0, 1, i
2
2
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1 = (2t + 1)(t − 1) = 0
2
1
=⇒ t = 1 or t = − .
2
Next evaluate the points on r(t) at these times to obtain the
2 points of intersection:
1
1
r(1) = h12 − 1, 12 , i = h0, 1, i
2
2
1
1
1 1 1
r(− ) = h(− )2 − 1, (− )2 , (− )i
2
2
2 2 2
Problem 9(b) - Spring 2008
For the space curve r(t) = ht 2 − 1, t 2 , t/2i,
(b) find all points where the particle with position vector r(t)
intersects the plane x + y − 2z = 0.
Solution:
Plug the x, y and z-coordinates of r(t) into the equation of
the plane and solve for t:
t
(t 2 − 1) + t 2 − 2( ) = 2t 2 − t − 1 = (2t + 1)(t − 1) = 0
2
1
=⇒ t = 1 or t = − .
2
Next evaluate the points on r(t) at these times to obtain the
2 points of intersection:
1
1
r(1) = h12 − 1, 12 , i = h0, 1, i
2
2
3 1 1
1
1
1 1 1
r(− ) = h(− )2 − 1, (− )2 , (− )i = h− , , − i.
4 4 4
2
2
2 2 2
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Solution:
Part (a)
Z
Z Z
1
Z
x
xy dy dx.
xy dA =
D
0
x2
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Solution:
Part (a)
Z
Z Z
D
Part (b)
1
Z
x
xy dy dx.
xy dA =
Z
1
√
Z
0
x2
y
xy dx dy .
0
y
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Solution:
Z
Z Z
Part (a)
D
Z
Part (b)
Z
x
xy dy dx.
1
√
Z
0
x2
y
xy dx dy .
0
Part (c)
1
xy dA =
Z
1
Z
y
x
xy dy dx
0
x2
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Solution:
Z
x
xy dy dx.
xy dA =
D
Z
Part (b)
1
√
Z
x2
0
y
xy dx dy .
0
Part (c)
1
Z
Z Z
Part (a)
Z
1
Z
y
x
Z
xy dy dx =
0
x2
0
1
1 2
xy
2
x
dx
x2
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Solution:
1
Z
Z Z
Part (a)
Z
x
xy dy dx.
xy dA =
D
1
Z
Part (b)
√
Z
x2
0
y
xy dx dy .
0
Part (c)
Z
1
Z
y
x
Z
xy dy dx =
0
Z
=
0
1
x2
1 3 1 5
x − x
2
2
0
dx
1
1 2
xy
2
x
dx
x2
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Solution:
1
Z
Z Z
Part (a)
Z
x
xy dy dx.
xy dA =
D
Z
Part (b)
1
√
Z
0
x2
y
xy dx dy .
0
y
x
1 2
xy
dx
2
0
x2
0
x2
1
Z 1
1 3 1 5
1 4
1 6 =
x − x
dx = x − x 2
2
8
12 0
0
Part (c)
Z
1
Z
x
Z
xy dy dx =
1
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Solution:
1
Z
Z Z
Part (a)
Z
x
xy dy dx.
xy dA =
D
Z
Part (b)
1
√
Z
0
x2
y
xy dx dy .
0
y
x
1 2
xy
dx
2
0
x2
0
x2
1
Z 1
1 3 1 5
1 4
1 6 1
1
=
x − x
dx = x − x = −
2
2
8
12
8
12
0
0
Part (c)
Z
1
Z
x
Z
xy dy dx =
1
Problem 10 - Spring 2008
Let D be the region of the xy -plane above the graph of y = x 2 and
below the line y = x.
(a) Determine
an iterated integral expression for the double integral
RR
xy
dA
D
(b) Find an equivalent iterated integral to the one found in part (a)
with the reversed order of integration.
(c) Evaluate one of the two iterated integrals in parts (a), (b).
Solution:
1
Z
Z Z
Part (a)
Z
x
xy dy dx.
xy dA =
D
Z
Part (b)
1
√
Z
0
x2
y
xy dx dy .
0
y
x
1 2
xy
dx
2
0
x2
0
x2
1
Z 1
1 3 1 5
1 4
1 6 1
1
1
=
x − x
dx = x − x = −
=
2
2
8
12
8
12
24
0
0
Part (c)
Z
1
Z
x
Z
xy dy dx =
1
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
∇f = h2x, 4y − 2i = 0
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
1
.
2
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f h2x, 4y − 2i = λ∇g = λh2x, 2y i.
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f h2x, 4y − 2i = λ∇g = λh2x, 2y i.
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
We get
2x = λ2x
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f h2x, 4y − 2i = λ∇g = λh2x, 2y i.
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
We get
2x = λ2x =⇒ λ = 1 or x = 0.
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f h2x, 4y − 2i = λ∇g = λh2x, 2y i.
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
We get
2x = λ2x =⇒ λ = 1 or x = 0.
If λ = 1, then 4y − 2 = 2y
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f h2x, 4y − 2i = λ∇g = λh2x, 2y i.
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
We get
2x = λ2x =⇒ λ = 1 or x = 0.
If λ = 1, then 4y − 2 = 2y =⇒ y = 1.
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f h2x, 4y − 2i = λ∇g = λh2x, 2y i.
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
We get
2x = λ2x =⇒ λ = 1 or x = 0.
If λ = 1, then 4y − 2 = 2y =⇒ y = 1.
√
Plugging in g (x, y ), gives (0, 21 ), (0, ±2) and (± 3, 1).
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f h2x, 4y − 2i = λ∇g = λh2x, 2y i.
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
We get
2x = λ2x =⇒ λ = 1 or x = 0.
If λ = 1, then 4y − 2 = 2y =⇒ y = 1.
√
Plugging in g (x, y ), gives (0, 21 ), (0, ±2) and (± 3, 1). We get
√
f (0, 12 ) = − 12 , f (0, 2) = 4, f (0, −2) = 12, f (± 3, 1) = 3.
Problem 11 - Spring 2008
Find the absolute maximum and absolute minimum values of
f (x, y ) = x 2 + 2y 2 − 2y in the set D = {(x, y ) : x 2 + y 2 ≤ 4}.
Solution:
Find critical points of f (x, y ):
1
.
2
Use Lagrange multipliers to study max and min values of f on the
circle g (x, y ) = x 2 + y 2 = 4:
∇f h2x, 4y − 2i = λ∇g = λh2x, 2y i.
∇f = h2x, 4y − 2i = 0 =⇒ x = 0 and y =
We get
2x = λ2x =⇒ λ = 1 or x = 0.
If λ = 1, then 4y − 2 = 2y =⇒ y = 1.
√
Plugging in g (x, y ), gives (0, 21 ), (0, ±2) and (± 3, 1). We get
√
f (0, 12 ) = − 12 , f (0, 2) = 4, f (0, −2) = 12, f (± 3, 1) = 3.
The maximum value of f (x, y ) is 12 and its minimum value is − 12 .
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
x = r cos θ
y = r sin θ
dA = r dr dθ;
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
x = r cos θ
Z Z
(3x + 3y )dA
D
y = r sin θ
dA = r dr dθ;
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
x = r cos θ
y = r sin θ
Z Z
π
2
Z
(3x + 3y )dA =
D
0
dA = r dr dθ;
3
Z
(3r cos θ + 3r sin θ)r dr dθ
2
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
x = r cos θ
y = r sin θ
Z Z
π
2
Z
(3x + 3y )dA =
D
π
2
Z
=
0
0
Z
2
dA = r dr dθ;
3
Z
(3r cos θ + 3r sin θ)r dr dθ
2
3
3r 2 (cos θ + sin θ) dr dθ
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
x = r cos θ
y = r sin θ
Z Z
π
2
Z
(3x + 3y )dA =
D
π
2
Z
=
0
0
Z
2
dA = r dr dθ;
3
Z
(3r cos θ + 3r sin θ)r dr dθ
2
3
3r 2 (cos θ + sin θ) dr dθ =
Z
0
π
2
h
i3
r 3 (cos θ + sin θ) dθ
2
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
x = r cos θ
y = r sin θ
Z Z
π
2
Z
(3x + 3y )dA =
D
π
2
Z
=
Z
=
0
π
2
0
0
Z
3
Z
(3r cos θ + 3r sin θ)r dr dθ
2
3
3r 2 (cos θ + sin θ) dr dθ =
2
(27 − 8)(cos θ + sin θ) dθ
dA = r dr dθ;
Z
0
π
2
h
i3
r 3 (cos θ + sin θ) dθ
2
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
x = r cos θ
y = r sin θ
Z Z
π
2
Z
(3x + 3y )dA =
D
π
2
Z
=
Z
=
0
π
2
0
0
Z
2
dA = r dr dθ;
3
Z
(3r cos θ + 3r sin θ)r dr dθ
2
3
3r 2 (cos θ + sin θ) dr dθ =
Z
π
2
h
i3
r 3 (cos θ + sin θ) dθ
2
0
π
2
(27 − 8)(cos θ + sin θ) dθ = 19(sin θ − cos θ)
0
Problem 12(a) - Spring 2008
Let D be the region in the first quadrant x, y ≥ 0 that lies between the
two circles x 2 + y 2 = 4 and x 2 + y 2 = 9.
(a) Describe the region D using polar coordinates.
RR
(b) Evaluate the double integral
3x + 3y dA.
D
Solution:
The domain is:
π
D = {(r , θ) | 2 ≤ r ≤ 3, 0 ≤ θ ≤ }.
2
Calculate the integral using the substitution:
x = r cos θ
y = r sin θ
Z Z
π
2
Z
(3x + 3y )dA =
D
π
2
Z
=
Z
=
0
π
2
0
0
Z
2
dA = r dr dθ;
3
Z
(3r cos θ + 3r sin θ)r dr dθ
2
3
3r 2 (cos θ + sin θ) dr dθ =
Z
π
2
h
i3
r 3 (cos θ + sin θ) dθ
2
0
π
2
(27 − 8)(cos θ + sin θ) dθ = 19(sin θ − cos θ) = 19 + 19 = 38.
0
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Since F(x, y , z) is constant on the surface, the Chain Rule
gives:
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Since F(x, y , z) is constant on the surface, the Chain Rule
gives:
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
=0
∂x ∂x
∂y ∂x
∂z ∂x
∂x
∂z ∂x
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
= 0.
∂y ∂y
∂y ∂y
∂z ∂y
∂y
∂z ∂y
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Since F(x, y , z) is constant on the surface, the Chain Rule
gives:
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
=0
∂x ∂x
∂y ∂x
∂z ∂x
∂x
∂z ∂x
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
= 0.
∂y ∂y
∂y ∂y
∂z ∂y
∂y
∂z ∂y
Thus,
− ∂F
∂z
= ∂F∂x
∂x
∂z
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Since F(x, y , z) is constant on the surface, the Chain Rule
gives:
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
=0
∂x ∂x
∂y ∂x
∂z ∂x
∂x
∂z ∂x
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
= 0.
∂y ∂y
∂y ∂y
∂z ∂y
∂y
∂z ∂y
Thus,
− ∂F
−3x 2 + 3yz
∂z
= ∂F∂x =
3z 2 − 3xy
∂x
∂z
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Since F(x, y , z) is constant on the surface, the Chain Rule
gives:
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
=0
∂x ∂x
∂y ∂x
∂z ∂x
∂x
∂z ∂x
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
= 0.
∂y ∂y
∂y ∂y
∂z ∂y
∂y
∂z ∂y
Thus,
− ∂F
−3x 2 + 3yz
−x 2 + yz
∂z
,
=
= ∂F∂x =
3z 2 − 3xy
z 2 − xy
∂x
∂z
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Since F(x, y , z) is constant on the surface, the Chain Rule
gives:
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
=0
∂x ∂x
∂y ∂x
∂z ∂x
∂x
∂z ∂x
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
= 0.
∂y ∂y
∂y ∂y
∂z ∂y
∂y
∂z ∂y
Thus,
− ∂F
−3x 2 + 3yz
−x 2 + yz
∂z
,
=
= ∂F∂x =
3z 2 − 3xy
z 2 − xy
∂x
∂z
− ∂F
∂z
∂y
= ∂F
∂y
∂z
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Since F(x, y , z) is constant on the surface, the Chain Rule
gives:
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
=0
∂x ∂x
∂y ∂x
∂z ∂x
∂x
∂z ∂x
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
= 0.
∂y ∂y
∂y ∂y
∂z ∂y
∂y
∂z ∂y
Thus,
− ∂F
−3x 2 + 3yz
−x 2 + yz
∂z
,
=
= ∂F∂x =
3z 2 − 3xy
z 2 − xy
∂x
∂z
− ∂F
∂z
−3y 2 + 3xz
∂y
= ∂F =
∂y
3z 2 − 3xy
∂z
Problem 13(a) - Spring 2008
(a) Find
∂z
∂x
and
∂z
∂y
at the point (1, 0, −1) for the implicit
function z
determined by the equation x 3 + y 3 + z 3 − 3xyz = 0.
Solution:
Consider the function F(x, y , z) = x 3 + y 3 + z 3 − 3yxz.
Since F(x, y , z) is constant on the surface, the Chain Rule
gives:
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
=0
∂x ∂x
∂y ∂x
∂z ∂x
∂x
∂z ∂x
∂F ∂x
∂F ∂y
∂F ∂z
∂F
∂F ∂z
+
+
=
+
= 0.
∂y ∂y
∂y ∂y
∂z ∂y
∂y
∂z ∂y
Thus,
− ∂F
−3x 2 + 3yz
−x 2 + yz
∂z
,
=
= ∂F∂x =
3z 2 − 3xy
z 2 − xy
∂x
∂z
− ∂F
∂z
−3y 2 + 3xz
−y 2 + xz
∂y
= ∂F =
=
.
2 − 3xy
2 − xy
∂y
3z
z
∂z
Problem 13(b) - Spring 2008
(b) Is the tangent plane to the surface x 3 + y 3 + z 3 − 3xyz = 0
at the point (1, 0, −1) perpendicular to the plane
2x + y − 3z = 2? Justify your answer with an appropriate
calculation.
Problem 13(b) - Spring 2008
(b) Is the tangent plane to the surface x 3 + y 3 + z 3 − 3xyz = 0
at the point (1, 0, −1) perpendicular to the plane
2x + y − 3z = 2? Justify your answer with an appropriate
calculation.
Solution:
Since F(x, y , z) = x 3 + y 3 + z 3 − 3xyz is constant along the
surface F(x, y , z) = 0, ∇F is normal (orthogonal) to the
surface.
Problem 13(b) - Spring 2008
(b) Is the tangent plane to the surface x 3 + y 3 + z 3 − 3xyz = 0
at the point (1, 0, −1) perpendicular to the plane
2x + y − 3z = 2? Justify your answer with an appropriate
calculation.
Solution:
Since F(x, y , z) = x 3 + y 3 + z 3 − 3xyz is constant along the
surface F(x, y , z) = 0, ∇F is normal (orthogonal) to the
surface.
Calculating, we obtain:
∇F = h3x 2 − 3yz, 3y 2 − 3xz, 3z 2 − 3xy i
∇F(1, 0, −1) = h3, 3, 3i,
Problem 13(b) - Spring 2008
(b) Is the tangent plane to the surface x 3 + y 3 + z 3 − 3xyz = 0
at the point (1, 0, −1) perpendicular to the plane
2x + y − 3z = 2? Justify your answer with an appropriate
calculation.
Solution:
Since F(x, y , z) = x 3 + y 3 + z 3 − 3xyz is constant along the
surface F(x, y , z) = 0, ∇F is normal (orthogonal) to the
surface.
Calculating, we obtain:
∇F = h3x 2 − 3yz, 3y 2 − 3xz, 3z 2 − 3xy i
∇F(1, 0, −1) = h3, 3, 3i,
which is the normal vector to the tangent plane of the
surface.
Problem 13(b) - Spring 2008
(b) Is the tangent plane to the surface x 3 + y 3 + z 3 − 3xyz = 0
at the point (1, 0, −1) perpendicular to the plane
2x + y − 3z = 2? Justify your answer with an appropriate
calculation.
Solution:
Since F(x, y , z) = x 3 + y 3 + z 3 − 3xyz is constant along the
surface F(x, y , z) = 0, ∇F is normal (orthogonal) to the
surface.
Calculating, we obtain:
∇F = h3x 2 − 3yz, 3y 2 − 3xz, 3z 2 − 3xy i
∇F(1, 0, −1) = h3, 3, 3i,
which is the normal vector to the tangent plane of the
surface.
Since the normal to the plane 2x + y − 3z = 2 is h2, 1, −3i
and h3, 3, 3i · h2, 1, −3i = 0, it is perpendicular.
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
∂y
=x
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x)
∂x
∂x
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x) = 4x 3 + 2xy
∂x
∂x
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x) = 4x 3 + 2xy
∂x
∂x
=⇒ g 0 (x) = 4x 3
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x) = 4x 3 + 2xy
∂x
∂x
=⇒ g 0 (x) = 4x 3 =⇒ g (x) = x 4 + constant.
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x) = 4x 3 + 2xy
∂x
∂x
=⇒ g 0 (x) = 4x 3 =⇒ g (x) = x 4 + constant.
Hence, F(x, y ) = x 4 + x 2 y is a potential function.
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x) = 4x 3 + 2xy
∂x
∂x
=⇒ g 0 (x) = 4x 3 =⇒ g (x) = x 4 + constant.
Hence, F(x, y ) = x 4 + x 2 y is a potential function.
By theZ fundamental
Z theorem of calculus for line integrals,
G · dr =
C
∇F · dr
C
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x) = 4x 3 + 2xy
∂x
∂x
=⇒ g 0 (x) = 4x 3 =⇒ g (x) = x 4 + constant.
Hence, F(x, y ) = x 4 + x 2 y is a potential function.
By theZ fundamental
Z theorem of calculus for line integrals,
∇F · dr = F(−1, 1) − F(2, −2)
G · dr =
C
C
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x) = 4x 3 + 2xy
∂x
∂x
=⇒ g 0 (x) = 4x 3 =⇒ g (x) = x 4 + constant.
Hence, F(x, y ) = x 4 + x 2 y is a potential function.
By theZ fundamental
Z theorem of calculus for line integrals,
∇F · dr = F(−1, 1) − F(2, −2) = 2 − 8
G · dr =
C
C
Problem 14(a) - Spring 2008
(a) Consider the vector field G(x, y ) = h4x 3 + 2xy , x 2 i. Show that G is
conservative (i.e. G is a potential or a gradient vector field), andRuse the
fundamental theorem for line integrals to determine the value of C G · dr,
where C is the contour consisting of the line starting at (2, −2) and
ending at (−1, 1).
Solution:
∂
∂
Since ∂y
(4x 3 + 2xy ) = 2x = ∂x
(x 2 ), there exists a potential
function F(x, y ), where ∇F = G.
Note that:
∂F
2
2
∂y
= x =⇒ F(x, y ) = x y + g (x),
where g (x) is some function of x.
Since
∂F
∂x
= 4x 3 + 2xy ,
∂(x 2 y + g (x))
∂F
=
= 2xy + g 0 (x) = 4x 3 + 2xy
∂x
∂x
=⇒ g 0 (x) = 4x 3 =⇒ g (x) = x 4 + constant.
Hence, F(x, y ) = x 4 + x 2 y is a potential function.
By theZ fundamental
Z theorem of calculus for line integrals,
∇F · dr = F(−1, 1) − F(2, −2) = 2 − 8 = −6.
G · dr =
C
C
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Parameterize these segments as follows:
C1 (t) = ht, 0i
C2 (t) = h1, ti
C2 (t) = h1 − t, 1 − ti
0≤t≤1
0≤t≤1
0≤t≤1
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Parameterize these segments as follows:
0≤t≤1
C1 (t) = ht, 0i
C2 (t) = h1, ti
0≤t≤1
C2 (t) = h1 − t, 1 − ti 0 ≤ t ≤ 1
Z
Z
Z
Z
F · dr =
F · dr +
F · dr +
F · dr
T
C1
C2
C3
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Parameterize these segments as follows:
0≤t≤1
C1 (t) = ht, 0i
C2 (t) = h1, ti
0≤t≤1
C2 (t) = h1 − t, 1 − ti 0 ≤ t ≤ 1
Z
Z
Z
Z
F · dr =
F · dr +
F · dr +
F · dr
T
C1
C2
C3
Z 1
Z 1
Z 1
1
1
(1 − t)2 −(1 − t), (1 − t)2 ·h−1,−1idt
= h0, 0i·h1, 0idt+ h t 2 −t, ti·h0, 1idt+
0 2
0 2
0
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Parameterize these segments as follows:
0≤t≤1
C1 (t) = ht, 0i
C2 (t) = h1, ti
0≤t≤1
C2 (t) = h1 − t, 1 − ti 0 ≤ t ≤ 1
Z
Z
Z
Z
F · dr =
F · dr +
F · dr +
F · dr
T
C1
C2
C3
Z 1
Z 1
Z 1
1
1
(1 − t)2 −(1 − t), (1 − t)2 ·h−1,−1idt
= h0, 0i·h1, 0idt+ h t 2 −t, ti·h0, 1idt+
0 2
0 2
0
Z 1
Z 1
3
1
=0+
t dt +
− t 2 + 2t − dt
2
2
0
0
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Parameterize these segments as follows:
0≤t≤1
C1 (t) = ht, 0i
C2 (t) = h1, ti
0≤t≤1
C2 (t) = h1 − t, 1 − ti 0 ≤ t ≤ 1
Z
Z
Z
Z
F · dr =
F · dr +
F · dr +
F · dr
T
C1
C2
C3
Z 1
Z 1
Z 1
1
1
(1 − t)2 −(1 − t), (1 − t)2 ·h−1,−1idt
= h0, 0i·h1, 0idt+ h t 2 −t, ti·h0, 1idt+
0 2
0 2
0
Z 1
Z 1
Z 1
3
1
3
1
=0+
t dt +
− t 2 + 2t − dt =
− t 2 + 3t − dt
2
2
2
2
0
0
0
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Parameterize these segments as follows:
0≤t≤1
C1 (t) = ht, 0i
C2 (t) = h1, ti
0≤t≤1
C2 (t) = h1 − t, 1 − ti 0 ≤ t ≤ 1
Z
Z
Z
Z
F · dr =
F · dr +
F · dr +
F · dr
T
C1
C2
C3
Z 1
Z 1
Z 1
1
1
(1 − t)2 −(1 − t), (1 − t)2 ·h−1,−1idt
= h0, 0i·h1, 0idt+ h t 2 −t, ti·h0, 1idt+
0 2
0 2
0
Z 1
Z 1
Z 1
3
1
3
1
=0+
t dt +
− t 2 + 2t − dt =
− t 2 + 3t − dt
2
2
2
2
0
0
0
1 3
3 2
1 1
= − t + t − t
2
2
2
0
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Parameterize these segments as follows:
0≤t≤1
C1 (t) = ht, 0i
C2 (t) = h1, ti
0≤t≤1
C2 (t) = h1 − t, 1 − ti 0 ≤ t ≤ 1
Z
Z
Z
Z
F · dr =
F · dr +
F · dr +
F · dr
T
C1
C2
C3
Z 1
Z 1
Z 1
1
1
(1 − t)2 −(1 − t), (1 − t)2 ·h−1,−1idt
= h0, 0i·h1, 0idt+ h t 2 −t, ti·h0, 1idt+
0 2
0 2
0
Z 1
Z 1
Z 1
3
1
3
1
=0+
t dt +
− t 2 + 2t − dt =
− t 2 + 3t − dt
2
2
2
2
0
0
0
1 3
3 2
1 1
1
3
1
= − t + t − t = − + −
2
2
2
2
2
2
0
Problem 14(b) - Spring 2008
(b) Now let T denote the closed contour consisting of the triangle with
vertices at (0,0),(1,0), and (1,1) with the counterclockwise
orientation,
R
and let F(x, y ) = h 12 y 2 − y , xy i. Compute T F · dr directly (from the
definition of line integral).
Solution:
The curve T is the union of the segment C1 from (0, 0) to (1, 0),
the segment C2 from (1, 0) to (1, 1) and the segment C3 from (1, 1)
to (0, 0).
Parameterize these segments as follows:
0≤t≤1
C1 (t) = ht, 0i
C2 (t) = h1, ti
0≤t≤1
C2 (t) = h1 − t, 1 − ti 0 ≤ t ≤ 1
Z
Z
Z
Z
F · dr =
F · dr +
F · dr +
F · dr
T
C1
C2
C3
Z 1
Z 1
Z 1
1
1
(1 − t)2 −(1 − t), (1 − t)2 ·h−1,−1idt
= h0, 0i·h1, 0idt+ h t 2 −t, ti·h0, 1idt+
0 2
0 2
0
Z 1
Z 1
Z 1
3
1
3
1
=0+
t dt +
− t 2 + 2t − dt =
− t 2 + 3t − dt
2
2
2
2
0
0
0
1 3
3 2
1 1
1
3
1
1
= − t + t − t = − + − = .
2
2
2
2
2
2
2
0
Problem 14(c) - Spring 2008
Let F(x, y ) = h 12 y 2 − y , xy i.
(c) Explain how
R Green’s theorem can be used to show that the
integral T F · dr in (b) must be equal to the area of the
region D interior to T.
Problem 14(c) - Spring 2008
Let F(x, y ) = h 12 y 2 − y , xy i.
(c) Explain how
R Green’s theorem can be used to show that the
integral T F · dr in (b) must be equal to the area of the
region D interior to T.
Solution:
By Green’s Theorem,
Z
Z
1
F · dr = ( y 2 − y ) dx + xy dy
T
T 2
Problem 14(c) - Spring 2008
Let F(x, y ) = h 12 y 2 − y , xy i.
(c) Explain how
R Green’s theorem can be used to show that the
integral T F · dr in (b) must be equal to the area of the
region D interior to T.
Solution:
By Green’s Theorem,
Z
Z
1
F · dr = ( y 2 − y ) dx + xy dy
T
T 2
Z Z
1 2
∂(xy ) ∂( 2 y − y )
=
−
dA
∂y
D ∂x
Problem 14(c) - Spring 2008
Let F(x, y ) = h 12 y 2 − y , xy i.
(c) Explain how
R Green’s theorem can be used to show that the
integral T F · dr in (b) must be equal to the area of the
region D interior to T.
Solution:
By Green’s Theorem,
Z
Z
1
F · dr = ( y 2 − y ) dx + xy dy
T
T 2
Z Z
Z Z
1 2
∂(xy ) ∂( 2 y − y )
=
−
dA =
(y − y + 1) dA
∂y
D ∂x
D
Problem 14(c) - Spring 2008
Let F(x, y ) = h 12 y 2 − y , xy i.
(c) Explain how
R Green’s theorem can be used to show that the
integral T F · dr in (b) must be equal to the area of the
region D interior to T.
Solution:
By Green’s Theorem,
Z
Z
1
F · dr = ( y 2 − y ) dx + xy dy
T
T 2
Z Z
Z Z
1 2
∂(xy ) ∂( 2 y − y )
=
−
dA =
(y − y + 1) dA
∂y
D ∂x
D
Z Z
=
dA.
D
Problem 14(c) - Spring 2008
Let F(x, y ) = h 12 y 2 − y , xy i.
(c) Explain how
R Green’s theorem can be used to show that the
integral T F · dr in (b) must be equal to the area of the
region D interior to T.
Solution:
By Green’s Theorem,
Z
Z
1
F · dr = ( y 2 − y ) dx + xy dy
T
T 2
Z Z
Z Z
1 2
∂(xy ) ∂( 2 y − y )
=
−
dA =
(y − y + 1) dA
∂y
D ∂x
D
Z Z
=
dA.
D
Since
1
2
=
then the
RR
D dA
integral 12
is the area of the triangle D,
in part (b) is equal to this area.
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
3
I=
e x dx dy
√
0
y
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
(c) We next evaluate the integral using part (b).
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
(c) We next evaluate the integral using part (b).
Z 2 Z x2
3
I=
e x dy dx
0
0
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
(c) We next evaluate the integral using part (b).
Z 2 Z x2
Z 2h
ix 2
3
x3
ex y
dx
I=
e dy dx =
0
0
0
0
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
(c) We next evaluate the integral using part (b).
Z 2 Z x2
Z 2h
Z 2
ix 2
3
x3
x3
e y
dx =
I=
e dy dx =
e x x 2 dx
0
0
0
0
0
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
(c) We next evaluate the integral using part (b).
Z 2 Z x2
Z 2h
Z 2
ix 2
3
x3
x3
e y
dx =
I=
e dy dx =
e x x 2 dx
0
=
1
3
Z
0
2
0
3
e x (3x 2 ) dx
0
0
0
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
(c) We next evaluate the integral using part (b).
Z 2 Z x2
Z 2h
Z 2
ix 2
3
x3
x3
e y
dx =
I=
e dy dx =
e x x 2 dx
0
=
1
3
Z
0
2
0
0
0
1 x 3 2
x3
2
e (3x ) dx = e 3
0
0
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
(c) We next evaluate the integral using part (b).
Z 2 Z x2
Z 2h
Z 2
ix 2
3
x3
x3
e y
dx =
I=
e dy dx =
e x x 2 dx
0
=
1
3
Z
0
2
0
0
0
1 x 3 2
1
x3
2
e (3x ) dx = e = (e 8 − e 0 )
3
3
0
0
Problem 15(a,b,c) - Fall 2007
Z 4Z 2
Let
3
e x dx dy .
I=
√
0
y
(a) Sketch the region of integration
(b) Write the integral I with the order of integration reversed.
(c) Evaluate the integral I. Show your work.
Solution:
(a) (b) Make your sketch and from the sketch, we see:
Z 4Z 2
Z 2 Z x2
3
x3
dx
dy
e x dy dx.
I=
e
=
√
0
y
0
0
(c) We next evaluate the integral using part (b).
Z 2 Z x2
Z 2h
Z 2
ix 2
3
x3
x3
e y
dx =
I=
e dy dx =
e x x 2 dx
0
=
1
3
Z
0
2
0
0
0
0
1 x 3 2
1
1
x3
2
e (3x ) dx = e = (e 8 − e 0 ) = (e 8 − 1).
3
3
3
0
Problem 16 - Fall 2007
Find the distance from the point (3, 2, −7) to the line L
x = 1 + t, y = 2 − t, z = 1 + 3t.
Problem 16 - Fall 2007
Find the distance from the point (3, 2, −7) to the line L
x = 1 + t, y = 2 − t, z = 1 + 3t.
Solution:
Note that the plane T passing through P = (3, 2, −7) with
normal vector the direction n = h1, −1, 3i of the line L must
intersect L in the point Q closest to P.
Problem 16 - Fall 2007
Find the distance from the point (3, 2, −7) to the line L
x = 1 + t, y = 2 − t, z = 1 + 3t.
Solution:
Note that the plane T passing through P = (3, 2, −7) with
normal vector the direction n = h1, −1, 3i of the line L must
intersect L in the point Q closest to P. We now find Q.
Problem 16 - Fall 2007
Find the distance from the point (3, 2, −7) to the line L
x = 1 + t, y = 2 − t, z = 1 + 3t.
Solution:
Note that the plane T passing through P = (3, 2, −7) with
normal vector the direction n = h1, −1, 3i of the line L must
intersect L in the point Q closest to P. We now find Q.
The equation of the plane T is:
(x − 3) − (y − 2) + 3(z + 7) = 0.
Problem 16 - Fall 2007
Find the distance from the point (3, 2, −7) to the line L
x = 1 + t, y = 2 − t, z = 1 + 3t.
Solution:
Note that the plane T passing through P = (3, 2, −7) with
normal vector the direction n = h1, −1, 3i of the line L must
intersect L in the point Q closest to P. We now find Q.
The equation of the plane T is:
(x − 3) − (y − 2) + 3(z + 7) = 0.
Substitute in this equation the parametric values of L and
solve for t:
Problem 16 - Fall 2007
Find the distance from the point (3, 2, −7) to the line L
x = 1 + t, y = 2 − t, z = 1 + 3t.
Solution:
Note that the plane T passing through P = (3, 2, −7) with
normal vector the direction n = h1, −1, 3i of the line L must
intersect L in the point Q closest to P. We now find Q.
The equation of the plane T is:
(x − 3) − (y − 2) + 3(z + 7) = 0.
Substitute in this equation the parametric values of L and
solve for t:
0 = (1 + t) − 3 − [(2 − t) − 2] + 3[(1 + 3t) + 7] = 11t + 22.
Problem 16 - Fall 2007
Find the distance from the point (3, 2, −7) to the line L
x = 1 + t, y = 2 − t, z = 1 + 3t.
Solution:
Note that the plane T passing through P = (3, 2, −7) with
normal vector the direction n = h1, −1, 3i of the line L must
intersect L in the point Q closest to P. We now find Q.
The equation of the plane T is:
(x − 3) − (y − 2) + 3(z + 7) = 0.
Substitute in this equation the parametric values of L and
solve for t:
0 = (1 + t) − 3 − [(2 − t) − 2] + 3[(1 + 3t) + 7] = 11t + 22.
Hence, t = −2 and Q = h−1, 4, −5i.
Problem 16 - Fall 2007
Find the distance from the point (3, 2, −7) to the line L
x = 1 + t, y = 2 − t, z = 1 + 3t.
Solution:
Note that the plane T passing through P = (3, 2, −7) with
normal vector the direction n = h1, −1, 3i of the line L must
intersect L in the point Q closest to P. We now find Q.
The equation of the plane T is:
(x − 3) − (y − 2) + 3(z + 7) = 0.
Substitute in this equation the parametric values of L and
solve for t:
0 = (1 + t) − 3 − [(2 − t) − 2] + 3[(1 + 3t) + 7] = 11t + 22.
Hence, t = −2 and Q = h−1, 4, −5i.
√
√
The distance from P and Q is d = 42 + 22 + 22 = 24,
and thus the distance from P to L.
Problem 17(a) - Fall 2007
Find the velocity and acceleration of a particle moving along the
curve
r(t) = ht, t 2 , t 3 i
at the point (2, 4, 8).
Problem 17(a) - Fall 2007
Find the velocity and acceleration of a particle moving along the
curve
r(t) = ht, t 2 , t 3 i
at the point (2, 4, 8).
Solution:
Recall that the velocity v(t) = r0 (t) and the acceleration
a(t) = r0 (t):
Problem 17(a) - Fall 2007
Find the velocity and acceleration of a particle moving along the
curve
r(t) = ht, t 2 , t 3 i
at the point (2, 4, 8).
Solution:
Recall that the velocity v(t) = r0 (t) and the acceleration
a(t) = r0 (t):
v(t) = r0 (t) = h1, 2t, 3t 2 i
Problem 17(a) - Fall 2007
Find the velocity and acceleration of a particle moving along the
curve
r(t) = ht, t 2 , t 3 i
at the point (2, 4, 8).
Solution:
Recall that the velocity v(t) = r0 (t) and the acceleration
a(t) = r0 (t):
v(t) = r0 (t) = h1, 2t, 3t 2 i
a(t) = v0 (t) = h0, 2, 6ti.
Problem 17(a) - Fall 2007
Find the velocity and acceleration of a particle moving along the
curve
r(t) = ht, t 2 , t 3 i
at the point (2, 4, 8).
Solution:
Recall that the velocity v(t) = r0 (t) and the acceleration
a(t) = r0 (t):
v(t) = r0 (t) = h1, 2t, 3t 2 i
a(t) = v0 (t) = h0, 2, 6ti.
As the point (2, 4, 8) corresponds to t = 2 on r(t),
Problem 17(a) - Fall 2007
Find the velocity and acceleration of a particle moving along the
curve
r(t) = ht, t 2 , t 3 i
at the point (2, 4, 8).
Solution:
Recall that the velocity v(t) = r0 (t) and the acceleration
a(t) = r0 (t):
v(t) = r0 (t) = h1, 2t, 3t 2 i
a(t) = v0 (t) = h0, 2, 6ti.
As the point (2, 4, 8) corresponds to t = 2 on r(t),
v(2) = h1, 4, 12i
a(2) = h0, 2, 12i.
Problem 17(b) - Fall 2007
Find all points where the curve in part (a) intersects the surface
z = 3x 3 + xy − x.
Problem 17(b) - Fall 2007
Find all points where the curve in part (a) intersects the surface
z = 3x 3 + xy − x.
Solution:
Plug the x, y and z-coordinates into the equation of the
surface and solve for t.
Problem 17(b) - Fall 2007
Find all points where the curve in part (a) intersects the surface
z = 3x 3 + xy − x.
Solution:
Plug the x, y and z-coordinates into the equation of the
surface and solve for t.
t 3 = 3t 3 + t 3 − t
Problem 17(b) - Fall 2007
Find all points where the curve in part (a) intersects the surface
z = 3x 3 + xy − x.
Solution:
Plug the x, y and z-coordinates into the equation of the
surface and solve for t.
t 3 = 3t 3 + t 3 − t
=⇒ 3t 3 − t = t(3t 2 − 1) = 0
Problem 17(b) - Fall 2007
Find all points where the curve in part (a) intersects the surface
z = 3x 3 + xy − x.
Solution:
Plug the x, y and z-coordinates into the equation of the
surface and solve for t.
t 3 = 3t 3 + t 3 − t
or
=⇒ 3t 3 − t = t(3t 2 − 1) = 0 =⇒ t = 0
1
t = ±√ .
3
Problem 17(b) - Fall 2007
Find all points where the curve in part (a) intersects the surface
z = 3x 3 + xy − x.
Solution:
Plug the x, y and z-coordinates into the equation of the
surface and solve for t.
t 3 = 3t 3 + t 3 − t
or
=⇒ 3t 3 − t = t(3t 2 − 1) = 0 =⇒ t = 0
1
t = ±√ .
3
Next plug these t values into r(t) to get the 3 points of
intersection:
Problem 17(b) - Fall 2007
Find all points where the curve in part (a) intersects the surface
z = 3x 3 + xy − x.
Solution:
Plug the x, y and z-coordinates into the equation of the
surface and solve for t.
t 3 = 3t 3 + t 3 − t
or
=⇒ 3t 3 − t = t(3t 2 − 1) = 0 =⇒ t = 0
1
t = ±√ .
3
Next plug these t values into r(t) to get the 3 points of
intersection:
r(0) = h0, 0, 0, i
1
1 1 1
r( √ ) = h √ , , √ i
3
3 3 3 3
1 1
1
1
r(− √ ) = h− √ , , − √ i.
3
3 3 3 3
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates the domain D ⊂ R2 for the
integral.
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
This gives the volume:
Z 2π Z √3 p
V=
4 − r 2 dA
0
0
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
This gives the volume:
Z
Z 2π Z √3 p
2
V=
4 − r dA =
0
0
0
2π
√
Z
0
3
p
4 − r 2 r dr dθ
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
This gives the volume:
Z
Z 2π Z √3 p
2
V=
4 − r dA =
0
1
=−
2
Z
0
2π
0
0
√
Z
0
3
1
(4−r 2 ) 2 (−2r ) dr dθ
2π
√
Z
0
3
p
4 − r 2 r dr dθ
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
This gives the volume:
Z
Z 2π Z √3 p
2
V=
4 − r dA =
0
1
=−
2
Z
0
2π
0
0
√
Z
0
3
2π
√
Z
1
2
p
4 − r 2 r dr dθ
2π
0
1
(4−r ) (−2r ) dr dθ = −
2
2
3
Z
0
3
2
(4 − r 2 ) 2
3
√3
dθ
0
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
This gives the volume:
Z
Z 2π Z √3 p
2
V=
4 − r dA =
0
1
=−
2
Z
0
2π
0
0
√
Z
3
1
=−
3
Z
1
2
2π
(1 − 8) dθ
0
√
Z
3
p
4 − r 2 r dr dθ
2π
0
1
(4−r ) (−2r ) dr dθ = −
2
2
0
2π
Z
0
3
2
(4 − r 2 ) 2
3
√3
dθ
0
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
This gives the volume:
Z
Z 2π Z √3 p
2
V=
4 − r dA =
0
1
=−
2
Z
0
2π
0
0
√
Z
3
1
=−
3
Z
2π
Z
0
2π
3
p
4 − r 2 r dr dθ
2π
0
1
2
(1 − 8) dθ =
0
√
Z
1
(4−r ) (−2r ) dr dθ = −
2
2
0
2π
7
dθ
3
Z
0
3
2
(4 − r 2 ) 2
3
√3
dθ
0
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
This gives the volume:
Z
Z 2π Z √3 p
2
V=
4 − r dA =
0
1
=−
2
Z
0
2π
0
0
√
Z
3
1
=−
3
Z
1
2
2π
Z
(1 − 8) dθ =
0
√
Z
0
2π
3
p
4 − r 2 r dr dθ
2π
0
1
(4−r ) (−2r ) dr dθ = −
2
2
0
2π
Z
0
3
2
(4 − r 2 ) 2
3
2π
7
7 dθ = θ
3
3 0
√3
dθ
0
Problem 18 - Fall 2007
Find the volume V of the solid which lies below the sphere
x 2 + y 2 + z 2 = 4, above the xy -plane, and inside the cylinder
x 2 + y 2 = 3.
Solution:
We first describe in polar coordinates
the domain D ⊂ R2 for the
√
integral. D = {(r , θ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π}.
p
√
The graphing function is z = 4 − x 2 − y 2 = 4 − r 2 .
This gives the volume:
Z
Z 2π Z √3 p
2
V=
4 − r dA =
0
1
=−
2
Z
0
2π
0
0
√
Z
3
1
=−
3
Z
1
2
2π
Z
(1 − 8) dθ =
0
√
Z
0
2π
3
p
4 − r 2 r dr dθ
2π
0
1
(4−r ) (−2r ) dr dθ = −
2
2
0
2π
Z
0
3
2
(4 − r 2 ) 2
3
2π
7
7 14π
dθ = θ =
.
3
3 0
3
√3
dθ
0
Problem 19(a) - Fall 2007
Consider the line integral
√
1 + x dx + 2xy dy ,
Z
C
where C is the triangular path starting from (0, 0), to (2, 0), to (2, 3), and back
to (0, 0).
(a) Evaluate this line integral directly, without using Green’s Theorem.
Problem 19(a) - Fall 2007
Consider the line integral
√
1 + x dx + 2xy dy ,
Z
C
where C is the triangular path starting from (0, 0), to (2, 0), to (2, 3), and back
to (0, 0).
(a) Evaluate this line integral directly, without using Green’s Theorem.
Solution:
The curve C is the union of the segment C1 from (0, 0) to (2, 0), the
segment C2 from (2, 0) to (2, 3) and the segment C3 from (2, 3) to (0, 0).
Problem 19(a) - Fall 2007
Consider the line integral
√
1 + x dx + 2xy dy ,
Z
C
where C is the triangular path starting from (0, 0), to (2, 0), to (2, 3), and back
to (0, 0).
(a) Evaluate this line integral directly, without using Green’s Theorem.
Solution:
The curve C is the union of the segment C1 from (0, 0) to (2, 0), the
segment C2 from (2, 0) to (2, 3) and the segment C3 from (2, 3) to (0, 0).
Parameterize these segments:
C1 (t) = h2t, 0i
C2 (t) = h2, 3ti
C3 (t) = h2 − 2t, 3 − 3ti
0≤t≤1
0≤t≤1
0 ≤ t ≤ 1.
Problem 19(a) - Fall 2007
Consider the line integral
√
1 + x dx + 2xy dy ,
Z
C
where C is the triangular path starting from (0, 0), to (2, 0), to (2, 3), and back
to (0, 0).
(a) Evaluate this line integral directly, without using Green’s Theorem.
Solution:
The curve C is the union of the segment C1 from (0, 0) to (2, 0), the
segment C2 from (2, 0) to (2, 3) and the segment C3 from (2, 3) to (0, 0).
Parameterize these segments:
Thus,
R √
C
C1 (t) = h2t, 0i
C2 (t) = h2, 3ti
C3 (t) = h2 − 2t, 3 − 3ti
1 + x dx + 2xy dy
0≤t≤1
0≤t≤1
0 ≤ t ≤ 1.
Problem 19(a) - Fall 2007
Consider the line integral
√
1 + x dx + 2xy dy ,
Z
C
where C is the triangular path starting from (0, 0), to (2, 0), to (2, 3), and back
to (0, 0).
(a) Evaluate this line integral directly, without using Green’s Theorem.
Solution:
The curve C is the union of the segment C1 from (0, 0) to (2, 0), the
segment C2 from (2, 0) to (2, 3) and the segment C3 from (2, 3) to (0, 0).
Parameterize these segments:
Thus,
R √
=
C1
1 + x dx + 2xy dy
Z
1 + x dx + 2xy dy +
0≤t≤1
0≤t≤1
0 ≤ t ≤ 1.
C
√
Z
C1 (t) = h2t, 0i
C2 (t) = h2, 3ti
C3 (t) = h2 − 2t, 3 − 3ti
C2
√
√
Z
1 + x dx + 2xy dy +
C3
1 + x dx + 2xy dy
Problem 19(a) - Fall 2007
Consider the line integral
√
1 + x dx + 2xy dy ,
Z
C
where C is the triangular path starting from (0, 0), to (2, 0), to (2, 3), and back
to (0, 0).
(a) Evaluate this line integral directly, without using Green’s Theorem.
Solution:
The curve C is the union of the segment C1 from (0, 0) to (2, 0), the
segment C2 from (2, 0) to (2, 3) and the segment C3 from (2, 3) to (0, 0).
Parameterize these segments:
Thus,
R √
=
1 + x dx + 2xy dy
Z
1 + x dx + 2xy dy +
C1
1
Z
=
0
√
0≤t≤1
0≤t≤1
0 ≤ t ≤ 1.
C
√
Z
C1 (t) = h2t, 0i
C2 (t) = h2, 3ti
C3 (t) = h2 − 2t, 3 − 3ti
√
C2
1
Z
1 + 2t(2) dt+
1 + x dx + 2xy dy
C3
1
Z
4·3t(3) dt+
0
√
Z
1 + x dx + 2xy dy +
0
p
1 + (2 − 2t)(−2)+2(2−2t)(3−3t)(−3)dt
Problem 19(a) - Fall 2007
Consider the line integral
√
1 + x dx + 2xy dy ,
Z
C
where C is the triangular path starting from (0, 0), to (2, 0), to (2, 3), and back
to (0, 0).
(a) Evaluate this line integral directly, without using Green’s Theorem.
Solution:
The curve C is the union of the segment C1 from (0, 0) to (2, 0), the
segment C2 from (2, 0) to (2, 3) and the segment C3 from (2, 3) to (0, 0).
Parameterize these segments:
Thus,
R √
=
1 + x dx + 2xy dy
Z
1 + x dx + 2xy dy +
C1
1
Z
=
√
0≤t≤1
0≤t≤1
0 ≤ t ≤ 1.
C
√
Z
C1 (t) = h2t, 0i
C2 (t) = h2, 3ti
C3 (t) = h2 − 2t, 3 − 3ti
√
C2
1
Z
1 + 2t(2) dt+
0
=
0
1 + x dx + 2xy dy
C3
1
Z
4·3t(3) dt+
0
1
Z
√
Z
1 + x dx + 2xy dy +
p
1 + (2 − 2t)(−2)+2(2−2t)(3−3t)(−3)dt
0
√
√
2 1 + 2t + 36t 2 + 3 − 2t − 36t + 72t − 36 dt.
Problem 19(a) - Fall 2007
Consider the line integral
√
1 + x dx + 2xy dy ,
Z
C
where C is the triangular path starting from (0, 0), to (2, 0), to (2, 3), and back
to (0, 0).
(a) Evaluate this line integral directly, without using Green’s Theorem.
Solution:
The curve C is the union of the segment C1 from (0, 0) to (2, 0), the
segment C2 from (2, 0) to (2, 3) and the segment C3 from (2, 3) to (0, 0).
Parameterize these segments:
Thus,
R √
=
1 + x dx + 2xy dy
Z
1 + x dx + 2xy dy +
C1
1
Z
=
√
0≤t≤1
0≤t≤1
0 ≤ t ≤ 1.
C
√
Z
C1 (t) = h2t, 0i
C2 (t) = h2, 3ti
C3 (t) = h2 − 2t, 3 − 3ti
√
C2
1
Z
1 + 2t(2) dt+
0
=
0
1 + x dx + 2xy dy
C3
1
Z
4·3t(3) dt+
0
1
Z
√
Z
1 + x dx + 2xy dy +
p
1 + (2 − 2t)(−2)+2(2−2t)(3−3t)(−3)dt
0
√
√
2 1 + 2t + 36t 2 + 3 − 2t − 36t + 72t − 36 dt.
This long straightforward integral is left to you the student to do.
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Solution:
Let D denote the dimensional triangle bounded by C.
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Solution:
Let D denote the dimensional triangle bounded by C.
Green’s Theorem gives:
Z
C
√
Z Z 1 + x dx + 2xy dy =
D
√
∂(2xy ) ∂( 1 + x)
−
dA
∂x
∂y
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Solution:
Let D denote the dimensional triangle bounded by C.
Green’s Theorem gives:
Z
√
Z Z 1 + x dx + 2xy dy =
C
D
Z Z
=
2y dA
D
√
∂(2xy ) ∂( 1 + x)
−
dA
∂x
∂y
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Solution:
Let D denote the dimensional triangle bounded by C.
Green’s Theorem gives:
Z
√
Z Z 1 + x dx + 2xy dy =
C
D
Z Z
=
Z
2
Z
2y dA =
D
√
∂(2xy ) ∂( 1 + x)
−
dA
∂x
∂y
3
2x
2y dy dx
0
0
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Solution:
Let D denote the dimensional triangle bounded by C.
Green’s Theorem gives:
Z
√
Z Z 1 + x dx + 2xy dy =
C
D
Z Z
=
Z
2
Z
2y dA =
D
√
∂(2xy ) ∂( 1 + x)
−
dA
∂x
∂y
3
2x
Z
2y dy dx =
0
0
0
2
h
y2
i 32 x
0
dx
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Solution:
Let D denote the dimensional triangle bounded by C.
Green’s Theorem gives:
Z
√
Z Z 1 + x dx + 2xy dy =
C
D
Z Z
=
2
Z
Z
3
2x
2y dA =
0
Z
0
Z
2y dy dx =
D
=
√
∂(2xy ) ∂( 1 + x)
−
dA
∂x
∂y
2
0
9 2
x dx
4
0
2
h
y2
i 32 x
0
dx
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Solution:
Let D denote the dimensional triangle bounded by C.
Green’s Theorem gives:
Z
√
Z Z 1 + x dx + 2xy dy =
C
D
Z Z
=
2
Z
Z
2y dA =
0
Z
0
3
2x
Z
2y dy dx =
D
=
√
∂(2xy ) ∂( 1 + x)
−
dA
∂x
∂y
2
0
0
2
9 2
3 3 x dx = x 4
4
0
2
h
y2
i 32 x
0
dx
Problem 19(b) - Fall 2007
Consider the line integral
Z
√
1 + x dx + 2xy dy ,
C
where C is the triangular path starting from (0, 0), to (2, 0), to
(2, 3), and back to (0, 0).
Evaluate this line integral using Green’s theorem.
Solution:
Let D denote the dimensional triangle bounded by C.
Green’s Theorem gives:
Z
√
Z Z 1 + x dx + 2xy dy =
C
D
Z Z
=
2
Z
Z
2y dA =
0
Z
0
3
2x
Z
2
2y dy dx =
D
=
√
∂(2xy ) ∂( 1 + x)
−
dA
∂x
∂y
2
0
0
2
9 2
3 3 x dx = x = 6.
4
4
0
h
y2
i 32 x
0
dx
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists.
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
y2
∂f
= 2
∂x
x
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
y2
∂f
= 2 =⇒ f (x, y ) =
∂x
x
Z
y2
dx
x2
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
y2
∂f
= 2 =⇒ f (x, y ) =
∂x
x
Z
y2
y2
=
−
dx
+ g (y ),
x2
x
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
y2
∂f
= 2 =⇒ f (x, y ) =
∂x
x
where g (y ) is a function of y .
Z
y2
y2
=
−
dx
+ g (y ),
x2
x
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
y2
∂f
= 2 =⇒ f (x, y ) =
∂x
x
where g (y ) is a function of y .
Then:
∂f
∂
y2
=
(− + g (y ))
∂y
∂y
x
Z
y2
y2
=
−
dx
+ g (y ),
x2
x
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
Z 2
y
y2
y2
∂f
=
−
dx
+ g (y ),
= 2 =⇒ f (x, y ) =
x2
x
∂x
x
where g (y ) is a function of y .
Then:
2y
∂f
∂
y2
+ g 0 (y )
=
(− + g (y )) = −
x
∂y
∂y
x
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
Z 2
y
y2
y2
∂f
=
−
dx
+ g (y ),
= 2 =⇒ f (x, y ) =
x2
x
∂x
x
where g (y ) is a function of y .
Then:
2y
2y
∂f
∂
y2
+ g 0 (y ) = − .
=
(− + g (y )) = −
x
x
∂y
∂y
x
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
Z 2
y
y2
y2
∂f
=
−
dx
+ g (y ),
= 2 =⇒ f (x, y ) =
x2
x
∂x
x
where g (y ) is a function of y .
Then:
2y
2y
∂f
∂
y2
+ g 0 (y ) = − .
=
(− + g (y )) = −
x
x
∂y
∂y
x
Hence,
g 0 (y ) = 0
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
Z 2
y
y2
y2
∂f
=
−
dx
+ g (y ),
= 2 =⇒ f (x, y ) =
x2
x
∂x
x
where g (y ) is a function of y .
Then:
2y
2y
∂f
∂
y2
+ g 0 (y ) = − .
=
(− + g (y )) = −
x
x
∂y
∂y
x
Hence,
g 0 (y ) = 0 =⇒ g (y ) is constant.
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
Z 2
y
y2
y2
∂f
=
−
dx
+ g (y ),
= 2 =⇒ f (x, y ) =
x2
x
∂x
x
where g (y ) is a function of y .
Then:
2y
2y
∂f
∂
y2
+ g 0 (y ) = − .
=
(− + g (y )) = −
x
x
∂y
∂y
x
Hence,
g 0 (y ) = 0 =⇒ g (y ) is constant.
Taking the constant g (y ) to be zero, we obtain:
Problem 20(a) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
Find a function f such that ∇f = F.
Solution:
Suppose f exists. Then:
Z 2
y
y2
y2
∂f
=
−
dx
+ g (y ),
= 2 =⇒ f (x, y ) =
x2
x
∂x
x
where g (y ) is a function of y .
Then:
2y
2y
∂f
∂
y2
+ g 0 (y ) = − .
=
(− + g (y )) = −
x
x
∂y
∂y
x
Hence,
g 0 (y ) = 0 =⇒ g (y ) is constant.
Taking the constant g (y ) to be zero, we obtain:
f (x, y ) = −
y2
.
x
Problem 20(b) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
π
3
Let C be the curve given by
R r(t) = ht , sin ti for 2 ≤ t ≤ π.
Evaluate the line integral C F · dr.
Problem 20(b) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
π
3
Let C be the curve given by
R r(t) = ht , sin ti for 2 ≤ t ≤ π.
Evaluate the line integral C F · dr.
Solution:
2
We will apply the potential function f (x, y ) = −yx in part
(a) and the fundamental theorem of calculus for line integrals.
Problem 20(b) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
π
3
Let C be the curve given by
R r(t) = ht , sin ti for 2 ≤ t ≤ π.
Evaluate the line integral C F · dr.
Solution:
2
We will apply the potential function f (x, y ) = −yx in part
(a) and the fundamental theorem of calculus for line integrals.
We get:
Z
π
F · dr = f (r(π)) − f (r( ))
2
C
Problem 20(b) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
π
3
Let C be the curve given by
R r(t) = ht , sin ti for 2 ≤ t ≤ π.
Evaluate the line integral C F · dr.
Solution:
2
We will apply the potential function f (x, y ) = −yx in part
(a) and the fundamental theorem of calculus for line integrals.
We get:
Z
π3
π
F · dr = f (r(π)) − f (r( )) = f (π 3 , 0) − f ( , 1)
8
2
C
Problem 20(b) - Fall 2007
Consider the vector field F = (y 2 /x 2 )i − (2y /x)j.
π
3
Let C be the curve given by
R r(t) = ht , sin ti for 2 ≤ t ≤ π.
Evaluate the line integral C F · dr.
Solution:
2
We will apply the potential function f (x, y ) = −yx in part
(a) and the fundamental theorem of calculus for line integrals.
We get:
Z
π3
8
π
F · dr = f (r(π)) − f (r( )) = f (π 3 , 0) − f ( , 1) = 3
8
π
2
C
Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Solution:
The direction vector v of the line L is parallel to both planes.
Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Solution:
The direction vector v of the line L is parallel to both planes.
Hence, v = n1 × n2 where n1 = h1, −2, 1i and n2 = h2, 1, 1i are the
normal vectors of the respective planes:
Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Solution:
The direction vector v of the line L is parallel to both planes.
Hence, v = n1 × n2 where n1 = h1, −2, 1i and n2 = h2, 1, 1i are the
normal vectors of the respective planes:
i
j k v = n1 × n2 = 1 −2 1 2 1 1 Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Solution:
The direction vector v of the line L is parallel to both planes.
Hence, v = n1 × n2 where n1 = h1, −2, 1i and n2 = h2, 1, 1i are the
normal vectors of the respective planes:
i
j k v = n1 × n2 = 1 −2 1 = h−3, 1, 5i.
2 1 1 Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Solution:
The direction vector v of the line L is parallel to both planes.
Hence, v = n1 × n2 where n1 = h1, −2, 1i and n2 = h2, 1, 1i are the
normal vectors of the respective planes:
i
j k v = n1 × n2 = 1 −2 1 = h−3, 1, 5i.
2 1 1 Next find the intersection point of L with the xy -plane by setting
z = 0:
Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Solution:
The direction vector v of the line L is parallel to both planes.
Hence, v = n1 × n2 where n1 = h1, −2, 1i and n2 = h2, 1, 1i are the
normal vectors of the respective planes:
i
j k v = n1 × n2 = 1 −2 1 = h−3, 1, 5i.
2 1 1 Next find the intersection point of L with the xy -plane by setting
z = 0:
x − 2y = 1
2x + y = 1
Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Solution:
The direction vector v of the line L is parallel to both planes.
Hence, v = n1 × n2 where n1 = h1, −2, 1i and n2 = h2, 1, 1i are the
normal vectors of the respective planes:
i
j k v = n1 × n2 = 1 −2 1 = h−3, 1, 5i.
2 1 1 Next find the intersection point of L with the xy -plane by setting
z = 0:
x − 2y = 1
2x + y = 1
=⇒ ( 35 , − 15 , 0) is the intersection point.
Problem 21 - Fall 2006
Find parametric equations for the line L in which the planes
x − 2y + z = 1 and 2x + y + z = 1 intersect.
Solution:
The direction vector v of the line L is parallel to both planes.
Hence, v = n1 × n2 where n1 = h1, −2, 1i and n2 = h2, 1, 1i are the
normal vectors of the respective planes:
i
j k v = n1 × n2 = 1 −2 1 = h−3, 1, 5i.
2 1 1 Next find the intersection point of L with the xy -plane by setting
z = 0:
x − 2y = 1
2x + y = 1
=⇒ ( 35 , − 15 , 0) is the intersection point.
Parametric equations are:
x = 35 − 3t
y = − 15 + t
z = 5t.
Problem 22 - Fall 2006
Consider the surface x 2 + y 2 − 2z 2 = 0 and the point P(1, 1, 1)
which lies on the surface.
(i) Find the equation of the tangent plane to the surface at P.
(ii) Find the equation of the normal line to the surface at P.
Problem 22 - Fall 2006
Consider the surface x 2 + y 2 − 2z 2 = 0 and the point P(1, 1, 1)
which lies on the surface.
(i) Find the equation of the tangent plane to the surface at P.
(ii) Find the equation of the normal line to the surface at P.
Solution:
Recall that the gradient of F(x, y , z) = x 2 + y 2 − 2z 2 is
normal (orthogonal) to the surface.
Problem 22 - Fall 2006
Consider the surface x 2 + y 2 − 2z 2 = 0 and the point P(1, 1, 1)
which lies on the surface.
(i) Find the equation of the tangent plane to the surface at P.
(ii) Find the equation of the normal line to the surface at P.
Solution:
Recall that the gradient of F(x, y , z) = x 2 + y 2 − 2z 2 is
normal (orthogonal) to the surface.
Calculating, we obtain:
∇F(x, y , z) = h2x, 2y , −4zi
∇F(1, 1, 1) = h2, 2, −4i.
Problem 22 - Fall 2006
Consider the surface x 2 + y 2 − 2z 2 = 0 and the point P(1, 1, 1)
which lies on the surface.
(i) Find the equation of the tangent plane to the surface at P.
(ii) Find the equation of the normal line to the surface at P.
Solution:
Recall that the gradient of F(x, y , z) = x 2 + y 2 − 2z 2 is
normal (orthogonal) to the surface.
Calculating, we obtain:
∇F(x, y , z) = h2x, 2y , −4zi
∇F(1, 1, 1) = h2, 2, −4i.
The equation of the tangent plane is:
h2, 2, −4i·hx −1, y −1, z −1i = 2(x −1)+2(y −1)−4(z −1) = 0
Problem 22 - Fall 2006
Consider the surface x 2 + y 2 − 2z 2 = 0 and the point P(1, 1, 1)
which lies on the surface.
(i) Find the equation of the tangent plane to the surface at P.
(ii) Find the equation of the normal line to the surface at P.
Solution:
Recall that the gradient of F(x, y , z) = x 2 + y 2 − 2z 2 is
normal (orthogonal) to the surface.
Calculating, we obtain:
∇F(x, y , z) = h2x, 2y , −4zi
∇F(1, 1, 1) = h2, 2, −4i.
The equation of the tangent plane is:
h2, 2, −4i·hx −1, y −1, z −1i = 2(x −1)+2(y −1)−4(z −1) = 0
The vector equation of the normal line is:
L(t) = h1, 1, 1, i + th2, 2, −4i = h1 + 2t, 1 + 2t, 1 − 4ti.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
2y = λ2y
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
2y = λ2y =⇒ y = 0 or λ = 1.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
2y = λ2y =⇒ y = 0 or λ = 1.
y =0
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
2y = λ2y =⇒ y = 0 or λ = 1.
y = 0 =⇒ x = ±2.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
2y = λ2y =⇒ y = 0 or λ = 1.
y = 0 =⇒ x = ±2.
λ=1
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
2y = λ2y =⇒ y = 0 or λ = 1.
y = 0 =⇒ x = ±2.
λ = 1 =⇒ 2x − 2 = 2x, which is impossible.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
2y = λ2y =⇒ y = 0 or λ = 1.
y = 0 =⇒ x = ±2.
λ = 1 =⇒ 2x − 2 = 2x, which is impossible.
Now check the values of f at 3 points:
f (1, 0) = −1, f (2, 0) = 0, f (−2, 0) = 8.
Problem 23 - Fall 2006
Find the maximum and minimum values of the function
f (x, y ) = x 2 + y 2 − 2x
on the disc x 2 + y 2 ≤ 4.
Solution:
We first find the critical points.
∇f = h2x − 2, 2y i = 0 =⇒ x = 1 and y = 0.
Next use Lagrange multipliers to study max and min of f on
the boundary circle g (x, y ) = x 2 + y 2 = 4:
∇f = h2x − 2, 2y i = λ∇g = λh2x, 2y i.
2y = λ2y =⇒ y = 0 or λ = 1.
y = 0 =⇒ x = ±2.
λ = 1 =⇒ 2x − 2 = 2x, which is impossible.
Now check the values of f at 3 points:
f (1, 0) = −1, f (2, 0) = 0, f (−2, 0) = 8.
The maximum value is 8 and the minimum value is −1.
Problem 24 - Fall 2006
Evaluate the iterated integral
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx.
0
0
Problem 24 - Fall 2006
Evaluate the iterated integral
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx.
0
0
Solution:
The domain of integration for the function described in polar
coordinates is:
π
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ }.
2
Problem 24 - Fall 2006
Evaluate the iterated integral
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx.
0
0
Solution:
The domain of integration for the function described in polar
coordinates is:
π
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ }.
2
Since
dA = dy dx
in polar coordinates is:
Problem 24 - Fall 2006
Evaluate the iterated integral
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx.
0
0
Solution:
The domain of integration for the function described in polar
coordinates is:
π
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ }.
2
Since
dA = dy dx
p
in polar coordinates is: r dr dθ and r = x 2 + y 2 ,
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx
0
0
Problem 24 - Fall 2006
Evaluate the iterated integral
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx.
0
0
Solution:
The domain of integration for the function described in polar
coordinates is:
π
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ }.
2
Since
dA = dy dx
p
in polar coordinates is: r dr dθ and r = x 2 + y 2 ,
Z 1 Z √1−x 2 p
Z πZ 1
2
2
2
x + y dy dx =
r 2 dr dθ
0
0
Z
0
0
π
2
r3
3
1
dθ
0
0
Problem 24 - Fall 2006
Evaluate the iterated integral
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx.
0
0
Solution:
The domain of integration for the function described in polar
coordinates is:
π
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ }.
2
Since
dA = dy dx
p
in polar coordinates is: r dr dθ and r = x 2 + y 2 ,
Z 1 Z √1−x 2 p
Z πZ 1
2
2
2
x + y dy dx =
r 2 dr dθ
0
0
Z
0
0
π
2
r3
3
1
0
1
dθ =
3
Z
0
π
2
dθ
0
Problem 24 - Fall 2006
Evaluate the iterated integral
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx.
0
0
Solution:
The domain of integration for the function described in polar
coordinates is:
π
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ }.
2
Since
dA = dy dx
p
in polar coordinates is: r dr dθ and r = x 2 + y 2 ,
Z 1 Z √1−x 2 p
Z πZ 1
2
2
2
x + y dy dx =
r 2 dr dθ
0
0
Z
0
0
π
2
r3
3
1
0
1
dθ =
3
Z
0
π
2
0
π
θ 2
dθ = 3 0
Problem 24 - Fall 2006
Evaluate the iterated integral
Z 1 Z √1−x 2 p
x 2 + y 2 dy dx.
0
0
Solution:
The domain of integration for the function described in polar
coordinates is:
π
D = {(r , θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ }.
2
Since
dA = dy dx
p
in polar coordinates is: r dr dθ and r = x 2 + y 2 ,
Z 1 Z √1−x 2 p
Z πZ 1
2
2
2
x + y dy dx =
r 2 dr dθ
0
0
Z
0
0
π
2
r3
3
1
0
1
dθ =
3
Z
0
π
2
0
π
θ 2
π
dθ = = .
3 0
6
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 4 − r 2 .
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 4 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 2
V=
(4 − r 2 )r dr dθ
0
1
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 4 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 2
Z 2π Z 2
2
V=
(4 − r )r dr dθ =
(4r − r 3 ) dr dθ
0
1
0
1
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 4 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 2
Z 2π Z 2
2
V=
(4 − r )r dr dθ =
(4r − r 3 ) dr dθ
0
Z
=
0
2π
1
1 4 2
2
2r − r
dθ
4
1
0
1
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 4 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 2
Z 2π Z 2
2
V=
(4 − r )r dr dθ =
(4r − r 3 ) dr dθ
0
Z
=
0
2π
1
0
1
Z 2π
1 4 2
1
2
2r − r
dθ =
(8 − 4) − (2 − )dθ
4
4
0
1
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 4 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 2
Z 2π Z 2
2
V=
(4 − r )r dr dθ =
(4r − r 3 ) dr dθ
0
Z
=
0
2π
1
0
1
Z 2π
1 4 2
1
2
2r − r
dθ =
(8 − 4) − (2 − )dθ
4
4
0
1
Z 2π
9
=
dθ
4 0
Problem 25(b) - Fall 2006
Find the volume V of the solid under the surface z = 4 − x 2 − y 2
and above the region in the xy -plane between the circles
x 2 + y 2 = 1 and x 2 + y 2 = 4.
Solution:
The domain of integration for the function z = 4 − x 2 − y 2
described in polar coordinates is:
D = {(r , θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
In polar coordinates r 2 = x 2 + y 2 and so z = 4 − r 2 .
Applying Fubini’s Theorem,
Z 2π Z 2
Z 2π Z 2
2
V=
(4 − r )r dr dθ =
(4r − r 3 ) dr dθ
0
Z
=
0
2π
1
0
1
Z 2π
1 4 2
1
2
2r − r
dθ =
(8 − 4) − (2 − )dθ
4
4
0
1
Z 2π
9
9π
=
dθ =
.
4 0
2
Problem 26(a) - Fall 2006
Determine whether the following vector fields are conservative or
not. Find a potential function for those which are indeed
conservative.
(a) F(x, y ) = (x 2 + xy )i + (xy − y 2 )j.
Problem 26(a) - Fall 2006
Determine whether the following vector fields are conservative or
not. Find a potential function for those which are indeed
conservative.
(a) F(x, y ) = (x 2 + xy )i + (xy − y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where
P(x, y ) = x 2 + xy and Q(x, y ) = xy − y 2 .
Problem 26(a) - Fall 2006
Determine whether the following vector fields are conservative or
not. Find a potential function for those which are indeed
conservative.
(a) F(x, y ) = (x 2 + xy )i + (xy − y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where
P(x, y ) = x 2 + xy and Q(x, y ) = xy − y 2 .
Since Py (x, y ) = x 6= y = Qx (x, y ), the vector field F(x, y ) is
not conservative.
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ).
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
∂f
= 3x 2 y + y 2
∂x
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
Z
∂f
2
2
= 3x y + y =⇒ f (x, y ) = 3x 2 y + y 2 dx
∂x
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
Z
∂f
2
2
= 3x y + y =⇒ f (x, y ) = 3x 2 y + y 2 dx = x 3 y + y 2 x + g (y ).
∂x
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
Z
∂f
2
2
= 3x y + y =⇒ f (x, y ) = 3x 2 y + y 2 dx = x 3 y + y 2 x + g (y ).
∂x
Since fy = Q, then
∂f
∂ 3
=
(x y + y 2 x + g (y ))
∂y
∂y
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
Z
∂f
2
2
= 3x y + y =⇒ f (x, y ) = 3x 2 y + y 2 dx = x 3 y + y 2 x + g (y ).
∂x
Since fy = Q, then
∂f
∂ 3
=
(x y + y 2 x + g (y )) = x 3 + 2xy + g 0 (y )
∂y
∂y
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
Z
∂f
2
2
= 3x y + y =⇒ f (x, y ) = 3x 2 y + y 2 dx = x 3 y + y 2 x + g (y ).
∂x
Since fy = Q, then
∂f
∂ 3
=
(x y + y 2 x + g (y )) = x 3 + 2xy + g 0 (y ) = x 3 + 2xy + 3y 2 .
∂y
∂y
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
Z
∂f
2
2
= 3x y + y =⇒ f (x, y ) = 3x 2 y + y 2 dx = x 3 y + y 2 x + g (y ).
∂x
Since fy = Q, then
∂f
∂ 3
=
(x y + y 2 x + g (y )) = x 3 + 2xy + g 0 (y ) = x 3 + 2xy + 3y 2 .
∂y
∂y
Hence, g 0 (y ) = 3y 2
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
Z
∂f
2
2
= 3x y + y =⇒ f (x, y ) = 3x 2 y + y 2 dx = x 3 y + y 2 x + g (y ).
∂x
Since fy = Q, then
∂f
∂ 3
=
(x y + y 2 x + g (y )) = x 3 + 2xy + g 0 (y ) = x 3 + 2xy + 3y 2 .
∂y
∂y
Hence, g 0 (y ) = 3y 2 =⇒ g (y ) = y 3 + C, for some constant C.
Problem 26(b) - Fall 2006
Determine whether the following vector fields are conservative or not.
Find a potential function for those which are indeed conservative.
(b) F(x, y ) = (3x 2 y + y 2 )i + (x 3 + 2xy + 3y 2 )j.
Solution:
Note that F(x, y ) = P(x, y )i + Q(x, y )j, where P(x, y ) = 3x 2 y + y 2
and Q(x, y ) = x 3 + 2xy + 3y 2 .
Since Py (x, y ) = 3x 2 + 2y = Qx (x, y ) and P(x, y ) and Q(x, y ) are
infinitely differentiable on the entire plane, F(x, y ) has a potential
function f (x, y ). Thus,
Z
∂f
2
2
= 3x y + y =⇒ f (x, y ) = 3x 2 y + y 2 dx = x 3 y + y 2 x + g (y ).
∂x
Since fy = Q, then
∂f
∂ 3
=
(x y + y 2 x + g (y )) = x 3 + 2xy + g 0 (y ) = x 3 + 2xy + 3y 2 .
∂y
∂y
Hence, g 0 (y ) = 3y 2 =⇒ g (y ) = y 3 + C, for some constant C.
Taking C = 0, gives:
f (x, y ) = x 3 y + y 2 x + y 3 .
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Solution:
Let C1 be the segment from (0, 0) to (1, 2) and let C2 be the
segment from (1, 2) to (0, 3).
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Solution:
Let C1 be the segment from (0, 0) to (1, 2) and let C2 be the
segment from (1, 2) to (0, 3).
Parameterizations for these segments are:
C1 (t) = ht, 2ti
0≤t≤1
C2 (t) = h1 − t, 2 + ti
0 ≤ t ≤ 1.
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Solution:
Let C1 be the segment from (0, 0) to (1, 2) and let C2 be the
segment from (1, 2) to (0, 3).
Parameterizations for these segments are:
C1 (t) = ht, 2ti
0≤t≤1
C2 (t) = h1 − t, 2 + ti
Now calculate:
Z
Z
(x 2 + y ) dx + (xy + 1) dy =
C
C1 ∪C2
0 ≤ t ≤ 1.
(x 2 + y ) dx + (xy + 1) dy
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Solution:
Let C1 be the segment from (0, 0) to (1, 2) and let C2 be the
segment from (1, 2) to (0, 3).
Parameterizations for these segments are:
C1 (t) = ht, 2ti
0≤t≤1
C2 (t) = h1 − t, 2 + ti
Now calculate:
Z
Z
(x 2 + y ) dx + (xy + 1) dy =
0 ≤ t ≤ 1.
(x 2 + y ) dx + (xy + 1) dy
C1 ∪C2
C
1
Z
(t 2 +2t) dt+(2t 2 +1)2 dt+
=
0
1
Z
[(1−t)2 +2+t](−1)+(1−t)(2+t)+1 dt
0
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Solution:
Let C1 be the segment from (0, 0) to (1, 2) and let C2 be the
segment from (1, 2) to (0, 3).
Parameterizations for these segments are:
C1 (t) = ht, 2ti
0≤t≤1
C2 (t) = h1 − t, 2 + ti
Now calculate:
Z
Z
(x 2 + y ) dx + (xy + 1) dy =
0 ≤ t ≤ 1.
(x 2 + y ) dx + (xy + 1) dy
C1 ∪C2
C
1
Z
(t 2 +2t) dt+(2t 2 +1)2 dt+
=
0
1
Z
[(1−t)2 +2+t](−1)+(1−t)(2+t)+1 dt
0
Z
=
0
1
3t 2 + 2t − 2 dt
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Solution:
Let C1 be the segment from (0, 0) to (1, 2) and let C2 be the
segment from (1, 2) to (0, 3).
Parameterizations for these segments are:
C1 (t) = ht, 2ti
0≤t≤1
C2 (t) = h1 − t, 2 + ti
Now calculate:
Z
Z
(x 2 + y ) dx + (xy + 1) dy =
0 ≤ t ≤ 1.
(x 2 + y ) dx + (xy + 1) dy
C1 ∪C2
C
1
Z
(t 2 +2t) dt+(2t 2 +1)2 dt+
=
0
1
Z
[(1−t)2 +2+t](−1)+(1−t)(2+t)+1 dt
0
Z
=
0
1
1
3t 2 + 2t − 2 dt = t 3 + t 2 + 2t 0
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Solution:
Let C1 be the segment from (0, 0) to (1, 2) and let C2 be the
segment from (1, 2) to (0, 3).
Parameterizations for these segments are:
C1 (t) = ht, 2ti
0≤t≤1
C2 (t) = h1 − t, 2 + ti
Now calculate:
Z
Z
(x 2 + y ) dx + (xy + 1) dy =
0 ≤ t ≤ 1.
(x 2 + y ) dx + (xy + 1) dy
C1 ∪C2
C
1
Z
(t 2 +2t) dt+(2t 2 +1)2 dt+
=
0
1
Z
[(1−t)2 +2+t](−1)+(1−t)(2+t)+1 dt
0
Z
=
0
1
1
3t 2 + 2t − 2 dt = t 3 + t 2 + 2t = 1 + 1 + 2
0
Problem 27 - Fall 2006
R
Evaluate the line integral C (x 2 + y )dx + (xy + 1)dy , where C is the
curve starting at (0, 0), traveling along a line segment to (1, 2) and then
traveling along a second line segment to (0, 3).
Solution:
Let C1 be the segment from (0, 0) to (1, 2) and let C2 be the
segment from (1, 2) to (0, 3).
Parameterizations for these segments are:
C1 (t) = ht, 2ti
0≤t≤1
C2 (t) = h1 − t, 2 + ti
Now calculate:
Z
Z
(x 2 + y ) dx + (xy + 1) dy =
0 ≤ t ≤ 1.
(x 2 + y ) dx + (xy + 1) dy
C1 ∪C2
C
1
Z
(t 2 +2t) dt+(2t 2 +1)2 dt+
=
0
1
Z
[(1−t)2 +2+t](−1)+(1−t)(2+t)+1 dt
0
Z
=
0
1
1
3t 2 + 2t − 2 dt = t 3 + t 2 + 2t = 1 + 1 + 2 = 4.
0
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Z
Z
F · dr =
C
C
(y 3 + sin 2x) dx + (2xy 2 + cos y ) dy .
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Z
Z
F · dr =
C
(y 3 + sin 2x) dx + (2xy 2 + cos y ) dy .
C
Recall and apply Green’s Theorem:
Z
Z Z
P dx + Q dy =
C
D
where D is the disk with boundary C:
∂Q
∂P
−
dA,
∂x
∂y
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Z
Z
F · dr =
C
(y 3 + sin 2x) dx + (2xy 2 + cos y ) dy .
C
Recall and apply Green’s Theorem:
Z
Z Z
P dx + Q dy =
C
D
∂Q
∂P
−
dA,
∂x
∂y
where D is the disk with boundary C:
Z
C
(y 3 +sin 2x) dx+(2xy 2 +cos y ) dy =
Z Z
2y 2 −3y 2 dA
D
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Z
Z
F · dr =
C
(y 3 + sin 2x) dx + (2xy 2 + cos y ) dy .
C
Recall and apply Green’s Theorem:
Z
Z Z
P dx + Q dy =
C
D
∂Q
∂P
−
dA,
∂x
∂y
where D is the disk with boundary C:
Z
C
(y 3 +sin 2x) dx+(2xy 2 +cos y ) dy =
Z Z
2y 2 −3y 2 dA = −
D
Z Z
y 2 dA.
D
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Z
Z
F · dr =
C
(y 3 + sin 2x) dx + (2xy 2 + cos y ) dy .
C
Recall and apply Green’s Theorem:
Z
Z Z
P dx + Q dy =
C
D
∂Q
∂P
−
dA,
∂x
∂y
where D is the disk with boundary C:
Z
C
(y 3 +sin 2x) dx+(2xy 2 +cos y ) dy =
Z Z
2y 2 −3y 2 dA = −
D
Next evaluate the integral using polar coordinates:
Z Z
y 2 dA.
D
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Z
Z
(y 3 + sin 2x) dx + (2xy 2 + cos y ) dy .
F · dr =
C
C
Recall and apply Green’s Theorem:
Z
Z Z
P dx + Q dy =
C
D
∂Q
∂P
−
dA,
∂x
∂y
where D is the disk with boundary C:
Z
(y 3 +sin 2x) dx+(2xy 2 +cos y ) dy =
C
Z Z
2y 2 −3y 2 dA = −
D
Next evaluate the integral using polar coordinates:
Z Z
y 2 dA = −
−
D
2π
Z
0
1
Z
(r sin θ)2 r dr dθ
0
Z Z
y 2 dA.
D
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Z
Z
F · dr =
C
(y 3 + sin 2x) dx + (2xy 2 + cos y ) dy .
C
Recall and apply Green’s Theorem:
Z
Z Z
P dx + Q dy =
C
D
∂Q
∂P
−
dA,
∂x
∂y
where D is the disk with boundary C:
Z
(y 3 +sin 2x) dx+(2xy 2 +cos y ) dy =
C
Z Z
2y 2 −3y 2 dA = −
Z Z
Next evaluate the integral using polar coordinates: Z Z
Z 2π Z 1
Z 2π
y 2 dA = −
−
D
(r sin θ)2 r dr dθ = −
0
0
y 2 dA.
D
D
0
r4
sin2 θ
4
1
dθ
0
Problem 28 - Fall 2006
R
Use Green’s Theorem to evaluate the line integral C F · dr where
F = hy 3 + sin 2x, 2xy 2 + cos y i and C is the unit circle x 2 + y 2 = 1 which
is oriented counterclockwise.
Solution:
R
First rewrite C F · dr in standard form:
Z
Z
F · dr =
C
(y 3 + sin 2x) dx + (2xy 2 + cos y ) dy .
C
Recall and apply Green’s Theorem:
Z
Z Z
P dx + Q dy =
C
D
∂Q
∂P
−
dA,
∂x
∂y
where D is the disk with boundary C:
Z
(y 3 +sin 2x) dx+(2xy 2 +cos y ) dy =
C
Z Z
2y 2 −3y 2 dA = −
Z Z
Next evaluate the integral using polar coordinates: Z Z
Z 2π Z 1
Z 2π
y 2 dA = −
−
D
0
(r sin θ)2 r dr dθ = −
0
Z
1 2π 2
=−
sin θ dθ.
4 0
y 2 dA.
D
D
0
r4
sin2 θ
4
1
dθ
0
Problem 29
RR 2
(a) Express the double integral
R x y − x dA as an iterated
integral and evaluate it, where R is the first quadrant region
enclosed by the curves y = 0, y = x 2 and y = 2 − x.
Problem 29
RR 2
(a) Express the double integral
R x y − x dA as an iterated
integral and evaluate it, where R is the first quadrant region
enclosed by the curves y = 0, y = x 2 and y = 2 − x.
Solution:
First rewrite the integral as an iterated integral.
Problem 29
RR 2
(a) Express the double integral
R x y − x dA as an iterated
integral and evaluate it, where R is the first quadrant region
enclosed by the curves y = 0, y = x 2 and y = 2 − x.
Solution:
First rewrite the integral as an iterated integral.
Z Z
2
Z
(x y − x) dA =
R
0
1
Z
2−y
√
(x 2 y − x) dx dy
y
Problem 29
RR 2
(a) Express the double integral
R x y − x dA as an iterated
integral and evaluate it, where R is the first quadrant region
enclosed by the curves y = 0, y = x 2 and y = 2 − x.
Solution:
First rewrite the integral as an iterated integral.
Z Z
Z
2
(x y − x) dA =
0
R
Z
=
0
1
1
Z
2−y
√
(x 2 y − x) dx dy
y
x2
x 3y 2
−
3
2
2−y
√
dy
y
Problem 29
RR 2
(a) Express the double integral
R x y − x dA as an iterated
integral and evaluate it, where R is the first quadrant region
enclosed by the curves y = 0, y = x 2 and y = 2 − x.
Solution:
First rewrite the integral as an iterated integral.
Z Z
Z
2
(x y − x) dA =
0
R
Z
=
0
Z
=
0
1
1
1
Z
2−y
√
(x 2 y − x) dx dy
y
x2
x 3y 2
−
3
2
2−y
√
dy
y
#
" 7
(2 − y )3 y 2
(2 − y )2
y2
y
−
−
−
dy
3
2
3
2
Problem 29
RR 2
(a) Express the double integral
R x y − x dA as an iterated
integral and evaluate it, where R is the first quadrant region
enclosed by the curves y = 0, y = x 2 and y = 2 − x.
Solution:
First rewrite the integral as an iterated integral.
Z Z
Z
2
(x y − x) dA =
0
R
Z
=
0
Z
=
0
1
1
1
Z
2−y
√
(x 2 y − x) dx dy
y
x2
x 3y 2
−
3
2
2−y
√
dy
y
#
" 7
(2 − y )3 y 2
(2 − y )2
y2
y
−
−
−
dy
3
2
3
2
The remaining straightforward integral is left to you the student to
do.
Problem 29
(b) Find an equivalent iterated integral expression for the double
integral in (a), where the order of integration is reversed from
the order used in part (a). (Do not evaluate this integral.
Problem 29
(b) Find an equivalent iterated integral expression for the double
integral in (a), where the order of integration is reversed from
the order used in part (a). (Do not evaluate this integral. )
Solution:
Z Z
(x 2 y − x) dA
R
Z
1 Z x2
2
Z
2 Z 2−x
(x y − x) dy dx +
=
0
0
1
0
(x 2 y − x) dy dx.
Problem 30
Calculate the line integral
2
Z
F · dr,
C
where F(x, y ) = y xi + xy j, and C is the path starting at (1, 2), moving
along a line segment to (3, 0) and then moving along a second line
segment to (0, 1).
Problem 30
Calculate the line integral
2
Z
F · dr,
C
where F(x, y ) = y xi + xy j, and C is the path starting at (1, 2), moving
along a line segment to (3, 0) and then moving along a second line
segment to (0, 1).
Solution:
Let C1 be the segment from (1, 2) to (3, 0) and let C2 be the
segment from (3, 0) to (0, 1).
Problem 30
Calculate the line integral
2
Z
F · dr,
C
where F(x, y ) = y xi + xy j, and C is the path starting at (1, 2), moving
along a line segment to (3, 0) and then moving along a second line
segment to (0, 1).
Solution:
Let C1 be the segment from (1, 2) to (3, 0) and let C2 be the
segment from (3, 0) to (0, 1).
Parameterizations for these curves are:
C1 (t) = h1 + 2t, 2 − 2ti
0≤t≤1
C2 (t) = h3 − 3t, ti
0≤t≤1
Problem 30
Calculate the line integral
Z
F · dr,
C
2
where F(x, y ) = y xi + xy j, and C is the path starting at (1, 2), moving
along a line segment to (3, 0) and then moving along a second line
segment to (0, 1).
Solution:
Let C1 be the segment from (1, 2) to (3, 0) and let C2 be the
segment from (3, 0) to (0, 1).
Parameterizations for these curves are:
C1 (t) = h1 + 2t, 2 − 2ti
0≤t≤1
C2 (t) = h3 − 3t, ti
0≤t≤1
Next calculate:
Z
Z
y 2 x dx + xy dy =
C
C1
y 2 x dx + xy dy +
Z
C2
y 2 x dx + xy dy =
Problem 30
Calculate the line integral
Z
F · dr,
C
2
where F(x, y ) = y xi + xy j, and C is the path starting at (1, 2), moving
along a line segment to (3, 0) and then moving along a second line
segment to (0, 1).
Solution:
Let C1 be the segment from (1, 2) to (3, 0) and let C2 be the
segment from (3, 0) to (0, 1).
Parameterizations for these curves are:
C1 (t) = h1 + 2t, 2 − 2ti
0≤t≤1
C2 (t) = h3 − 3t, ti
0≤t≤1
Next calculate:
Z
Z
y 2 x dx + xy dy =
1
Z
C
y 2 x dx + xy dy +
C1
(2−2t)2 (1+2t)2 dt+(1+2t)(2−2t)(−2) dt+
0
Z
0
1
Z
y 2 x dx + xy dy =
C2
t 2 (3−3t)(−3) dt+(3−3t)t dt.
Problem 30
Calculate the line integral
Z
F · dr,
C
2
where F(x, y ) = y xi + xy j, and C is the path starting at (1, 2), moving
along a line segment to (3, 0) and then moving along a second line
segment to (0, 1).
Solution:
Let C1 be the segment from (1, 2) to (3, 0) and let C2 be the
segment from (3, 0) to (0, 1).
Parameterizations for these curves are:
C1 (t) = h1 + 2t, 2 − 2ti
0≤t≤1
C2 (t) = h3 − 3t, ti
0≤t≤1
Next calculate:
Z
Z
y 2 x dx + xy dy =
1
Z
C
y 2 x dx + xy dy +
C1
(2−2t)2 (1+2t)2 dt+(1+2t)(2−2t)(−2) dt+
0
Z
1
Z
y 2 x dx + xy dy =
C2
t 2 (3−3t)(−3) dt+(3−3t)t dt.
0
I leave the remaining long but straightforward calculation to you the
student.
Problem 31
Evaluate the integral
Z Z
y
R
p
x 2 + y 2 dA
with R
the region {(x, y ) : 1 ≤ x 2 + y 2 ≤ 2, 0 ≤ y ≤ x.
Problem 31
Evaluate the integral
Z Z
y
p
x 2 + y 2 dA
R
with R
the region {(x, y ) : 1 ≤ x 2 + y 2 ≤ 2, 0 ≤ y ≤ x.}
Solution:
First describe the domain R in polar coordinates:
√
π
R = {1 ≤ r ≤ 2, 0 ≤ θ ≤ }.
4
Problem 31
Evaluate the integral
Z Z
y
p
x 2 + y 2 dA
R
with R
the region {(x, y ) : 1 ≤ x 2 + y 2 ≤ 2, 0 ≤ y ≤ x.}
Solution:
First describe the domain R in polar coordinates:
√
π
R = {1 ≤ r ≤ 2, 0 ≤ θ ≤ }.
4
Rewrite the integral in polar coordinates:
Z π4 Z √2
Z Z p
y x 2 + y 2 dA =
(r sin θ r )r dr dθ.
R
0
1
Problem 31
Evaluate the integral
Z Z
y
p
x 2 + y 2 dA
R
with R
the region {(x, y ) : 1 ≤ x 2 + y 2 ≤ 2, 0 ≤ y ≤ x.}
Solution:
First describe the domain R in polar coordinates:
√
π
R = {1 ≤ r ≤ 2, 0 ≤ θ ≤ }.
4
Rewrite the integral in polar coordinates:
Z π4 Z √2
Z Z p
y x 2 + y 2 dA =
(r sin θ r )r dr dθ.
0
R
Now evaluating the integral:
Z π4 Z √2
sin θ r 3 dr dθ
0
1
1
Problem 31
Evaluate the integral
Z Z
y
p
x 2 + y 2 dA
R
with R
the region {(x, y ) : 1 ≤ x 2 + y 2 ≤ 2, 0 ≤ y ≤ x.}
Solution:
First describe the domain R in polar coordinates:
√
π
R = {1 ≤ r ≤ 2, 0 ≤ θ ≤ }.
4
Rewrite the integral in polar coordinates:
Z π4 Z √2
Z Z p
y x 2 + y 2 dA =
(r sin θ r )r dr dθ.
0
R
1
Now evaluating the integral:
Z π4 Z √2
Z
3
sin θ r dr dθ =
0
1
0
π
4
1
sin θ r 4
4
√2
dθ
1
Problem 31
Evaluate the integral
Z Z
y
p
x 2 + y 2 dA
R
with R
the region {(x, y ) : 1 ≤ x 2 + y 2 ≤ 2, 0 ≤ y ≤ x.}
Solution:
First describe the domain R in polar coordinates:
√
π
R = {1 ≤ r ≤ 2, 0 ≤ θ ≤ }.
4
Rewrite the integral in polar coordinates:
Z π4 Z √2
Z Z p
y x 2 + y 2 dA =
(r sin θ r )r dr dθ.
0
R
1
Now evaluating the integral:
Z π4 Z √2
Z
3
sin θ r dr dθ =
0
=
3
4
1
Z
0
π
4
sin θ dθ
0
π
4
1
sin θ r 4
4
√2
dθ
1
Problem 31
Evaluate the integral
Z Z
y
p
x 2 + y 2 dA
R
with R
the region {(x, y ) : 1 ≤ x 2 + y 2 ≤ 2, 0 ≤ y ≤ x.}
Solution:
First describe the domain R in polar coordinates:
√
π
R = {1 ≤ r ≤ 2, 0 ≤ θ ≤ }.
4
Rewrite the integral in polar coordinates:
Z π4 Z √2
Z Z p
y x 2 + y 2 dA =
(r sin θ r )r dr dθ.
0
R
1
Now evaluating the integral:
Z π4 Z √2
Z
3
sin θ r dr dθ =
0
=
3
4
1
Z
0
π
4
sin θ dθ =
0
π
4
π4
3
(− cos θ)
4
0
1
sin θ r 4
4
√2
dθ
1
Problem 31
Evaluate the integral
Z Z
y
p
x 2 + y 2 dA
R
with R
the region {(x, y ) : 1 ≤ x 2 + y 2 ≤ 2, 0 ≤ y ≤ x.}
Solution:
First describe the domain R in polar coordinates:
√
π
R = {1 ≤ r ≤ 2, 0 ≤ θ ≤ }.
4
Rewrite the integral in polar coordinates:
Z π4 Z √2
Z Z p
y x 2 + y 2 dA =
(r sin θ r )r dr dθ.
R
0
Now evaluating the integral:
Z π4 Z √2
Z
3
sin θ r dr dθ =
1
√2
1
sin θ r 4
dθ
4
0
1
0
1
Z π
π4
3 4
3
3
1
=
sin θ dθ = (− cos θ) =
1− √
.
4 0
4
4
0
2
π
4
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Then: fx (x, y ) =
1
y
+ 2x
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Then: fx (x, y ) = y1 + 2x =⇒ f (x, y ) =
g (y ) is a function of y .
x
y
+ x 2 + g (y ), where
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Then: fx (x, y ) = y1 + 2x =⇒ f (x, y ) =
g (y ) is a function of y .
Then:
fy (x, y ) =
−x
+ g 0 (y )
y2
x
y
+ x 2 + g (y ), where
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Then: fx (x, y ) = y1 + 2x =⇒ f (x, y ) =
g (y ) is a function of y .
Then:
fy (x, y ) =
x
y
+ x 2 + g (y ), where
−x
−x
+ g 0 (y ) = 2 + 1
y2
y
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Then: fx (x, y ) = y1 + 2x =⇒ f (x, y ) =
g (y ) is a function of y .
Then:
fy (x, y ) =
x
y
+ x 2 + g (y ), where
−x
−x
+ g 0 (y ) = 2 + 1
y2
y
=⇒ g 0 (y ) = 1
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Then: fx (x, y ) = y1 + 2x =⇒ f (x, y ) =
g (y ) is a function of y .
Then:
fy (x, y ) =
x
y
+ x 2 + g (y ), where
−x
−x
+ g 0 (y ) = 2 + 1
y2
y
=⇒ g 0 (y ) = 1 =⇒ g (y ) = y + K,
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Then: fx (x, y ) = y1 + 2x =⇒ f (x, y ) =
g (y ) is a function of y .
Then:
fy (x, y ) =
x
y
+ x 2 + g (y ), where
−x
−x
+ g 0 (y ) = 2 + 1
y2
y
=⇒ g 0 (y ) = 1 =⇒ g (y ) = y + K,
where K is a constant.
Problem 32
D
E
(a) Show that the vector field F(x, y ) = y1 + 2x, − yx2 + 1 is
conservative by finding a potential function f (x, y ).
Solution:
Suppose f (x, y ) is the desired potential function.
Then: fx (x, y ) = y1 + 2x =⇒ f (x, y ) =
g (y ) is a function of y .
Then:
fy (x, y ) =
x
y
+ x 2 + g (y ), where
−x
−x
+ g 0 (y ) = 2 + 1
y2
y
=⇒ g 0 (y ) = 1 =⇒ g (y ) = y + K,
where K is a constant.
Taking K = 0, we obtain: f (x, y ) =
x
y
+ x2 + y.
Problem 32
(b) Let C be the path described by the parametric curve
2
r(t) = h1 + 2t,
R 1 + t i for (a) to determine the value of the
line integral C F · dr.
Problem 32
(b) Let C be the path described by the parametric curve
2
r(t) = h1 + 2t,
R 1 + t i for (a) to determine the value of the
line integral C F · dr.
Solution:
By the Fundamental Theorem of Calculus for line integrals,
Z
F · dr = f (x2 , y2 ) − f (x1 , y1 ),
C
where (x1 , y1 ) is the beginning point for r(t) and (x2 , y2 ) is the
ending point for r(t).
Problem 32
(b) Let C be the path described by the parametric curve
2
r(t) = h1 + 2t,
R 1 + t i for (a) to determine the value of the
line integral C F · dr.
Solution:
By the Fundamental Theorem of Calculus for line integrals,
Z
F · dr = f (x2 , y2 ) − f (x1 , y1 ),
C
where (x1 , y1 ) is the beginning point for r(t) and (x2 , y2 ) is the
ending point for r(t). Note that the person who made this problem
forgot to give the beginning and ending times, so this is all we can
do.
Problem 33
(a) Find the equation of the tangent plane at the point
P = (1, 1 − 1) in the level surface
f (x, y , z) = 3x 2 + xyz + z 2 = 1.
Problem 33
(a) Find the equation of the tangent plane at the point
P = (1, 1 − 1) in the level surface
f (x, y , z) = 3x 2 + xyz + z 2 = 1.
Solution:
Recall that the gradient field ∇f (x, y , z) is orthogonal to the
level set surfaces of f (x, y ).
Problem 33
(a) Find the equation of the tangent plane at the point
P = (1, 1 − 1) in the level surface
f (x, y , z) = 3x 2 + xyz + z 2 = 1.
Solution:
Recall that the gradient field ∇f (x, y , z) is orthogonal to the
level set surfaces of f (x, y ). Hence n = ∇f (1, 1, −1) is a
normal vector for the tangent plane.
Problem 33
(a) Find the equation of the tangent plane at the point
P = (1, 1 − 1) in the level surface
f (x, y , z) = 3x 2 + xyz + z 2 = 1.
Solution:
Recall that the gradient field ∇f (x, y , z) is orthogonal to the
level set surfaces of f (x, y ). Hence n = ∇f (1, 1, −1) is a
normal vector for the tangent plane.
Calculating:
∇f = h6x + yz, xz, xy + 2zi
∇f (1, 1, −1) = h6 − 1, −1, 1 − 2i = h5, −1, −1i.
Problem 33
(a) Find the equation of the tangent plane at the point
P = (1, 1 − 1) in the level surface
f (x, y , z) = 3x 2 + xyz + z 2 = 1.
Solution:
Recall that the gradient field ∇f (x, y , z) is orthogonal to the
level set surfaces of f (x, y ). Hence n = ∇f (1, 1, −1) is a
normal vector for the tangent plane.
Calculating:
∇f = h6x + yz, xz, xy + 2zi
∇f (1, 1, −1) = h6 − 1, −1, 1 − 2i = h5, −1, −1i.
The equation of the tangent plane is:
h5, −1, −1i·hx −1, y −1, z +1i = 5(x −1)−(y −1)−(z +1) = 0.
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
The unit vector u in the direction of v is:
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
The unit vector u in the direction of v is:
v
h3, −2, −4i
√
u=
.
=
|v|
29
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
The unit vector u in the direction of v is:
v
h3, −2, −4i
√
u=
.
=
|v|
29
By part (a), ∇f at (1, 1, −1) is:
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
The unit vector u in the direction of v is:
v
h3, −2, −4i
√
u=
.
=
|v|
29
By part (a), ∇f at (1, 1, −1) is:
∇f (1, 1, −1) = h5, −1, −1i.
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
The unit vector u in the direction of v is:
v
h3, −2, −4i
√
u=
.
=
|v|
29
By part (a), ∇f at (1, 1, −1) is:
∇f (1, 1, −1) = h5, −1, −1i.
The directional derivative is:
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
The unit vector u in the direction of v is:
v
h3, −2, −4i
√
u=
.
=
|v|
29
By part (a), ∇f at (1, 1, −1) is:
∇f (1, 1, −1) = h5, −1, −1i.
The directional derivative is:
1
Du f (1, 1, −1) = ∇f (1, 1, −1) · √ h3, −2, −4i
29
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
The unit vector u in the direction of v is:
v
h3, −2, −4i
√
u=
.
=
|v|
29
By part (a), ∇f at (1, 1, −1) is:
∇f (1, 1, −1) = h5, −1, −1i.
The directional derivative is:
1
Du f (1, 1, −1) = ∇f (1, 1, −1) · √ h3, −2, −4i
29
1
= √ (h5, −1, −1i · h3, −2, −4i
29
Problem 33
(b) Find the directional derivative of the function f (x, y , z) at
P = (1, 1, −1) in the direction of the tangent vector to the space
curve r(t) = h2t 2 − t, t −2 , t 2 − 2t 3 i at t = 1.
Solution:
First find the tangent vector v to r(t) at t = 1:
r0 (t) = h4t − 1, −2t −3 , 2t − 6t 2 i
v = r0 (1) = h3, −2, −4i.
The unit vector u in the direction of v is:
v
h3, −2, −4i
√
u=
.
=
|v|
29
By part (a), ∇f at (1, 1, −1) is:
∇f (1, 1, −1) = h5, −1, −1i.
The directional derivative is:
1
Du f (1, 1, −1) = ∇f (1, 1, −1) · √ h3, −2, −4i
29
1
21
= √ (h5, −1, −1i · h3, −2, −4i = √ .
29
29
Problem 34
Find the absolute maxima and minima of the function
f (x, y ) = x 2 − 2xy + 2y 2 − 2y .
in the region bounded by the lines x = 0, y = 0 and x + y = 7.
Problem 34
Find the absolute maxima and minima of the function
f (x, y ) = x 2 − 2xy + 2y 2 − 2y .
in the region bounded by the lines x = 0, y = 0 and x + y = 7.
Solution:
This problem is left to you the student to do.
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in
the direction of the point Q = (3, 2).
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in
the direction of the point Q = (3, 2).
Solution:
The unit vector u in the direction h3, 2i is:
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in
the direction of the point Q = (3, 2).
Solution:
The unit vector u in the direction h3, 2i is:
−→
u=
1
h2, 2i
= √ h1, 1i.
=√
4+4
2
|PQ|
PQ
−→
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in
the direction of the point Q = (3, 2).
Solution:
The unit vector u in the direction h3, 2i is:
−→
1
h2, 2i
= √ h1, 1i.
=√
4+4
2
|PQ|
Calculate ∇f (1, 0):
u=
PQ
−→
∇f = he xy + xye xy , x 2 e xy i
∇f (1, 0) = h1, 1i.
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in
the direction of the point Q = (3, 2).
Solution:
The unit vector u in the direction h3, 2i is:
−→
1
h2, 2i
= √ h1, 1i.
=√
4+4
2
|PQ|
Calculate ∇f (1, 0):
u=
PQ
−→
∇f = he xy + xye xy , x 2 e xy i
∇f (1, 0) = h1, 1i.
The directional derivative is:
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in
the direction of the point Q = (3, 2).
Solution:
The unit vector u in the direction h3, 2i is:
−→
1
h2, 2i
= √ h1, 1i.
=√
4+4
2
|PQ|
Calculate ∇f (1, 0):
u=
PQ
−→
∇f = he xy + xye xy , x 2 e xy i
∇f (1, 0) = h1, 1i.
The directional derivative is:
Du f (0, 1) = ∇f (1, 0) · u
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in
the direction of the point Q = (3, 2).
Solution:
The unit vector u in the direction h3, 2i is:
−→
1
h2, 2i
= √ h1, 1i.
=√
4+4
2
|PQ|
Calculate ∇f (1, 0):
u=
PQ
−→
∇f = he xy + xye xy , x 2 e xy i
∇f (1, 0) = h1, 1i.
The directional derivative is:
1
Du f (0, 1) = ∇f (1, 0) · u = h1, 1i · √ h1, 1i
2
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(a) Find the rate of change of the function f at the point P in
the direction of the point Q = (3, 2).
Solution:
The unit vector u in the direction h3, 2i is:
−→
1
h2, 2i
= √ h1, 1i.
=√
4+4
2
|PQ|
Calculate ∇f (1, 0):
u=
PQ
−→
∇f = he xy + xye xy , x 2 e xy i
∇f (1, 0) = h1, 1i.
The directional derivative is:
1
2
Du f (0, 1) = ∇f (1, 0) · u = h1, 1i · √ h1, 1i = √ .
2
2
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi = h1, 1i · ha, bi
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi = h1, 1i · ha, bi = a + b.
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi = h1, 1i · ha, bi = a + b.
Setting Dv f (1, 0) = 0 means a + b = 0
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi = h1, 1i · ha, bi = a + b.
Setting Dv f (1, 0) = 0 means a + b = 0 =⇒ a = −b.
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi = h1, 1i · ha, bi = a + b.
Setting Dv f (1, 0) = 0 means a + b = 0 =⇒ a = −b.
Then a2 + b 2 = (−b)2 + b 2 = 2b 2 = 1
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi = h1, 1i · ha, bi = a + b.
Setting Dv f (1, 0) = 0 means a + b = 0 =⇒ a = −b.
Then a2 + b 2 = (−b)2 + b 2 = 2b 2 = 1 =⇒ b = ± √12 .
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi = h1, 1i · ha, bi = a + b.
Setting Dv f (1, 0) = 0 means a + b = 0 =⇒ a = −b.
Then a2 + b 2 = (−b)2 + b 2 = 2b 2 = 1 =⇒ b = ± √12 .
Hence, the two possibilities for v are:
Problem 35
Consider the function f (x, y ) = xe xy . Let P be the point (1, 0).
(b) Give a direction in terms of a unit vector (there are two
possibilities) for which the rate of change of f at P in that
direction is zero.
Solution:
Let v = ha, bi be a vector with a2 + b 2 = 1 and suppose
Dv f (1, 0) = 0.
Calculating, we get:
Dv f (1, 0) = ∇f (0, 1) · ha, bi = h1, 1i · ha, bi = a + b.
Setting Dv f (1, 0) = 0 means a + b = 0 =⇒ a = −b.
Then a2 + b 2 = (−b)2 + b 2 = 2b 2 = 1 =⇒ b = ± √12 .
Hence, the two possibilities for v are: h √12 , − √12 i and
h− √12 , √12 i.
Problem 36
(a) Find the work done by the vector field F(x, y ) = hx − y , xi
over the circle r(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
Problem 36
(a) Find the work done by the vector field F(x, y ) = hx − y , xi
over the circle r(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
Solution:
The work W doneZ is:
W=
F · dr
C
Problem 36
(a) Find the work done by the vector field F(x, y ) = hx − y , xi
over the circle r(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
Solution:
The work W doneZ is:
Z
W=
F · dr = (x − y ) dx + x dy .
C
C
Problem 36
(a) Find the work done by the vector field F(x, y ) = hx − y , xi
over the circle r(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
Solution:
The work W doneZ is:
Z
W=
F · dr = (x − y ) dx + x dy .
C
C
Note r(t) = hcos t, sin ti parameterizes C and
r0 (t) = h− sin t, cos ti.
Problem 36
(a) Find the work done by the vector field F(x, y ) = hx − y , xi
over the circle r(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
Solution:
The work W doneZ is:
Z
W=
F · dr = (x − y ) dx + x dy .
C
C
Note r(t) = hcos t, sin ti parameterizes C and
r0 (t) = h− sin t, cos ti.
So, the work done is:
Z 2π
Z
W=
(cos t − sin t)(− sin t) dt +
0
0
2π
cos t cos t dt
Problem 36
(a) Find the work done by the vector field F(x, y ) = hx − y , xi
over the circle r(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
Solution:
The work W doneZ is:
Z
W=
F · dr = (x − y ) dx + x dy .
C
C
Note r(t) = hcos t, sin ti parameterizes C and
r0 (t) = h− sin t, cos ti.
So, the work done is:
Z 2π
Z 2π
W=
(cos t − sin t)(− sin t) dt +
cos t cos t dt
0
0
Z 2π
Z 2π
=−
sin(t) cos(t) dt +
sin2 (t) + cos2 (t) dt
0
0
Problem 36
(a) Find the work done by the vector field F(x, y ) = hx − y , xi
over the circle r(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
Solution:
The work W doneZ is:
Z
W=
F · dr = (x − y ) dx + x dy .
C
C
Note r(t) = hcos t, sin ti parameterizes C and
r0 (t) = h− sin t, cos ti.
So, the work done is:
Z 2π
Z 2π
W=
(cos t − sin t)(− sin t) dt +
cos t cos t dt
0
0
Z 2π
Z 2π
=−
sin(t) cos(t) dt +
sin2 (t) + cos2 (t) dt
0
0
2π
2
sin (t) =−
+ 2π
2 0
Problem 36
(a) Find the work done by the vector field F(x, y ) = hx − y , xi
over the circle r(t) = hcos t, sin ti, 0 ≤ t ≤ 2π.
Solution:
The work W doneZ is:
Z
W=
F · dr = (x − y ) dx + x dy .
C
C
Note r(t) = hcos t, sin ti parameterizes C and
r0 (t) = h− sin t, cos ti.
So, the work done is:
Z 2π
Z 2π
W=
(cos t − sin t)(− sin t) dt +
cos t cos t dt
0
0
Z 2π
Z 2π
=−
sin(t) cos(t) dt +
sin2 (t) + cos2 (t) dt
0
0
2π
2
sin (t) =−
+ 2π = 2π.
2 0
Problem 36
(b) Use
Theorem to calculate the line integral
R Green’s
2
C (−y ) dx + xy dy , over the positively (counterclockwise)
oriented closed curve C defined by x = 1, y = 1 and the
coordinate axes.
Problem 36
(b) Use
Theorem to calculate the line integral
R Green’s
2
C (−y ) dx + xy dy , over the positively (counterclockwise)
oriented closed curve C defined by x = 1, y = 1 and the
coordinate axes.
Solution:
Recall that Green’s Theorem is:
Z
Z Z ∂Q
∂P
P dx + Q dy =
−
dA,
∂x
∂y
C
D
where D is the square bounded by C.
Problem 36
(b) Use
Theorem to calculate the line integral
R Green’s
2
C (−y ) dx + xy dy , over the positively (counterclockwise)
oriented closed curve C defined by x = 1, y = 1 and the
coordinate axes.
Solution:
Recall that Green’s Theorem is:
Z
Z Z ∂Q
∂P
P dx + Q dy =
−
dA,
∂x
∂y
C
D
where D is the square bounded by C.
Z
Applying Green’s Theorem, we obtain:
Z 1Z 1
(−y 2 ) dx +(xy ) dy =
(y +2y ) dy dx
C
0
0
Problem 36
(b) Use
Theorem to calculate the line integral
R Green’s
2
C (−y ) dx + xy dy , over the positively (counterclockwise)
oriented closed curve C defined by x = 1, y = 1 and the
coordinate axes.
Solution:
Recall that Green’s Theorem is:
Z
Z Z ∂Q
∂P
P dx + Q dy =
−
dA,
∂x
∂y
C
D
where D is the square bounded by C.
Z
Applying Green’s Theorem, we obtain:
Z 1Z 1
Z
(−y 2 ) dx +(xy ) dy =
(y +2y ) dy dx = 3
C
0
0
0
1Z 1
y dy dx
0
Problem 36
(b) Use
Theorem to calculate the line integral
R Green’s
2
C (−y ) dx + xy dy , over the positively (counterclockwise)
oriented closed curve C defined by x = 1, y = 1 and the
coordinate axes.
Solution:
Recall that Green’s Theorem is:
Z
Z Z ∂Q
∂P
P dx + Q dy =
−
dA,
∂x
∂y
C
D
where D is the square bounded by C.
Z
Applying Green’s Theorem, we obtain:
Z 1Z 1
Z
(−y 2 ) dx +(xy ) dy =
(y +2y ) dy dx = 3
C
0
Z
=3
0
0
1 2 1
y
2
0
0
dx
1Z 1
y dy dx
0
Problem 36
(b) Use
Theorem to calculate the line integral
R Green’s
2
C (−y ) dx + xy dy , over the positively (counterclockwise)
oriented closed curve C defined by x = 1, y = 1 and the
coordinate axes.
Solution:
Recall that Green’s Theorem is:
Z
Z Z ∂Q
∂P
P dx + Q dy =
−
dA,
∂x
∂y
C
D
where D is the square bounded by C.
Z
Applying Green’s Theorem, we obtain:
Z 1Z 1
Z
(−y 2 ) dx +(xy ) dy =
(y +2y ) dy dx = 3
C
0
Z
=3
0
0
1 2 1
y
2
0
0
Z
dx = 3
0
1
1
dx
2
1Z 1
y dy dx
0
Problem 36
(b) Use
Theorem to calculate the line integral
R Green’s
2
C (−y ) dx + xy dy , over the positively (counterclockwise)
oriented closed curve C defined by x = 1, y = 1 and the
coordinate axes.
Solution:
Recall that Green’s Theorem is:
Z
Z Z ∂Q
∂P
P dx + Q dy =
−
dA,
∂x
∂y
C
D
where D is the square bounded by C.
Z
Applying Green’s Theorem, we obtain:
Z 1Z 1
Z
(−y 2 ) dx +(xy ) dy =
(y +2y ) dy dx = 3
C
0
Z
=3
0
0
1 2 1
y
2
0
1Z 1
0
Z
dx = 3
0
1
1
3
dx = .
2
2
y dy dx
0
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ).
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
fx (x, y ) = x 2 y
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
Z
x 3y
2
fx (x, y ) = x y =⇒ f (x, y ) = x 2 y dx =
+ g (y ),
3
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
Z
x 3y
2
fx (x, y ) = x y =⇒ f (x, y ) = x 2 y dx =
+ g (y ),
3
where g (y ) is a function of y .
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
Z
x 3y
2
fx (x, y ) = x y =⇒ f (x, y ) = x 2 y dx =
+ g (y ),
3
where g (y ) is a function of y .
Then:
∂ x 3y
fy (x, y ) =
(
+ g (y ))
∂y 3
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
Z
x 3y
2
fx (x, y ) = x y =⇒ f (x, y ) = x 2 y dx =
+ g (y ),
3
where g (y ) is a function of y .
Then:
∂ x 3y
1
fy (x, y ) =
(
+ g (y )) = x 3 + g 0 (y )
∂y 3
3
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
Z
x 3y
2
fx (x, y ) = x y =⇒ f (x, y ) = x 2 y dx =
+ g (y ),
3
where g (y ) is a function of y .
Then:
∂ x 3y
1
1
fy (x, y ) =
(
+ g (y )) = x 3 + g 0 (y ) = x 3
∂y 3
3
3
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
Z
x 3y
2
fx (x, y ) = x y =⇒ f (x, y ) = x 2 y dx =
+ g (y ),
3
where g (y ) is a function of y .
Then:
∂ x 3y
1
1
fy (x, y ) =
(
+ g (y )) = x 3 + g 0 (y ) = x 3
∂y 3
3
3
=⇒ g 0 (y ) = 0
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
Z
x 3y
2
fx (x, y ) = x y =⇒ f (x, y ) = x 2 y dx =
+ g (y ),
3
where g (y ) is a function of y .
Then:
∂ x 3y
1
1
fy (x, y ) =
(
+ g (y )) = x 3 + g 0 (y ) = x 3
∂y 3
3
3
=⇒ g 0 (y ) = 0 =⇒ g (y ) = C,
for some constant C.
Problem 37
(a) Show that the vector field F(x, y ) = hx 2 y , 13 x 3 i is
conservative and find a function f such that F = ∇f .
Solution:
Since
∂ 2
∂ 1 3
(x y ) = x 2 =
( x ),
∂y
∂x 3
the vector field is conservative.
Suppose f (x, y ) is a potential function for F(x, y ). Then:
Z
x 3y
2
fx (x, y ) = x y =⇒ f (x, y ) = x 2 y dx =
+ g (y ),
3
where g (y ) is a function of y .
Then:
∂ x 3y
1
1
fy (x, y ) =
(
+ g (y )) = x 3 + g 0 (y ) = x 3
∂y 3
3
3
=⇒ g 0 (y ) = 0 =⇒ g (y ) = C,
for some constant C.
3
Setting C = 0, we get: f (x, y ) = x y .
3
Problem 37
(b) Using
the result in part (a), calculate the line integral
R
4
C F · dr, along the curve C which is the arc of y = x from
(0, 0) to (2, 16).
Problem 37
(b) Using
the result in part (a), calculate the line integral
R
4
C F · dr, along the curve C which is the arc of y = x from
(0, 0) to (2, 16).
Solution:
By the Fundamental Theorem of Calculus,
Z
F · dr = f (2, 16) − f (0, 0)
C
Problem 37
(b) Using
the result in part (a), calculate the line integral
R
4
C F · dr, along the curve C which is the arc of y = x from
(0, 0) to (2, 16).
Solution:
By the Fundamental Theorem of Calculus,
Z
8 · 16
F · dr = f (2, 16) − f (0, 0) =
−0
3
C
Problem 37
(b) Using
the result in part (a), calculate the line integral
R
4
C F · dr, along the curve C which is the arc of y = x from
(0, 0) to (2, 16).
Solution:
By the Fundamental Theorem of Calculus,
Z
8 · 16
F · dr = f (2, 16) − f (0, 0) =
− 0 = 128.
3
C
Problem 38
Consider the surface x 2 + y 2 − 14 z 2 = 0 and the point
√
P(1, 2, −2 5) which lies on the surface.
(a) Find the equation of the tangent plane to the surface at the
point P.
(b) Find the equation of the normal line to the surface at the
point P.
Problem 38
Consider the surface x 2 + y 2 − 14 z 2 = 0 and the point
√
P(1, 2, −2 5) which lies on the surface.
(a) Find the equation of the tangent plane to the surface at the
point P.
(b) Find the equation of the normal line to the surface at the
point P.
Solution:
Let F(x, y , z) = x 2 + y 2 − 41 z 2 and note that ∇F is normal to
the level set surface x 2 + y 2 − 14 z 2 = 0.
Problem 38
Consider the surface x 2 + y 2 − 14 z 2 = 0 and the point
√
P(1, 2, −2 5) which lies on the surface.
(a) Find the equation of the tangent plane to the surface at the
point P.
(b) Find the equation of the normal line to the surface at the
point P.
Solution:
Let F(x, y , z) = x 2 + y 2 − 41 z 2 and note that ∇F is normal to
the level set surface x 2 + y 2 − 14 z 2 = 0.
√
Calculating ∇F at (1, 2, −2 5), we obtain:
Problem 38
Consider the surface x 2 + y 2 − 14 z 2 = 0 and the point
√
P(1, 2, −2 5) which lies on the surface.
(a) Find the equation of the tangent plane to the surface at the
point P.
(b) Find the equation of the normal line to the surface at the
point P.
Solution:
Let F(x, y , z) = x 2 + y 2 − 41 z 2 and note that ∇F is normal to
the level set surface x 2 + y 2 − 14 z 2 = 0.
√
Calculating ∇F at (1, 2, −2 5), we obtain:
√
√
∇F = h2x, 2y , − 12 zi
∇F(1, 2, −2 5) = h2, 4, 5i.
Problem 38
Consider the surface x 2 + y 2 − 14 z 2 = 0 and the point
√
P(1, 2, −2 5) which lies on the surface.
(a) Find the equation of the tangent plane to the surface at the
point P.
(b) Find the equation of the normal line to the surface at the
point P.
Solution:
Let F(x, y , z) = x 2 + y 2 − 41 z 2 and note that ∇F is normal to
the level set surface x 2 + y 2 − 14 z 2 = 0.
√
Calculating ∇F at (1, 2, −2 5), we obtain:
√
√
∇F = h2x, 2y , − 12 zi
∇F(1, 2, −2 5) = h2, 4, 5i.
The equation of the tangent plane is:
√
√
√
√
h2, 4, 5i·hx −1, y −2, z +2 5i = 2(x −1)+4(y −2)+ 5(z +2 5) = 0.
Problem 38
Consider the surface x 2 + y 2 − 14 z 2 = 0 and the point
√
P(1, 2, −2 5) which lies on the surface.
(a) Find the equation of the tangent plane to the surface at the
point P.
(b) Find the equation of the normal line to the surface at the
point P.
Solution:
Let F(x, y , z) = x 2 + y 2 − 41 z 2 and note that ∇F is normal to
the level set surface x 2 + y 2 − 14 z 2 = 0.
√
Calculating ∇F at (1, 2, −2 5), we obtain:
√
√
∇F = h2x, 2y , − 12 zi
∇F(1, 2, −2 5) = h2, 4, 5i.
The equation of the tangent plane is:
√
√
√
√
h2, 4, 5i·hx −1, y −2, z +2 5i = 2(x −1)+4(y −2)+ 5(z +2 5) = 0.
The vector equation of the normal line is:
√
√
L(t) = h1, 2, −2 5i+th2, 4, 5i
Problem 38
Consider the surface x 2 + y 2 − 14 z 2 = 0 and the point
√
P(1, 2, −2 5) which lies on the surface.
(a) Find the equation of the tangent plane to the surface at the
point P.
(b) Find the equation of the normal line to the surface at the
point P.
Solution:
Let F(x, y , z) = x 2 + y 2 − 41 z 2 and note that ∇F is normal to
the level set surface x 2 + y 2 − 14 z 2 = 0.
√
Calculating ∇F at (1, 2, −2 5), we obtain:
√
√
∇F = h2x, 2y , − 12 zi
∇F(1, 2, −2 5) = h2, 4, 5i.
The equation of the tangent plane is:
√
√
√
√
h2, 4, 5i·hx −1, y −2, z +2 5i = 2(x −1)+4(y −2)+ 5(z +2 5) = 0.
The vector equation of the normal line is:
√
√
√ √
L(t) = h1, 2, −2 5i+th2, 4, 5i = h1+2t, 2+4t, −2 5+ 5ti.
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
∇T = h2x − 1, 4y i
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
∇T = h2x − 1, 4y i = h0, 0i
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
∇T = h2x − 1, 4y i = h0, 0i =⇒ x =
1
and y = 0.
2
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
=⇒ 4y = λ2y
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
=⇒ 4y = λ2y =⇒ λ = 2 or
y = 0.
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
=⇒ 4y = λ2y =⇒ λ = 2 or
y = 0 =⇒ x = ±1.
y = 0.
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
=⇒ 4y = λ2y =⇒ λ = 2 or
y = 0 =⇒ x = ±1.
λ=2
y = 0.
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
=⇒ 4y = λ2y =⇒ λ = 2 or
y = 0 =⇒ x = ±1.
λ = 2 =⇒ 2x − 1 = 2(2x) = 4x
y = 0.
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
=⇒ 4y = λ2y =⇒ λ = 2 or
y = 0.
y = 0 =⇒ x = ±1.
√
λ = 2 =⇒ 2x − 1 = 2(2x) = 4x =⇒ x = − 12 and y = ± 23 .
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
=⇒ 4y = λ2y =⇒ λ = 2 or
y = 0.
y = 0 =⇒ x = ±1.
√
λ = 2 =⇒ 2x − 1 = 2(2x) = 4x =⇒ x = − 12 and y = ± 23 .
Checking values:
√
f ( 21 , 0) = − 41 , f (1, 0) = 0, f (−1, 0) = 2, f (− 12 , ± 23 ) = 94 .
Problem 39
A flat circular plate has the shape of the region x 2 + y 2 ≤ 1. The plate
(including the boundary x 2 + y 2 = 1) is heated so that the temperature
at any point (x, y ) on the plate is given by T(x, y ) = x 2 + 2y 2 − x. Find
the temperatures at the hottest and the coldest points on the plate,
including the boundary x 2 + y 2 = 1.
Solution:
First find the critical points:
1
∇T = h2x − 1, 4y i = h0, 0i =⇒ x = and y = 0.
2
Apply Lagrange multipliers with the constraint function
g (x, y ) = x 2 + y 2 = 1:
∇T = h2x − 1, 4y i = λ∇g = λh2x, 2y i,
=⇒ 4y = λ2y =⇒ λ = 2 or
y = 0.
y = 0 =⇒ x = ±1.
√
λ = 2 =⇒ 2x − 1 = 2(2x) = 4x =⇒ x = − 12 and y = ± 23 .
Checking values:
√
f ( 21 , 0) = − 41 , f (1, 0) = 0, f (−1, 0) = 2, f (− 12 , ± 23 ) = 94 .
The maximum value is
9
4
and the minimum value is − 14 .
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3, 0i:
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
v(t) =
a(t) dt
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i,
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i, we get
h−3 sin(0) + x0 , 3 cos(0) + y0 , 2 · 0 + z0 i
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i, we get
h−3 sin(0) + x0 , 3 cos(0) + y0 , 2 · 0 + z0 i = hx0 , 3 + y0 , z0 i
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i, we get
h−3 sin(0) + x0 , 3 cos(0) + y0 , 2 · 0 + z0 i = hx0 , 3 + y0 , z0 i = h0, 3, 0i,
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i, we get
h−3 sin(0) + x0 , 3 cos(0) + y0 , 2 · 0 + z0 i = hx0 , 3 + y0 , z0 i = h0, 3, 0i,
=⇒ x0 = y0 = z0 = 0.
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i, we get
h−3 sin(0) + x0 , 3 cos(0) + y0 , 2 · 0 + z0 i = hx0 , 3 + y0 , z0 i = h0, 3, 0i,
=⇒ x0 = y0 = z0 = 0.
Hence,
v(t) = h−3 sin t, 3 cos t, 2ti.
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i, we get
h−3 sin(0) + x0 , 3 cos(0) + y0 , 2 · 0 + z0 i = hx0 , 3 + y0 , z0 i = h0, 3, 0i,
=⇒ x0 = y0 = z0 = 0.
Hence,
v(t) = h−3 sin t, 3 cos t, 2ti.
Take dot products and solve for t:
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i, we get
h−3 sin(0) + x0 , 3 cos(0) + y0 , 2 · 0 + z0 i = hx0 , 3 + y0 , z0 i = h0, 3, 0i,
=⇒ x0 = y0 = z0 = 0.
Hence,
v(t) = h−3 sin t, 3 cos t, 2ti.
Take dot products and solve for t:
a(t) · v(t) = 9 cos t sin t − 9 sin t cos t + 4t = 4t = 0
Problem 40
The acceleration of a particle at any time t is given by
a(t) = h−3 cos t, −3 sin t, 2i,
while its initial velocity is v(0) = h0, 3, 0i. At what times, if any are the
velocity and the acceleration of the particle orthogonal?
Solution:
First find the velocity v(t) by integrating a(t) and using the initial
value v(0) = h0, 3,Z0i:
Z
v(t) = a(t) dt = h−3 cos t, −3 sin t, 2i dt
= h−3 sin t + x0 , 3 cos t + y0 , 2t + z0 i.
Since v(0) = h0, 3, 0i, we get
h−3 sin(0) + x0 , 3 cos(0) + y0 , 2 · 0 + z0 i = hx0 , 3 + y0 , z0 i = h0, 3, 0i,
=⇒ x0 = y0 = z0 = 0.
Hence,
v(t) = h−3 sin t, 3 cos t, 2ti.
Take dot products and solve for t:
a(t) · v(t) = 9 cos t sin t − 9 sin t cos t + 4t = 4t = 0
=⇒ t = 0.