Homework 07 Solutions
Math 21a
Spring, 2014
1. Evaluate the iterated integral.
Z 1Z zZ y
2
(a) (Stewart 12.7 # 6 )
ze−y dx dy dz
0
0
0
Solution: We perform the iterated integral:
Z 1Z z
Z 1Z z
Z 1Z zZ y
y
2
2
2
yze−y dy dz
ze−y · x dy dz =
ze−y dx dy dz =
0
0
0
0
0
0
0
0
Z 1
Z
Z
1 −y2 z
1 1 −z2
1 1 −z2
z·
=
dz = −
e
z e
− 1 dz = −
ze
− z dz
−2
2 0
2 0
0
0
1
1
1 −z2
1
1 −z2
1
1
e
− z2 =
e
+z =
=− ·
e−1 + 1 − e0 + 0
2
−2
2
4
4
0
0
1
1 1
+1−1 =
.
=
4 e
4e
Note that we’ve twice done an integral like
Z
2
xe−x dx =
1 −x2
e
+C
−2
using the substitution u = −x2 . (We did this for both the y integral and the z integral.)
Z √π Z x Z xz
(b) (Stewart 12.7 # 8 )
0
0
x2 sin y dy dz dx
0
Solution: Again we integrate:
Z √π Z x Z xz
Z √π Z x
Z √π Z x
xz
x2 sin y dy dz dx =
x2 · − cos y dz dx =
x2 (1 − cos(xz)) dz dx
0
0
0
0
Z √π
0
0
0
0
x
Z √π Z √π
1
1
=
x2 z − sin(xz) dx =
x2 x − sin(x2 ) dx =
x3 − x sin(x2 ) dx
x
x
0
0
0
0
√π √
1 4 1
1 4 1
1 √ 4 1
=
x + cos(x2 )
( π) + cos(( π)2 ) −
(0) + cos(02 )
=
4
2
4
2
4
2
0
=
This could also be written as
π2
1
1
π2
+ (−1) − 0 − =
− 1.
4
2
2
4
π2 − 4
.
4
2. Evaluate the triple integral:
ZZZ
(a) (Stewart 12.7 #10 )
yz cos(x5 ) dV , where E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ 2x}.
E
Solution:
By the given description of E, we can write the triple integral as an iterated integral:
ZZZ
Z 1 Z x Z 2x
yz cos(x5 ) dz dy dx.
yz cos(x5 ) dV =
E
0
0
0
This we integrate in the usual way:
Z 1 Z x Z 2x
Z 1Z x
Z 1Z x
1 2x
1
yz cos(x5 ) dz dy dx =
y cos(x5 ) · z 2 dy dx =
y cos(x5 ) · · 4x2 dy dx
2
2
0
0
0
0
0
0
0
0
Z 1Z x
=2
x2 y cos(x5 ) dy dx
0
0
Z 1
Z 1
1 2x
1
y
dx = 2
x2 cos(x5 ) · x2 dx =
x4 cos(x5 ) dx
2 0
2
0
0
0
1
sin(1)
1
1
= sin(x5 ) =
sin(1) − sin(0) =
.
5
5
5
0
Z 1
=2
x2 cos(x5 ) ·
ZZZ
xy dV , where E is bounded by the parabolic cylinders y = x2 , x = y 2 and the planes
(b) (Stewart 12.7 # 14 )
E
z = 0 and z = x + y.
Solution:
One easy way to describe the region E is as
E = {(x, y, z) : 0 ≤ z ≤ x + y, (x, y) ∈ D} ,
where D is the region in the xy-plane bounded by the two parabolas y = x2 and x = y 2 . This region we can draw:
y = x2
y
y=
√
x
1
x
1
Thus we can write our integral as either
Z 1 Z √x Z x+y
Z Z Z x+y
ZZZ
xy dV =
E
D
Z 1 Z √y Z x+y
xy dz dy dx
xy dz dA =
0
0
x2
or
xy dz dx dy.
0
y2
0
0
These last two iterated integrals are clearly equivalent (because of the symmetry in x and y), and we’ll compute
only the first:
Z 1 Z √x Z x+y
ZZZ
xy dz dy dx
xy dV =
E
0
x2
Z 1 Z √x
0
xy · z
=
0
x2
Z 1 Z √x
x+y
dy dx =
0
Z 1 Z √x
xy(x + y) dy dx =
0
x2
0
(x2 y + xy 2 ) dy dx
x2
y=√x
Z 1
Z 1
1
1
1 3/2
1
=
x2 · y 2 + x · y 3
x − x4 + x ·
x
− x6
dx
dx =
x2 ·
2
3
2
3
0
0
y=x2
Z
1 1 3
=
3x − 3x6 + 2x5/2 − 2x7 dx
6 0
1
2 7/2 2 8
1 3
3
4
1
1 9
3
1 3 4 3 7
x − x +
x
− x
=
− + −
= ·
=
.
=
6 4
7
7/2
8
6 4
7
7
4
6 14
28
0
3. (Based on Stewart 12.7# 40 ) Let E be the tetrahedron bounded by the planes x = 0, y = 0 z = 0 and x + y + z = 1.
(a) Use a triple integral to find the volume of E.
Solution: Here this tetrahedron is the volume between z = 0 and z = 1 − x − y over the region in the xy-plane
bounded by the positive axes and the line 1 − x − y = 0, or x + y = 1:
y
1
1
x
x+y =1
This volume is then given by integrating the constant 1 over this solid E:
Z 1 Z 1−x Z 1−x−y
ZZZ
1 dz dy dx
1 dV =
Volume of E =
0
E
0
0
Z 1 Z 1−x
(1 − x − y) dy dx
=
0
0
Z 1
(1 − x)y −
=
0
1−x
Z
1 2
1 1
dx =
y
(1 − x)2 dx
2
2 0
y=0
1
1
1
1
= − (1 − x)3 = − (0 − 1) = .
6
6
6
0
Thus the volume of E is 1/6.
(b) Suppose that the density of the tetrahedron at any point (x, y, z) is given by ρ(x, y, z) = y (in lbs/in3 ). Find the
total mass of the tetrahedron.
Solution:
The mass of E is given by integrating the density function:
Z 1 Z 1−x Z 1−x−y
ZZZ
ρ(x, y, z) dV =
Mass of E =
y dz dy dx
E
0
0
0
Z 1 Z 1−x
Z 1 Z 1−x 1
=
y (1 − x − y) dy dx =
(1 − x)y − y 2 dy dx
2
0
0
0
0
1−x
Z 1
Z 1
1
1
1
dx =
=
(1 − x)y 2 − y 3
(1 − x)3 dx
2
3
6 0
0
y=0
=−
1
1
1
1
(1 − x)4 = − (0 − 1) =
.
24
24
24
0
Thus the mass of E is 1/24 lbs.
4. In this problem we practice with switching the order of triple integration:
(a) (Stewart 12.7 #36 ) Write five other iterated integrals that are equal to the given iterated integral:
Z 1 Z x2 Z y
f (x, y, z) dz dy dx.
0
0
0
Solution: We switch the order of integration in pairs, starting with dz dy dx: The original outer limits for dy dx
gave us the region
y = x2
y
1
1
x
in the xy-plane. We write this outer pair of integrals as dx dy by re-writing y = x2 as x =
√
y. We get
Z 1Z 1 Z y
0
√
f (x, y, z) dz dx dy.
y
(1)
0
Now we do an example of switching the order of integration for the inner two integrals. In (1), the inner two
integrals are evaluated over a region Dy that depends on the value of y. This region is
z
1
y
√
y1
x
and so the triple integral can be written as
Z 1Z yZ 1
0
0
f (x, y, z) dx dz dy.
√
y
(2)
Now let’s switch the outer integrals, aiming for the order dx dy dz. This involves the region
z=y
z
1
1
y
in the yz-plane, and we get the triple integral
Z 1Z 1Z 1
0
z
√
f (x, y, z) dx dy dz.
(3)
y
To get the order dy dx dz, we’ll switch the inner integrals. For fixed z, the region in the xy-plane is
y = x2
y
1
z
√
z
1
x
This gives us the triple integral
Z 1 Z 1 Z x2
0
√
z
f (x, y, z) dy dx dz.
(4)
z
Finally, the last order is dy dz dx, which we get by switching the outer integrals in (4). This gives us the region
z
z = x2
1
1
x
in the xz-plane, which in turn gives us the integral
Z 1 Z x2 Z x2
f (x, y, z) dy dz dx.
0
0
(5)
z
(b) (Optional: For 4 points Extra Credit – Stewart 12.7 #34 ) The figure on page 881 in the text shows the region of
Z 1 Z 1−x2 Z 1−x
f (x, y, z) dy dz dx. Rewrite the integral in the five other orders.
integration for the triple integral
0
0
0
Solution: We do this as the previous part: switching in pairs, with a sketch for each switch. We otherwise
proceed without comment:
Z 1 Z 1−x2 Z 1−x
f (x, y, z) dy dz dx
0
0
0
z
1
Z 1 Z 1−x Z 1−x2
=
1 − x2
f (x, y, z) dz dy dx
0
0
0
1−x
1
y
y
1
Z 1 Z 1−y Z 1−x2
=
f (x, y, z) dz dx dy
0
0
0
1
xy = 1 − x
z
1
Z 1 Z 2y−y2 Z 1−y
=
Z √1−z
Z 1Z 1
f (x, y, z) dx dz dy
f (x, y, z) dx dz dy +
0
0
0
0
2y−y 2
2y − y 2
0
x
1 − y 1 z = 1 − x2
To get to dx dy dz and dy dx dz, we start again with the initial integral:
Z 1 Z 1−x2 Z 1−x
f (x, y, z) dy dz dx
0
0
0
z
1
Z 1 Z √1−z Z 1−x
f (x, y, z) dy dx dz
=
0
0
0
x
1 z = 1 − x2
y
1
Z 1 Z 1−√1−z Z √1−z
=
Z 1Z 1
f (x, y, z) dx dy dz +
0
0
0
0
√
1− 1−z
Z 1−y
f (x, y, z) dx dy dz
1−
√
1−z
0
√
1−z 1
xy = 1 − x
5. (Stewart 12.7 #52 ) The average value of a function f (x, y, z) over a solid region E is defined to be
ZZZ
1
fave =
f (x, y, z) dV
V (E)
E
where V (E) is the volume of E. Find the average value of the function f (x, y, z) = x2 z + y 2 z over the region enclosed
by the paraboloid z = 1 − x2 − y 2 and the plane z = 0.
Solution: One easy way to describe the region E is as the set of points between z = 0 and z = 1 − x2 − y 2 over the
disk D given by x2 + y 2 ≤ 1 in the xy-plane. Thus
Z Z Z 1−x2 −y2
V (E) =
ZZ
1 − x2 − y 2 dA
1 dz dA =
D
0
D
and
Z Z Z 1−x2 −y2
ZZZ
f (x, y, z) dV =
E
D
0
1
x2 z + y 2 z dz dA =
2
ZZ
x2 + y 2 (1 − x2 − y 2 )2 dA.
D
Both the resulting double integrals cry out for polar coordinates:
ZZ
Z 2π Z 1
π
V (E) =
1 − x2 − y 2 dA =
1 − r2 r dr dθ = .
2
D
0
0
and
ZZZ
f (x, y, z) dV =
E
1
2
ZZ
D
1
x2 + y 2 (1 − x2 − y 2 )2 dA =
2
Z 2π Z 1
0
r2 1 − r2
0
2
r dr dθ =
π
.
24
This last integral is easiest with the substitution u = 1 − r2 , so du = −2r dr and r2 = 1 − u. So the r-integral becomes
Z 1
Z
Z 0
2
1 1 2
1
−1
du =
(u − u3 ) du =
.
r2 1 − r2 r dr =
(1 − u)u2 ·
2
2
24
0
0
1
In any case, the average value of f is
fave =
1
V (E)
ZZZ
f (x, y, z) dV =
E
1
π
1
·
=
.
π/2 24
12