Homework 05 Solutions
Math 21a
Spring, 2014
1. In each part, DRAW the region D, and evaluate the integral.
ZZ
y
(a) (Stewart 12.3 #8 )
dA, where D is the region D = {(x, y) | 0 ≤ x ≤ 1 , 0 ≤ y ≤ x2 }.
5
D x +1
Solution:
Here is a sketch of D:
We integrate as follows:
ZZ
y
D
y
dA =
x5 + 1
1
Z
0
1
Z
0
1
=
2
x2
y
dy dx
x5 + 1
0
=
1
Z
2
1
y 2 y=x
dx
·
x5 + 1 2 y=0
1
Z
0
1
x4
1 1
dx = · ln |x5 + 1|
+1
2 5
0
x5
1
1
=
(ln 2 − ln 1) =
ln 2
10
10
x
1
ZZ
x3 dA, where D is the region D = {(x, y) | 1 ≤ x ≤ e , 0 ≤ y ≤ ln x}.
(b) (Stewart 12.3 #10 )
D
Solution:
Here is a sketch of D:
We integrate as follows:
ZZ
Z e Z ln x
x3 dA =
x3 dy dx
y
D
1
0
e
Z
=
1
y=ln x
x3 · y dx
y=0
1
e
Z
x3 ln x dx
=
1
(integrate by parts with u = ln x, dv = x3 dx)
h1
1 4 ie
x4 ln x −
x
=
4
16
1
1 4
1 4
1
= e −
e −0+
4
16
16
3 4
1
=
e +
16
16
0.5
0
1
2
e
x
2. (Stewart 12.3 #16 ) DRAW the region D. Set up the iterated integrals for both orders of integration. Then evaluate
the double integral using the easier order and explain why it’s easier.
ZZ
y 2 exy dA
where D is bounded by y = x, y = 4, and x = 0
D
Solution:
Here is a sketch of the region D:
y
4
3
2
1
1
2
3
4
x
As a type I region, D = {(x, y) | 0 ≤ x ≤ 4 , x ≤ y ≤ 4} and
ZZ
Z 4Z
y 2 exy dA =
D
0
4
y 2 exy dy dx.
x
As a type II region, D = {(x, y) | 0 ≤ y ≤ 4 , 0 ≤ x ≤ y} and
ZZ
Z 4Z
y 2 exy dA =
D
0
y
y 2 exy dx dy.
0
R
R
Evaluating y 2 exy dy requires integration by parts whereas y 2 exy dx does not, so the iterated integral corresponding
to D as a type II region appears easier to evaluate.
Z 4
ZZ
Z 4Z y
Z 4h
ix=y
2
yey − y dy
dy =
y 2 exy dA =
y 2 exy dx dy =
yexy
D
=
3.
0
0
h1
2
2
y
e
x=0
0
0
1
1 i4 1 16
1
17
− y2 =
e −8 −
− 0 = e16 −
2
2
2
2
2
0
(a) (Stewart 12.3 #36 ) Find the volume of the solid enclosed by the parabolic cylinder y = x2 and the planes z = 3y,
z = 2 + y.
Solution: The two planes intersect in the line y = 1, z = 3, so the region of integration is the plane region
enclosed by the parabola y = x2 and the line y = 1. We have 2 + y ≥ 3y for 0 ≤ y ≤ 1, so the solid region is
bounded above by z = 2 + y and bounded below by z = 3y.
Z 1 Z 1
Z 1 Z 1
Z 1 Z 1
Z 1 Z 1
V =
(2 + y) dy dx −
(3y) dy dx =
(2 + y − 3y) dy dx =
(2 − 2y) dy dx
−1
Z
1
=
x2
−1
Z
h
iy=1
2y − y 2
dx =
y=x2
−1
x2
−1
1
x2
(1 − 2x2 + x4 ) dx = x −
−1
x2
−1
2 3 1 5
x + x
3
5
1
=
−1
16
15
1
Z
Z
1−x2
(1 − x) dy dx.
(b) (Stewart 12.3 #38 ) Sketch the solid whose volume is given by the iterated integral
0
0
Solution: The solid lies below the plane z = 1−x and above the region D = {(x, y) | 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1−x2 }
in the xy-plane. Here are sketches of the region D (left) and the solid (right):
y
1
1
x
4. Sketch the region of integration and change the order of integration.
Z 1Z 4
(a) (Stewart 12.3 #42 )
f (x, y) dy dx
0
4x
Solution: Because the region of integration is D = {(x, y) | 4x ≤ y ≤ 4 , 0 ≤ x ≤ 1} = {(x, y) | 0 ≤ x ≤
Z 1Z 4
ZZ
Z 4 Z y/4
y
, 0 ≤ y ≤ 4} (see Figure 1) we have
f (x, y) dy dx =
f (x, y) dA =
f (x, y) dx dy.
4
0
4x
D
0
0
y
y
4
y
1
3
3
2
2
1
1
x
1
1
2
(a)
3
x
1
(b)
x
(c)
Figure 1: The regions for Problem 4
3
Z
√
Z
9−y
(b) (Stewart 12.3 #44 )
f (x, y) dx dy
0
0
Solution:
To reverse the order, we must break the region into two separate type I regions. (See Figure 1.)
Because the region of integration is
p
D = {(x, y) | 0 ≤ x ≤ 9 − y , 0 ≤ y ≤ 3}
√
√
= {(x, y) | 0 ≤ y ≤ 3 , 0 ≤ x ≤ 6} ∪ {(x, y) | 0 ≤ y ≤ 9 − x2 , 6 ≤ x ≤ 3}
we have
3
Z
√
Z
9−y
ZZ
f (x, y) dx dy =
0
f (x, y) dA
0
D
√
6
Z
3
Z
=
0
1
Z
Z
0
3
√
6
9−x2
Z
f (x, y) dy dx
0
π/4
(c) (Stewart 12.3 #46 )
f (x, y) dy dx
0
Solution:
Z
f (x, y) dy dx +
arctan x
Because the region of integration is
π
, 0 ≤ x ≤ 1}
4
π
= {(x, y) | 0 ≤ x ≤ tan y , 0 ≤ y ≤ }
4
ZZ
Z π/4 Z tan y
f (x, y) dy dx =
f (x, y) dA =
f (x, y) dx dy.
D = {(x, y) | arctan x ≤ y ≤
1
Z
Z
π/4
we have
0
arctan x
D
0
0
5. Evaluate the integral by reversing the order of integration.
Z 1Z 1
(a) (Stewart 12.3 #50 )
ex/y dy dx
0
Solution:
x
The region over which we are integrating is shown in Figure 2.
Z 1Z y
Z 1h
Z 1Z 1
ix=y
ex/y dy dx =
ex/y dx dy =
yex/y
dy
0
x
0
0
1
Z
=
0
0
i1
1
(e − 1)y dy = (e − 1)y 2
2
0
1
= (e − 1)
2
x=0
y
y
1
8
6
4
2
1
x
1
(a)
2
(b)
Figure 2: The regions for Problem 5
8
Z
(b) (Stewart 12.3 #52 )
0
Solution:
Z
2
√
3y
4
ex dx dy
Again the region of integration is shown in Figure 2.
8
Z
Z
0
2
√
3y
4
ex dx dy =
2
Z
0
0
=
6.
4
ex dy dx
0
2
Z
=
x3
Z
Z
y=x3
4
dx =
ex · y y=0
2
4
x3 ex dx
0
1 x 4 2
1
e = (e16 − 1)
4
4
0
(c) The half angle formulas for sin2 x and for cos2 x are
sin2 x =
1
1
− cos(2x)
2
2
and
cos2 x =
1
1
+ cos(2x).
2
2
x