Review for Exam 2.
I
Sections 13.1, 13.3. 14.1-14.7.
I
50 minutes.
I
5 problems, similar to homework problems.
I
No calculators, no notes, no books, no phones.
I
No green book needed.
Section 14.7
Example
(a) Find all the critical points of f (x, y ) = 12xy − 2x 3 − 3y 2 .
(b) For each critical point of f , determine whether f has a local
maximum, local minimum, or saddle point at that point.
Solution:
(a) ∇f (x, y ) = h12y − 6x 2 , 12x − 6y i = h0, 0i, then,
x 2 = 2y ,
y = 2x,
⇒
x(x − 4) = 0.
There are two solutions, x = 0 ⇒ y = 0, and x = 4 ⇒ y = 8.
That is, there are two critical points, (0, 0) and (4, 8).
Section 14.7
Example
(a) Find all the critical points of f (x, y ) = 12xy − 2x 3 − 3y 2 .
(b) For each critical point of f , determine whether f has a local
maximum, local minimum, or saddle point at that point.
Solution:
(b) Recalling ∇f (x, y ) = h12y − 6x 2 , 12x − 6y i, we compute
fxx = −12x,
fyy = −6,
2
fxy = 12.
D(x, y ) = fxx fyy − (fxy ) = 144
x
−1 ,
2
Since D(0, 0) = −144 < 0, the point (0, 0) is a saddle point of f .
Since D(4, 8) = 144(2 − 1) > 0, and fxx (4, 8) = (−12)4 < 0, the
point (4, 8) is a local maximum of f .
C
Section 14.7
Example
Find the absolute maximum and absolute minimum of
1
f (x, y ) = 2 + xy − 2x − y 2 in the closed triangular region with
4
vertices given by (0, 0), (1, 0), and (0, 2).
Solution:
We start finding the critical points inside the triangular region.
1 E
∇f (x, y ) = y − 2, x − y = h0, 0i,
2
D
⇒
y = 2,
y = 2x.
The solution is (1, 2). This point is outside in the triangular region
given by the problem, so there is no critical point inside the region.
Section 14.7
Example
Find the absolute maximum and absolute minimum of
1
f (x, y ) = 2 + xy − 2x − y 2 in the closed triangular region with
4
vertices given by (0, 0), (1, 0), and (0, 2).
Solution:
We now find the candidates for absolute maximum and minimum
on the borders of the triangular region. We first record the
boundary vertices:
(0, 0)
⇒
f (0, 0) = 2,
(1, 0)
⇒
f (1, 0) = 0,
(0, 2)
⇒
f (0, 2) = 1.
Section 14.7
Example
Find the absolute maximum and absolute minimum of
1
f (x, y ) = 2 + xy − 2x − y 2 in the closed triangular region with
4
vertices given by (0, 0), (1, 0), and (0, 2).
Solution:
I
The horizontal side of the triangle, y = 0, x ∈ (0, 1). Since
g (x) = f (x, 0) = 2 − 2x,
⇒
g 0 (x) = −2 6= 0.
there are no candidates in this part of the boundary.
I
The vertical side of the triangle is x = 0, y ∈ (0, 2). Then,
1
g (y ) = f (0, y ) = 2 − y 2 ,
4
⇒
1
g 0 (y ) = − y = 0,
2
so y = 0 and we recover the point (0, 0).
Section 14.7
Example
Find the absolute maximum and absolute minimum of
1
f (x, y ) = 2 + xy − 2x − y 2 in the closed triangular region with
4
vertices given by (0, 0), (1, 0), and (0, 2).
Solution:
I
The hypotenuse of the triangle y = 2 − 2x, x ∈ (0, 1). Then,
1
g (x) = f (x, 2 − 2x) = 2 + x(2 − 2x) − 2x − (2 − 2x)2 ,
4
= 2 + 2x − 2x 2 − 2x − (x 2 − 2x + 1),
= 1 + 2x − 3x 2 .
Then, g 0 (x) = 2 − 6x = 0 implies x = 13 , hence y = 43 . The
candidate is 13 , 43 .
Section 14.7
Example
Find the absolute maximum and absolute minimum of
1
f (x, y ) = 2 + xy − 2x − y 2 in the closed triangular region with
4
vertices given by (0, 0), (1, 0), and (0, 2).
Solution:
I
Recall that we have obtained a candidate point
evaluate f at this point,
1 4
3, 3
. We
1 4
4 2 1 16
4
f
,
=2+ − −
= .
3 3
9 3 4 9
3
Recalling that f (0, 0) = 2, f (1, 0) = 0, and f (0, 2) = 1, the
absolute maximum is at (0, 0), and the minimum is at (1, 0).
C
Section 14.6
Example
(a) Find the linear approximation L(x, y ) of the function
f (x, y ) = sin(2x + 3y ) + 1 at the point (−3, 2).
(b) Use the approximation above to estimate the value of
f (−2.8, 2.3).
Solution:
(a) L(x, y ) = fx (−3, 2) (x + 3) + fy (−3, 2) (y − 2) + f (−3, 2).
Since fx (x, y ) = 2 cos(2x + 3y ) and fy (x, y ) = 3 cos(2x + 3y ),
fx (−3, 2) = 2 cos(−6 + 6) = 2,
fy (−3, 2) = 3 cos(−6 + 6) = 3,
f (−3, 2) = sin(−6 + 6) + 1 = 1.
the linear approximation is L(x, y ) = 2(x + 3) + 3(y − 2) + 1.
Section 14.6
Example
(a) Find the linear approximation L(x, y ) of the function
f (x, y ) = sin(2x + 3y ) + 1 at the point (−3, 2).
(b) Use the approximation above to estimate the value of
f (−2.8, 2.3).
Solution:
(b) Recall: L(x, y ) = 2(x + 3) + 3(y − 2) + 1.
Now, the linear approximation of f (−2.8, 2.3) is L(−2.8, 2.3), and
L(−2.8, 2.3) = 2(−2.8 + 3) + 3(2.3 − 2) + 2
= 2(0.2) + 3(0.3) + 1
= 2.3.
We conclude L(−2.8, 2.3) = 2.3.
C
Section 14.5
Example
(a) Find the gradient of f (x, y , z) =
√
x + 2yz.
(b) Find the directional derivative of f at (0, 2, 1) in the direction
given by h0, 3, 4i.
(c) Find the maximum rate of change of f at the point (0, 2, 1).
Solution:
1
(a) ∇f (x, y , z) = √
h1, 2z, 2y i.
2 x + 2yz
(b) We evaluate the gradient above at (0, 2, 1),
1
1
∇f (0, 2, 1) = √
h1, 2, 4i = h1, 2, 4i.
4
2 0+4
Section 14.5
Example
(a) Find the gradient of f (x, y , z) =
√
x + 2yz.
(b) Find the directional derivative of f at (0, 2, 1) in the direction
given by h0, 3, 4i.
(c) Find the maximum rate of change of f at the point (0, 2, 1).
Solution:
(b) Recall: ∇f (0, 2, 1) = 14 h1, 2, 4i.
We now need a unit vector parallel to h0, 3, 4i,
u= √
1
1
h0, 3, 4i = h0, 3, 4i.
5
9 + 16
Then, Du f (0, 2, 1) = 14 h1, 2, 4i · 15 h0, 3, 4i =
Therefore, Du f (0, 2, 1) = 11/10.
1
20 (6
+ 16) =
11
10 .
Section 14.5
Example
(a) Find the gradient of f (x, y , z) =
√
x + 2yz.
(b) Find the directional derivative of f at (0, 2, 1) in the direction
given by h0, 3, 4i.
(c) Find the maximum rate of change of f at the point (0, 2, 1).
Solution:
(c) The maximum rate of change of f at a point is the magnitude
of its gradient at that point, that is,
√
1
1√
21
|∇f (0, 2, 1)| = |h1, 2, 4i| =
1 + 4 + 16 =
.
4
4
4
Therefore, the maximum rate
√ of change of the function f at the
point (0, 2, 1) is given by 21/4.
C
Section 14.3
Example
Find any value of the constant a such that the function
f (x, y ) = e −ax cos(y ) − e −y cos(x) is solution of Laplace’s
equation fxx + fyy = 0.
Solution:
fx = −ae −ax cos(y ) + e −y sin(x),
fy = −e −ax sin(y ) + e −y cos(x),
fxx = a2 e −ax cos(y ) + e −y cos(x),
fyy = −e −ax cos(y ) − e −y cos(x),
then
fxx + fyy = a2 e −ax cos(y ) + e −y cos(x)
+ −e −ax cos(y ) − e −y cos(x) ,
= (a2 − 1)e −ax cos(y ).
Function f is solution of fxx + fyy = 0 iff a = ±1.
C
Section 14.2
Example
x 2 sin2 (y )
.
Compute the limit
lim
(x,y )→(0,0) 2x 2 + 3y 2
Solution:
x2
6 1, the non-negative
that is, 2
Since
6
+
2x + 3y 2
x 2 sin2 (y )
satisfies the bounds
function f (x, y ) = 2
2x + 3y 2
x2
2x 2
3y 2 ,
0 6 f (x, y ) 6 sin2 (y ).
Since limy →0 sin2 (y ) = 0, the Sandwich Theorem implies that
x 2 sin2 (y )
lim
= 0.
(x,y )→(0,0) 2x 2 + 3y 2
C
Section 13.3
Example
E
3
2
Reparametrize the curve r(t) =
cos(t ) with
2
2
respect to its arc length measured from t = 1 in the direction of
increasing t.
D3
sin(t 2 ), 2t 2 ,
Solution:
We first compute the arc length function. We start with the
derivative
r0 (t) = h3t cos(t 2 ), 4t, −3 sin(t 2 )i,
We now need its magnitude,
q
0
|r (t)| = 9t 2 cos2 (t 2 ) + 16t 2 + 9 sin2 (t 2 ),
p
= 9t 2 + 16t 2 ,
√
= 9 + 16t,
= 5t.
Section 13.3
Example
D3
E
3
2
Reparametrize the curve r(t) =
cos(t ) with
2
2
respect to its arc length measured from t = 1 in the direction of
increasing t.
sin(t 2 ), 2t 2 ,
Solution:
Recall: |r0 (t)| = 5t. Then, the arc length function is
Z t
5 2 t 5 2
5τ dτ =
s(t) =
τ = (t − 1).
2
2
1
1
2
Inverting this function for t 2 , we obtain t 2 = s + 1. The
5
reparametrization of r(t) is given by
D3
2
2
3
2
E
sin s + 1 , 2 s + 1 , cos s + 1 .
r(s) =
2
5
5
2
5
C
Double integrals (Sect. 15.1)
I
Review: Integral of a single variable function.
I
Double integral on rectangles.
I
Fubini Theorem on rectangular domains.
I
Examples.
Next class:
I Double integrals over non-rectangular regions.
I Fubini Theorem on non-rectangular domains.
I Finding the limits of integration.
Review: Integral of a single variable function.
Definition
The definite integral of a function f : [a, b] → R, in the interval
[a, b] is the number
Z
b
f (x)dx = lim
n→∞
a
n
X
f (xi∗ ) ∆x.
i=0
y
xi∗
where
∈ [xi , xi+1 ] is called a
sample point, while {xi } is a
partition in [a, b], i = 0, · · · , n,
and with xi = a + i∆x, and
(b − a)
∆x =
.
n
f(x)
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x0
x1
x3
x2
n=4
x
x
x4
x*i = x i
The integral as an area.
y
The sum Sn =
n
X
f(x)
f (xi∗ ) ∆x
is
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i=0
called
Z b a Riemann sum. Then,
f (x) dx = lim Sn .
n→∞
a
x0
x1
n=4
x3
x2
x
x
x4
x*i = x i
y
Z
The integral
f(x)
b
f (x) dx is the
a
area in between the graph of f
and the horizontal axis.
a
b
x
Double integrals (Sect. 15.1)
I
Review: Integral of a single variable function.
I
Double integral on rectangles.
I
Fubini Theorem on rectangular domains.
I
Examples.
Double integrals on rectangles
Definition
The double integral of a function f : R ⊂ R2 → R in the rectangle
R = [a, b] × [c, d] is the number
ZZ
f (x, y ) dx dy = lim
R
n→∞
n X
n
X
f (xi∗ , yj∗ ) ∆x∆y .
i=0 j=0
where xi∗ ∈ [xi , xi+1 ], yj∗ = [yj , yj+1 ],
are sample points, while {xi } and {yj },
i, j = 0, · · · , n are partitions of the
intervals [a, b] and [c, d], and
(b − a)
(d − c)
∆x =
, ∆y =
.
n
n
z
y0
x4
x3
x
x2
x1
y1
y2
x0
y3
y
4
y
R
The double integral as a volume.
The sum
Sn = lim
n→∞
n X
n
X
z
f (xi∗ , yj∗ ) ∆x∆y is
y0
i=0 j=0
called
Z Z a Riemann sum. Then,
f (x, y ) dx dy = lim Sn .
x4
n→∞
R
x3
x2
x1
y1
y2
y3
y
4
x0
y
R
x
z
f(x,y)
ZZ
The integral
f (x, y ) dx dy is
R
y
the volume above R and below
the graph of f .
x
Double integrals (Sect. 15.1)
I
Review: Integral of a single variable function.
I
Double integral on rectangles.
I
Fubini Theorem on rectangular domains.
I
Examples.
R
Fubini Theorem on rectangular domains.
Theorem
If f : R ⊂ R2 → R is continuous in R = [x0 , x1 ] × [y0 , y1 ], then
Z y1 hZ x1
ZZ
i
f (x, y ) dx dy =
f (x, y ) dx dy ,
y
R
x
Z 0x1 hZ 0y1
i
=
f (x, y ) dy dx.
x0
y0
Remark: Fubini’s Theorem: The order of integration can be
switched in double integrals of continuous functions on a rectangle.
Notation: The double integral is also written as
ZZ
Z
y1
Z
x1
f (x, y ) dx dy =
R
f (x, y ) dx dy .
y0
x0
Fubini Theorem on rectangular domains.
Example
Use
Z Z Fubini’s Theorem to compute the double integral
f (x, y ) dx dy , where f (x, y ) = xy 2 + 2x 2 y 3 , and
R
R = [0, 2] × [1, 3]. Integrate first in x, then in y .
Solution: Since x ∈ [0, 2] and y ∈ [1, 3],
Z
ZZ
3Z 2
f (x, y ) dx dy =
1
R
=
0
Z 3 hZ
1
(xy 2 + 2x 2 y 3 )dx dy
2
2
2 3
i
(xy + 2x y )dx dy .
0
We compute the interior integral in x first, keeping y constant.
After that we compute the integral in y .
Fubini Theorem on rectangular domains.
Example
Use
Z Z Fubini’s Theorem to compute the double integral
f (x, y ) dx dy , where f (x, y ) = xy 2 + 2x 2 y 3 , and
R
R = [0, 2] × [1, 3]. Integrate first in x, then in y .
Solution: We compute the integral in x first, keeping y constant.
ZZ
Z 3 hZ 2
i
2
2 3
(xy + 2x y )dx dy
f (x, y ) dx dy =
1
0
R
Z 3h 2 2
y
2y 3 3 2 i
2
=
x +
x dy ,
2
3
0
0
1
Z 3h
16 3 i
2
=
2y + y dy .
3
1
We now compute the integral in y ,
Fubini Theorem on rectangular domains.
Example
Use
Z Z Fubini’s Theorem to compute the double integral
f (x, y ) dx dy , where f (x, y ) = xy 2 + 2x 2 y 3 , and
R
R = [0, 2] × [1, 3]. Integrate first in x, then in y .
Solution: We now compute the integral in y ,
ZZ
Z 3h
16 3 i
2
f (x, y ) dx dy =
2y + y dy ,
3
1
R
y 3 3 16 y 4 3
=2 +
.
3 1
3 4 1
ZZ
26 4
+ 80 = 372/3.
We conclude:
f (x, y ) dx dy = 2
3
3
R
C
Fubini Theorem on rectangular domains.
Example
Use
Z Z Fubini’s Theorem to compute the double integral
f (x, y ) dx dy , where f (x, y ) = xy 2 + 2x 2 y 3 , and
R
R = [0, 2] × [1, 3]. Integrate first in y , then in x.
Solution:
ZZ
Z
3Z 2
f (x, y ) dx dy =
R
1
=
0
Z 2 hZ
0
(xy 2 + 2x 2 y 3 )dx dy
3
2
2 3
i
(xy + 2x y )dy dx.
1
Z 2h x 3 3
2x 2 4 3 i
f (x, y ) dx dy =
y +
y dx.
3
4
1
1
R
0
ZZ
Fubini Theorem on rectangular domains.
Example
Use
Z Z Fubini’s Theorem to compute the double integral
f (x, y ) dx dy , where f (x, y ) = xy 2 + 2x 2 y 3 , and
R
R = [0, 2] × [1, 3]. Integrate first in x, then in y .
Z 2h x 3 3
2x 2 4 3 i
y +
y dx.
f (x, y ) dx dy =
3
4
1
1
0
R
ZZ
Solution:
Z 2h
i
26
2
f (x, y ) dx dy =
x + 40 x dx
3
R
0
26 x 2 2
x 3 2
=
+ 40 3 2 0
3 0Z Z
26
8
=
(2) + 40 ⇒
f (x, y ) dx dy = 372/3.
3
3
R
ZZ
Fubini Theorem on rectangular domains.
Example
Use
Z Z Fubini’s Theorem to compute the double integral
y
x
f (x, y ) dx dy , where f (x, y ) = + , and R = [1, 4] × [1, 2].
y
x
R
Solution: We choose to first integrate in y and then in x.
ZZ
Z 4 Z 2
x
y
f (x, y ) dx dy =
+
dy dx,
y
x
1
1
R
Z 4 hZ 2 x
y i
=
+
dy dx,
y
x
1
1
Z 4h 2 1 y 2 2 i
x ln(y ) +
=
dx,
x 2 1
1
1
Z 4h
3 1i
=
ln(2) x +
dx.
2x
1
Fubini Theorem on rectangular domains.
Example
Use
Z Z Fubini’s Theorem to compute the double integral
y
x
f (x, y ) dx dy , where f (x, y ) = + , and R = [1, 4] × [1, 2].
y
x
R
Solution: We compute the integral in x,
ZZ
Z 4h
3 1i
f (x, y ) dx dy =
ln(2) x +
dx
2x
R
1
4 x 2 4 3 = ln(2)
ln(x) ,
+
2 1
2
1
15
3
=
ln(2) + ln(4),
2
2
15
=
+ 3 ln(2).
2
ZZ
We conclude:
f (x, y ) dx dy = (21/2) ln(2).
R
C
A particular case of Fubini’s Theorem.
Corollary
If the continuous function f : R ⊂ R 2 → R satisfies that
f (x, y ) = g (x)h(y ), then the double integral of function f in the
rectangle R = [x0 , x1 ] × [y0 , y1 ] is given by
Z x1 Z y1
Z x1
Z y1
g (x)h(y )dy dx =
g (x)dx
h(y )dy .
x0
y0
x0
y0
Remark: In the case that f (x, y ) is a product of two functions g ,
h, with g (x) and h(y ), then the double integral of f is also a
product of the integral of g times the integral of h.
A particular case of Fubini’s Theorem.
Example
1 + x2
, in the
Compute the double integral of f (x, y ) =
1 + y2
rectangular region R = [0, 2] × [0, 1].
Solution:
ZZ
2Z 1
1 + x2
f (x, y ) dx dy =
dy dx,
2
R
0
0 1+y
i
hZ 2
ihZ 1 1
2
dy ,
=
(1 + x )dx
2
1
+
y
0
0
1 2 1 2 = x + x
arctan(y ) ,
3 0
0
0
π
8
=
2+
.
4
3
ZZ
We conclude
f (x, y ) dx dy = (14/3)(π/4) = (7/6)π.
Z
R
C
Double integrals on regions (Sect. 15.1)
I
I
Review: Fubini’s Theorem on rectangular domains.
Fubini’s Theorem on non-rectangular domains.
I
I
I
Type I: Domain functions y (x).
Type II: Domain functions x(y ).
Finding the limits of integration.
Review: Fubini’s Theorem on rectangular domains.
Theorem
If f : R ⊂ R2 → R is continuous in R = [a, b] × [c, d], then
ZZ
Z
b
d
Z
f (x, y ) dx dy =
R
f (x, y ) dy dx,
a
Z
c
d
Z
=
b
f (x, y ) dx dy .
c
a
z
f(x,y)
Remark: Fubini’s Theorem: It is
simple to switch the order of
integration in double integrals of
continuous functions on a
rectangle.
y
x
R
Double integrals on regions (Sect. 15.1)
I
I
Review: Fubini’s Theorem on rectangular domains.
Fubini’s Theorem on non-rectangular domains.
I
I
I
Type I: Domain functions y (x).
Type II: Domain functions x(y ).
Finding the limits of integration.
Fubini’s Theorem on Type I domains, y (x).
Theorem
If f : D ⊂ R2 → R is continuous in D, then hold (Type
I):
2
If D = (x, y ) ∈ R : x ∈ [a, b], y ∈ [g1 (x), g2 (x)] , with g1 , g2
continuous functions on [a, b], then
ZZ
Z
b
Z
g2 (x)
f (x, y ) dx dy =
D
f (x, y ) dy dx.
a
g1 (x)
z
y
f(x,y)
y = g2(x)
f(x,g1(x))
f(x,g (x))
2
y
a
y = g (x)
1
a
x
b
b
x
x
g (x)
1
g (x)
2
Double integrals on regions (Sect. 15.1)
I
I
Review: Fubini’s Theorem on rectangular domains.
Fubini’s Theorem on non-rectangular domains.
I
I
I
Type I: Domain functions y (x).
Type II: Domain functions x(y ).
Finding the limits of integration.
Fubini’s Theorem on Type II domains, x(y ).
Theorem
If f : D ⊂ R2 → R is continuous in D, then hold (Type
II):
2
If D = (x, y ) ∈ R : x ∈ [h1 (y ), h2 (y )], y ∈ [c, d] , with h1 , h2
continuous functions on [c, d], then
ZZ
Z
d
Z
h2 (y )
f (x, y ) dx dy =
D
f (x, y ) dx dy .
c
h1 (y )
z
y
d
x = h1(y)
f(x,y)
x = h2(y)
f(h1(y),y)
f(h2(y),y)
y
d
c
h 1(y)
c
h 2(y)
x
x
y
Summary: Fubini’s Theorem on non-rectangular domains.
Theorem
If f : D ⊂ R2 → R is continuous in D, then hold:
(a) (Type I) If D = (x, y ) ∈ R2 : x ∈ [a, b], y ∈ [g1 (x), g2 (x)] ,
with g1 , g2 continuous functions on [a, b], then
ZZ
Z
b
g2 (x)
Z
f (x, y ) dy dx.
f (x, y ) dx dy =
a
D
g1 (x)
(b) (Type II) If D = (x, y ) ∈ R2 : x ∈ [h1 (y ), h2 (y )], y ∈ [c, d] ,
with h1 , h2 continuous functions on [c, d], then
ZZ
Z
d
Z
h2 (y )
f (x, y ) dx dy =
D
f (x, y ) dx dy .
c
h1 (y )
A double integral on a Type I domain.
Example
Find the integral of f (x, y ) = x 2 + y 2 , on the domain
D = {(x, y ) ∈ R2 : 0 6 x 6 1, x 2 6 y 6 x}.
Solution:
This is a Type I domain,
with lower boundary
y
y = g (x) = x
2
2
y = g1 (x) = x ,
and upper boundary
y = g 1(x) = x2
y = g2 (x) = x.
0
1
x
A double integral on a Type I domain.
Example
Find the integral of f (x, y ) = x 2 + y 2 , on the domain
D = {(x, y ) ∈ R2 : 0 6 x 6 1, x 2 6 y 6 x}.
ZZ
Z b Z g2 (x)
f (x, y ) dy dx
Solution: Recall:
f (x, y ) dx dy =
a
D
with g1 (x) =
x2
g1 (x)
and g2 (x) = x, we obtain
ZZ
1Z x
Z
f (x, y ) dx dy =
D
(x 2 + y 2 )dy dx,
x2
0
Z 1 h 3 i
y x
x
x2 y 2 +
=
dx.
3 x2
x
0
ZZ
f (x, y ) dx dy =
D
Z 1h
2
x x −x
2
0
i
1 3
6
+ x −x
dx.
3
A double integral on a Type I domain.
Example
Find the integral of f (x, y ) = x 2 + y 2 , on the domain
D = {(x, y ) ∈ R2 : 0 6 x 6 1, x 2 6 y 6 x}.
ZZ
Z 1h
1 3
i
2
2
6
Solution:
f (x, y ) dx dy =
x x −x + x −x
dx.
3
D
0
ZZ
Z 1h
1 3 1 6i
f (x, y ) dx dy =
x − x + x − x dx,
3
3
D
0
hx4 x5 x4
x 7 i1
=
−
+
−
,
4
5
12 21 0
1 1
1
9
=
.
= − −
3 5 21
(3)(5)(7)
ZZ
3
We conclude:
f (x, y ) dx dy = .
35
D
3
4
C
Summary: Fubini’s Theorem on non-rectangular domains.
Theorem
If f : D ⊂ R2 → R is continuous in D, then hold:
(a) (Type I) If D = (x, y ) ∈ R2 : x ∈ [a, b], y ∈ [g1 (x), g2 (x)] ,
with g1 , g2 continuous functions on [a, b], then
ZZ
b
Z
g2 (x)
Z
f (x, y ) dy dx.
f (x, y ) dx dy =
a
D
g1 (x)
(b) (Type II) If D = (x, y ) ∈ R2 : x ∈ [h1 (y ), h2 (y )], y ∈ [c, d] ,
with h1 , h2 continuous functions on [c, d], then
ZZ
d
Z
Z
h2 (y )
f (x, y ) dx dy =
f (x, y ) dx dy .
D
c
h1 (y )
A double integral on a Type II domain.
Example
Find the integral of f (x, y ) = x 2 + y 2 on the domain
√
D = {(x, y ) ∈ R2 : y 6 x 6 y , 0 6 y 6 1}.
Solution:
This is a Type II domain,
with left boundary
x = h1 (y ) = y ,
y
1
x = h1(y) = y
and right boundary
x = g2 (y ) =
√
y.
x = h2(y) = y
0
Remark:
This domain is both Type I and Type II: y = x 2 ⇔ x =
1
√
x
y.
A double integral on a Type I domain.
Example
Find the integral of f (x, y ) = x 2 + y 2 , on the domain
√
D = {(x, y ) ∈ R2 : y 6 x 6 y , 0 6 y 6 1}.
ZZ
Z d Z h2 (y )
Solution: Recall:
f (x, y ) dx dy =
f (x, y ) dx dy
D
c
h1 (y )
√
with h1 (y ) = y and h2 (y ) = y , we obtain
ZZ
Z
1Z
f (x, y ) dx dy =
D
0
√
y
x 2 + y 2 dx dy ,
y
Z 1 h 3 √ √y i
x y
2
+ y x
dy .
=
3 y
y
0
Z 1h
i
1 3/2
3
2 1/2
f (x, y ) dx dy =
y
−y +y y
− y dy .
D
0 3
ZZ
A double integral on a Type I domain.
Example
Find the integral of f (x, y ) = x 2 + y 2 , on the domain
√
D = {(x, y ) ∈ R2 : y 6 x 6 y , 0 6 y 6 1}.
Solution:
Z
Z
Z 1h
i
1 3/2
3
2 1/2
f (x, y ) dx dy =
y
−y +y y
− y dy .
D
0 3
Z 1h
i
1 3/2 1 3
5/2
3
f (x, y ) dx dy =
y
− y +y
− y dy ,
3
D
0 3
h1 2
1 y 4 2 7/2 y 4 i1
5/2
=
y
−
+ y
−
,
35
3 4
7
4 0
2
1
2 1
9
=
−
+ − =
.
15 12 7 4
(3)(5)(7)
ZZ
3
We conclude
f (x, y ) dx dy = .
C
35
D
ZZ
Domains Type I and Type II.
Summary: We have shown that a double integral of a function f
on the domain D given in the pictures below holds,
ZZ
Z
1Z x
f (x, y )dx dy =
D
Z
1Z
x2
y
f (x, y )dx dy .
f (x, y )dy dx =
0
√
0
y
y
y
1
y = g (x) = x
2
x = h1(y) = y
y = g 1(x) = x2
0
1
x = h2(y) = y
x
0
1
Double integrals on regions (Sect. 15.1)
I
I
Review: Fubini’s Theorem on rectangular domains.
Fubini’s Theorem on non-rectangular domains.
I
I
I
Type I: Domain functions y (x).
Type II: Domain functions x(y ).
Finding the limits of integration.
x
Domains Type I and Type II.
Example
RR
Find the limits of integration of
D f (x, y ) dxdy in the domain
2
2
x
y
D = (x, y ) ∈ R2 :
+
6 1 when D is considered first as
9
4
Type I and then as Type II.
x2 y2
+
= 1.
Solution: The boundary is the ellipse
9
4
So, the boundary as Type I is given by
r
r
2
x
x2
y = −2 1 −
= g1 (x), y = 2 1 −
= g2 (x).
9
9
The boundary as Type II is given by
r
r
2
y2
y
= h1 (y ), x = 3 1 −
= h2 (y ).
x = −3 1 −
4
4
C
Domains Type I and Type II.
Example
Z
1 Z ex
dy dx.
Reverse the order of integration in
0
1
Solution:
This integral is written as Type I, since we first integrate on vertical
intervals [1, e x ], with boundaries y = e x , y = 1, while x ∈ [0, 1].
y
y = ex
By inverting the first equation and looking at
the figure we get the left and right boundaries:
x = ln(y)
e
x=1
x = ln(y ),
x = 1,
y ∈ [1, e].
with
1
y=1
x
1
Z
1 Z ex
Therefore, we conclude that
Z
e
Z
1
dy dx =
0
1
dx dy .
1
ln(y )
C