# SOLUTIONS TO HOMEWORK ASSIGNMENT #9, Math 253 ```SOLUTIONS TO HOMEWORK ASSIGNMENT #9, Math 253
1. For each of the following regions E, express the triple integral
iterated integral in cartesian coordinates.
RRR
E
f (x, y, z) dV as an
(a) E is the box [0, 2] &times; [−1, 1] &times; [3, 5];
Solution:
ZZZ
Z
Z
2
1
Z
5
f (x, y, z) dV =
f (x, y, z) dz dy dx
E
0
−1
3
(b) E is the pyramid with vertices (0, 0, 0), (1, 1, 1), (1, 1, −1), (−1, 1, 1), and (−1, 1, −1);
Solution:
x
–1 –0.6 –0.2 0.2
0.8
0.4
z 0
–0.4
–0.8
–1
0 x
–0.8
0 0.4 0.8 1
y
0.6
1
1
0.8
0.6
0.4
0.2
y
0
–0.2
–0.4
–0.6
–0.8
–1
Figure 1: Q1(b): Left: The solid E; Right: The image of E on xy-plane
Top function: z = y (plane passing through (0, 0, 0), (1, 1, 1), and (−1, 1, 1))
Bottom function: z = −y (plane passing through (0, 0, 0), (1, 1, −1), and (−1, 1, −1))
ZZZ
ZZ Z y
f (x, y, z) dV =
f (x, y, z) dz dA
E
Z
D
1
Z
−y
y
Z
y
=
f (x, y, z) dz dx dy
0
−y
−y
(c) E is the region in the first octant above the plane y = z and bounded by the
cylinder x2 + z 2 = 1.
Solution:
Left function: y = 0
Right function: y = z
ZZZ
ZZ Z z
f (x, y, z) dV =
f (x, y, z) dy dA
E
D
Z
1Z
0
√
1−x2
Z
z
=
f (x, y, z) dy dz dx
0
1
0
0
1
1
0.8
0.8
0.6
z
0.4
0.6
z
0.4
0.2
0.2
0
0
0.4
0 0.2 0.4
0.6 0.8 10.8 x
y
0
0
0.2
0.4
0.6
x
0.8
1
Figure 2: Q1(c): Left: The solid E; Right: The image of E on xz-plane
x
–1 –0.6 –0.2 0.2
1.2
0.8
z
0.4
0
–0.4
–1
–0.8
0 0.4
y
0x
1
0.6
1
1
0.8
0.6
0.4
0.2
y
0
–0.2
–0.4
–0.6
–0.8
–1
Figure 3: Q1(d): Left: The solid E; Right: The image of E on xy-plane
(d) E is the region inside the sphere x2 + y 2 + z 2 = 2 and above the elliptic paraboloid
z = x2 + y 2 .
Solution:
p
Top function: z = 2 − x2 − y 2
Bottom function: z = x2 + y 2
The boundary of image D on xy-plane is the intersection of top and bottom
function, which is x2 + y 2 = 1
Z Z Z √2−x2 −y2
ZZZ
f (x, y, z) dV =
f (x, y, z) dz dA
E
x2 +y 2
√
1 Z
1−x2
D
Z
=
f (x, y, z) dz dy dx
√
− 1−x2
−1
2. Consider the integral
ZZZ
Z √2−x2 −y2
Z
2
Z
4
x2 +y 2
Z
y
f (x, y, z) dV =
f (x, y, z) dz dy dx
E
−2
(a) Sketch the region E.
Solution:
x2
0
RRR
(b) Write the other five iterated integrals which represent
f (x, y, z) dV .
E
Solution:
If we project E onto xy-plane, then the top function is z = y, and the bottom
function is z = 0, as given in the question.
2
4
3
z2
1
0
0
1
2
y
3
4
–2
–1
0 x
1
2
Figure 4: Q2(a): The solid E
–2
–1
x0
1
2
4
4
3
3
z2
2y
1
1
0
0
0
1
2
y
3
4
Figure 5: Q2(b): Left: The image of E on xy-plane; Right: The image of E on yz-plane
In the order of dz dx dy,
ZZZ
Z
4
Z
f (x, y, z) dV =
Z
y
y
f (x, y, z) dz dx dy
√
− y
0
E
√
0
If we project E onto yz-plane, then the front function is x =
√
function is x = − y.
In the order of dx dy dz,
ZZZ
Z 4 Z 4 Z √y
f (x, y, z) dV =
f (x, y, z) dx dy dz
√
E
0
In the order of dx dz dy,
ZZZ
Z
f (x, y, z) dV =
z
4
Z
0
E
√
y, and back
− y
y
Z
√
y
√
− y
0
f (x, y, z) dx dy dz
If we project E onto xz-plane, we can think of it as a parabolic cylinder
E1 : {(x, y, z) : x2 &lt; y &lt; 4, −2 &lt; x &lt; 2, 0 &lt; x &lt; 4} (whole body) with the solid
E2 : {(x, y, z) : x2 &lt; y &lt; z, x2 &lt; z &lt; 4, −2 &lt; x &lt; 2} (the transparent part)
removed. So
ZZZ
ZZZ
ZZZ
f (x, y, z) dV =
f (x, y, z) dV −
f (x, y, z) dV
E
E1
E2
In the order of dy dx dz,
ZZZ
Z
f (x, y, z) dV =
E1
4
2
Z
4
f (x, y, z) dy dx dz
0
3
Z
−2
x2
4
3
4
3
z2
1
0
0
z2
1
0
–2
–1
0x
1
–2
–1
1
2
2
y
3
1
0 x
4 2
Figure 6: Q2(b): Left: The image of E on xz-plane; Right: solid: E, transparent: E2 , solid
+ transparent: E1
ZZZ
Z
4
Z
f (x, y, z) dV =
E2
Z
⇒
4
Z
2
Z
4
f (x, y, z) dy dx dz
x2
Z
4
f (x, y, z) dV =
0
x2
−2
E1
Z
2
Z
4
Z
⇒
Z
2
Z
4
Z
x2
4
f (x, y, z) dy dx dz
x2
2
Z
0
4
Z
x2
4
x2
x2
Z
4
2
Z
4
Z
4
f (x, y, z) dy dz dx
f (x, y, z) dy dz dx −
−2
−2
Z
f (x, y, z) dy dz dx
0
x2
−2
(c) Find the volume of E.
Solution:
The volume of E is given by
ZZZ
Z 2 Z 4Z y
Z
dV =
dz dy dx =
E
z
f (x, y, z) dy dz dx
−2
−2
f (x, y, z) dV =
E
√
4
f (x, y, z) dV =
ZZZ
Z
√
− z
0
In the order of dy dz dx,
ZZZ
Z
f (x, y, z) dV =
E2
4
f (x, y, z) dy dx dz −
E
ZZZ
√
− z
0
ZZZ
Z
√
z
0
2
Z
Z
4
2
y dy dx =
−2
x2
−2
&middot;
x2
x2
&cedil;
128
1 4
8 − x dx =
2
5
(d) Find the centre of mass of E when the density of E is constant.
Solution:
Let the constant density be ρ(x, y, z) = c. Then
ZZZ
128c
m=
c dV = c(volume of E) =
5
E
ZZZ
Z 2 Z 4Z y
1
1
5
x̄ =
cx dV =
cx dz dy dx =
(0) = 0
m
m −2 x2 0
128c
E
ZZZ
Z Z Z
1
1 2 4 y
5 512
20
ȳ =
cy dV =
cy dz dy dx =
=
m
m −2 x2 0
128c 7
7
E
ZZZ
Z 2 Z 4Z y
1
1
5 256
10
z̄ =
cz dV =
cz dz dy dx =
=
m
m −2 x2 0
128c 7
7
E
The centre of mass is (0, 20/7, 10/7) .
4
3. Let E be the solid bounded by z =
p
p
x2 + y 2 and z = 1 − x2 − y 2 ,
(a) Use cylindrical coordinates to find the volume of E.
Solution:
x
–0.2 0 0.2 0.4 0.6
–0.6
0.6
1.2
1
0.8
z 0.6
0.4
0.2
0
–0.2
0.4
0.2
0
y
–0.2
–0.4 0
y 0.4
0.4
–0.4
–0.4
x
–0.6
Figure 7: Q3: Left: The solid E; Right: The image of E on xy-plane
p
√
Top function: z = 1 p
− x2 − y 2 = 1 − r 2
Bottom function: z = x2 + y 2 = r
ZZ Z
ZZZ
dV =
V =
D
E
√
1−r2
dz dA
r
D is the circular image on xy-plane. The boundary of D is given by p
the intersection of the top and bottom function, which is x2 + y 2 = 1/2, or r = 1/2.
Z 2π Z √1/2 Z √1−r2
√
π
r dz dr dθ = (2 − 2)
V =
3
0
0
r
(b) Use spherical coordinates to find the volume of E.
Solution:
p
Write
the
function
in
spherical
coordinates:
z
=
x2 + y 2 ⇒ φ = π/4; z =
p
1 − x2 − y 2 ⇒ ρ = 1. So
ZZZ
Z π/4 Z 2π Z 1
√
π
ρ2 sin φ dρ dθ dφ = (2 − 2)
V =
dV =
3
0
0
0
E
2
2
4. Find the volume of the solid above
√ the xy-plane, under the surface z = 1 − x − y ,
and within the wedge x ≤ y ≤ 3x.
Solution:
Top function: z = 1 − x2 − y 2
Bottom function: z = 0
ZZZ
ZZ Z
V =
1−x2 −y 2
dV =
E
dz dA
D
0
√
Since on the xy-plane, 0 = 1 − x2 − y 2 ⇒ r = 1, y = x ⇒ θ = π/4 and y = 3x ⇒ θ =
π/3, D is the region on the xy-plane defined by D : {(r, θ) : 0 &lt; r &lt; 1, π/4 &lt; θ &lt; π/3}.
So
Z π/3 Z 1 Z 1−r2
π
V =
r dz dr dθ =
48
π/4
0
0
5
0.2
0
x
0.4 0.6
0.8
1
1
1
0.8
0.8
0.6
z
0.4
0.6
0
0.4
x
0.8
0.2
0
0 0.2 0.4
0.6 0.8 1
y
y
0.4
0.2
0
Figure 8: Q4: Left: The solid E; Right: The image of E on xy-plane
5. Find the volume remaining in a sphere of radius a after a hole of radius b is drilled
through the centre. Assume 0 &lt; b &lt; a.
Solution:
x
–1 –0.6 –0.2 0.2
0.8
0.4
z 0
–0.4
–0.8
–1
–0.8
0 x
0 0.4
y
0.8 1
0.6
1
1
0.8
0.6
0.4
0.2
y
0
–0.2
–0.4
–0.6
–0.8
–1
Figure 9: Q5: Left: The solid E; Right: The image of E on xy-plane
x2 + y 2 + z 2 = a2√
2
2
2
2
Top function: z = a2 −
px − y = a − r√
Bottom function: z = − a2 − x2 − y 2 = − a2 − r2
Cylindrical hole of radius b: x2 + y 2 = b2 ⇒ r = b
ZZZ
Z
V =
2π
Z
a
dV =
E
0
b
Z
√
a2 −r2
√
− a2 −r2
r dz dr dθ =
&cent;3/2
4 &iexcl; 2
π a − b2
3
2
1
z 0
–1
1
–2
–1 0 y
–2
2
1
0
x
–2
Figure 10: Q6: The solid E
6. Find the mass of the solid between the spheres x2 + y 2 + z 2 = 1 and x2 + y 2 + z 2 = 4
above the xy-plane when the density is ρ(x, y, z) = z.
6
Solution:
Outer function: x2 + y 2 + z 2 = 4 ⇒ ρ = 2
Inner function: x2 + y 2 + z 2 = 1 ⇒ ρ = 1
xy-plane: φ = π/2
ZZZ
Z
m=
π/2
Z
π/2
Z
2π
Z
2π
Z
2
z dV =
E
Z
0
0
1
2
=
0
=
0
15
π
4
7
(ρ cos φ)ρ2 sin φ dρ dθ dφ
1
ρ3 cos φ sin φ dρ dθ dφ
```