Module MA2E02 (Frolov), Multivariable Calculus, 2017
Homework Sheet 6
Due: by 12:00 on the first Friday after the corresponding tutorial session
Name and student number:
1.
(a) Express rectangular coordinates in terms of spherical coordinates.
Draw the corresponding picture.
Compute the volume of a ball of radius R.
(b) Consider the solid G bounded by the surfaces x2 + y 2 + z 2 = 4 and x2 + y 2 + z 2 = 9 and
below by the surface z = 0.
1. What is the surface x2 + y 2 + z 2 = 4?
2. What is the surface x2 + y 2 + z 2 = 9?
3. What is the surface z = 0?
4. Sketch the part of the boundary of the solid G which belongs to the surface z = 0.
5. Use triple integral and spherical coordinates to compute the volume V of the solid G.
6. Use triple integral and spherical coordinates to find the mass M of the solid G if its
density is
√
2
2
2
e x +y +z − 1
.
δ(x, y, z) = p
x2 + y 2 + z 2
7. What is the limiting value of δ(x, y, z) at x = y = z = 0?
Show the details of your work.
Solution :
(a) We have
x = r cos θ sin φ ,
y = r sin θ sin φ ,
z = r cos φ ,
r ≥ 0 , 0 ≤ φ ≤ π , 0 ≤ θ ≤ 2π ,
where r is the radial coordinate, φ is the angle between the radius-vector and the z-axis, and
θ is the angle between the projection of the radius-vector onto the xy-plane and the x-axis.
1
We use the spherical coordinates to get the volume of a ball of radius R
ZZZ
Z 2π Z π Z R
4
2
r dr sin φ dφ dθ = πR3 .
V =
dV =
3
0
0
0
V
(b)
(b)
(b)
(b)
1.
2.
3.
4.
Sphere of radius 2.
Sphere of radius 3.
The xy-plane.
The plot is shown below
3
2
1
�������
0
-1
-2
-3
-3
-2
0
-1
1
2
3
(b) 5. We use the spherical coordinates to get
ZZZ
2π
Z
V =
Z
π/2
Z
3
dV =
G
0
0
!
r2 dr sin φ dφ dθ =
2
38π
2π 3
(3 − 23 ) =
≈ 39.7935
3
3
(b) 6. We use the spherical coordinates to get
Z
ZZZ
2π
π/2
Z
Z
δ(x, y, z)dV =
M =
G
Z
0
0
3
2
Z
3
r
!
e −1 2
r dr sin φ dφ dθ
r
3
(er r − r)dr = −5π + 2π
d(er )r
2
Z 3 2
= −5π + 2π(3e3 − 2e2 ) − 2π
er dr = −5π + 2π(2e3 − e2 ) ≈ 190.268
= 2π
2
(b) 7.
er − 1
= 1.
r→0
r
δ(0, 0, 0) = lim
2.
(a) Express rectangular coordinates in terms of cylindrical coordinates.
Draw the corresponding picture.
(b) Consider the solid G bounded by the surfaces x2 + y 2 = 4 and x2 + y 2 = 16, above by
the surface z = 15 − x2 − y 2 , and below by the surface z = 0.
i. What are the surfaces x2 + y 2 = 4 and x2 + y 2 = 16?
2
ii. What is the surface z = 15 − x2 − y 2 ?
iii. What is the surface z = 0? Sketch the part of the boundary of the solid G which belongs
to the surface z = 0.
iv. Use triple integral to compute the volume V of the solid G.
v. Use triple integral to find the mass M of the solid G if its density is
δ(x, y, z) = e15−x
Show the details of your work.
Solution :
(a) We have
x = r cos θ ,
2 −y 2 −z
y = r sin θ ,
.
z=z
(b) i. Cylinder of radius 2, and cylinder of radius 4.
(b) ii. Paraboloid.
(b) iii. The xy-plane. The plot is shown below
4
2
0
-2
-4
-4
-2
0
3
2
4
(b) iv. We use the cylindrical coordinates to get
# !
ZZZ
Z
Z "Z
2
2π
V =
4
15−r
Z
rdz dr dθ = 2π
dV =
0
G
2
4
r(15 − r2 )dr = 2π(
2
0
15r2 r4 4
− )|2
2
4
4
= 2π(
4
15
16 −
− 30 + 4) = 2π(120 − 64 − 26) = 60π ≈ 188.496 .
2
4
(b) v. We use the cylindrical coordinates to get
ZZZ
Z
Z "Z
2π
M =
15−r2
4
e
δ(x, y, z)dV =
0
G
Z
= 2π
4
#
15−r2 −z
2
(0.1)
!
rdz dr dθ
0
2
2
(e15−r − 1)rdr = π(−e15−r − r2 )|42 = π(e11 −
2
1
− 12) ≈ 188061.
e
3. Let r = xi + yj + zk, let r = ||r||, let f be a differentiable function of one variable, and
let F(r) = f (r)r. Show that
(a) curl r = 0 ,
(b) ∇ r =
r
,
r
(c) div r = 3 ,
(d) ∇
1
r
=− 3,
r
r
f 0 (r)
r,
(c0 ) div F = 3f (r) + rf 0 (r) .
r
Solution: These formulas are derived just by using the definitions of curl , ∇ and div.
(a0 ) curl F = 0 ,
(b0 ) ∇ f (r) =
4