Math 227 Problem Set VIII Solutions
1. Let F = (x − yz)ı̂ı + (y + xz)̂ + (z + 2xy)k̂ and let
S1 be the portion of the cylinder x2 + y 2 = 2 that lies inside the sphere x2 + y 2 + z 2 = 4
S2 be the portion of the sphere x2 + y 2 + z 2 = 4 that lies outside the cylinder x2 + y 2 = 2
V be the volume bounded by S1 and S2
Find
RR
with n̂ pointing inward
(a) S1 F · n̂ dS
RRR
(b) V ∇ · F dV
RR
with n̂ pointing outward.
(c) S2 F · n̂ dS
Use the divergence theorem to answer at least one of parts (a), (b) and (c).
Solution. Observe that ∇ · F = 3. So
ZZZ
∇ · F dV =
V
ZZZ
3 dV
V
√
The horizontal cross-section of V at height z is a washer with outer radius 4 − z 2 (determined by the
√
equation of the sphere) and inner radius 2 (determined by the equation of the cylinder). So the cross
section has area π 4 − z 2 − π2 = π 2 − z 2 . On the intersection of the sphere and cylinder z 2 = 4 − 2 = 2
so
Z √2
Z √2
ZZZ
√
√
3/2 2
∇ · F dV = 3 √ π 2 − z dz = 6π
2 − z 2 dz = 6π 2 2 − 2 3 = 8 2π
− 2
V
0
√
√
On the cylindrical surface, using (surprise!) cylindrical coordinates, x = 2 cos θ, y = 2 sin θ, z = z,
so that
n̂ = − cos θı̂ı + sin θ̂
√
dS = 2 dθ dz
√
√
√
F · n̂ = 2 cos θ − z sin θ − cos θ + 2 sin θ + z cos θ − sin θ = − 2
so
ZZ
S1
By the divergence theorem
ZZ
S2
F · n̂ dS = −2
F · n̂ dS =
ZZZ
V
Z
√
2
√
− 2
Z
dz
2π
0
∇ · F dV −
ZZ
√
dθ = −8 2π
S1
√
F · n̂ dS = 16 2π
2. Show that the centroid (x̄, ȳ, z̄) (where, for example, x̄ is the average value of x) of a solid V with volume
|V | is given by
ZZ
(x̄, ȳ, z̄) =
1
2|V |
(x2 + y 2 + z 2 ) n̂ dS
∂V
Solution. Recall that one variant of the divergence theorem is
with f = x2 + y 2 + z 2 , gives
1
2|V |
ZZ
∂V
(x2 + y 2 + z 2 ) n̂ dS =
1
2|V |
1
ZZZ
V
RR
∂V
f n̂ dS =
RRR
V
∇f dV . Applying this,
(2xı̂ı + 2y̂ + 2z k̂) dV = (x̄, ȳ, z̄)
3. Let V be the solid in 3-space defined by
0≤z≤
9 − x2 − y 2
9 + x2 + y 2
Let S be the curved portion of the boundary of V oriented with outward normal and let
F = zy 3 ı̂ı + yx ̂ + (2z + y 2 )k̂
Assume that the volume of V is α and compute
pointing normal.
RR
F · n̂ dS in terms of α. Here n̂ is the outward
S
Solution. Observe that ∇ · F = x + 2. Since V is invariant under x → −x,
ZZZ
ZZZ
ZZ Z
∇ · F dV =
(x + 2) dV = 2
dV = 2α
V
V
V
Let S ′ be the bottom surface of V . It is z = 0, x2 + y 2 ≤ 9. On S ′
n̂ = −k̂
so
ZZ
S′
F · n̂ dS = −
ZZ
F · n̂ = −y 2
dS = dx dy
y 2 dx dy = −
Z
3
dr r
0
Z
2π
dθ r sin θ
0
2
= −π
Z
3
0
dr r3 = − 81
4 π
R 2π
Here, we used the following trick to compute 0 sin2 θ dθ. Since the graphs of cos θ and sin θ are just
translates of each other, the integral of cos2 θ over a full period and the integral of sin2 θ over a full
R 2π
R 2π
period are the same. That is 0 sin2 θ dθ = 0 cos2 θ dθ so that
Z
2π
2
sin θ dθ =
0
1
2
Z
2π
2
sin θ dθ +
0
1
2
Z
2π
2
cos θ dθ =
0
1
2
Z
2π
0
2
2
sin θ + cos θ dθ =
Now back to the main problem. By the divergence theorem
ZZ
ZZZ
ZZ
F · n̂ dS =
∇ · F dV −
F · n̂ dS = 2α +
S
V
S′
1
2
Z
2π
dθ = π
0
81
4 π
4. Find the flux of F = (y + xz)ı̂ı + (y + yz)̂ − (2x + z 2 )k̂ upward through the first octant part of the sphere
x2 + y 2 + z 2 = a2 .
Solution. Let V = (x, y, z) x2 + y 2 + z 2 ≤ a2 , x ≥ 0, y ≥ 0, z ≥ 0 . Then ∂V consists of an x = 0
face, a y = 0 face, a z = 0 face and the first octant part of the sphere. Call the latter S. Then
ZZZ
ZZZ
ZZZ
z + 1 + z − 2z dV =
dV = 81 34 πa3 = 16 πa3
∇ · F dV =
V
ZZ
x=0
Z Z face
y=0
face
ZZ
z=0
face
V
V
F · (−ı̂ı) dy dz =
ZZ
x=0
face
(−y) dy dz = −
Z
F · (−̂) dx dz = 0
F · (−k̂) dx dy =
ZZ
z=0
face
(2x) dx dy =
2
2a3
3
0
a
dr r
Z
0
π/2
dθ r sin θ = −
Z
0
a
3
r2 dr = − a3
By the divergence theorem
ZZ
ZZ
ZZ
ZZZ
ZZ
F·(−k̂) dx dy = π6 − 13 a3
F·(−̂) dx dz−
F·(−ı̂ı) dy dz−
∇ ·F dV −
F·n̂ dx dy =
z=0
face
y=0
face
x=0
face
V
s
5. Let E(r) be the electric field due to a charge configuration that has density ρ(r). Gauss’ law states that,
if V is any solid in IR3 with surface ∂V , then the electric flux
ZZZ
ZZ
ρ dV
E · n̂ dS = 4πQ
where
Q=
V
∂V
is the total charge in V . Here, as usual, n̂ is the outward pointing unit normal to ∂V . Show that
∇ · E(r) = 4πρ(r)
for all r in IR3 . This is one of Maxwell’s equations. Assume that ∇ · E(r) and ρ(r) are well–defined and
continuous everywhere.
Solution. By the divergence theorem
ZZ
∂V
So by Gauss’ law
ZZZ
V
∇ · E dV = 4π
ZZZ
E · n̂ dS =
ρ dV
ZZZ
V
⇒
V
∇ · E dV
ZZZ
V
∇ · E − 4πρ dV = 0
This is true for all solids V for which the divergence theorem applies. If there were some point in IR3 for
which ∇ ·E−4πρ were, say, strictly bigger than zero, then, by continuity, we could find a ball Bǫ centered
RRR on that point with ∇ · E − 4πρ > 0 everywhere on Bǫ . This would force Bǫ ∇ · E − 4πρ dV > 0, which
RRR violates V ∇ · E − 4πρ dV = 0 with V set equal to Bǫ . Hence ∇ · E − 4πρ must be zero everywhere.
r−a
6. Fix any point a ∈ IR3 and define the vector field E(r) = |r−a|
3.
(a) Compute ∇ · E(r).
RR
(b) Let Sǫ be the sphere of radius ǫ centered on a. Compute Sǫ E · n̂ dS where n̂ is the outward pointing
unit normal to Sǫ .
RR
(c) Let V be a solid in IR3 with surface ∂V . Suppose that a ∈
/ V and a ∈
/ ∂V . Evaluate ∂V E · n̂ dS
where n̂ is the outward pointing unit normal to ∂V .
RR
(d) Let V be a solid in IR3 with surface ∂V . Suppose that a ∈ V but a ∈
/ ∂V . Evaluate ∂V E · n̂ dS
where n̂ is the outward pointing unit normal to ∂V .
Solution. (a) Since r − a = (x − a1 )ı̂ı + (y − a2 )̂ + (z − a3 )k̂, we have
∂
∂x ı̂ı ·
(r − a) = 1
and
r−a
∂
∂x ı̂ı · |r−a|3
=
1
|r−a|3
∂
∂x |r
− a|2 = 2(x − a1 ) = 2ı̂ı · (r − a)
·(r−a)
− 32 ı̂ı|r−a|
ı · (r − a) =
5 2ı̂
3
1
|r−a|3
− 3 [ı̂ı·(r−a)]
|r−a|5
2
Similarly,
=
1
|r−a|3
−3
∂
∂z k̂
=
1
|r−a|3
− 3 [k̂·(r−a)]
|r−a|5
Adding
3
|r−a|3
∇ · E(r) =
[̂
·(r−a)]2
|r−a|5
∂
r−a
 · |r−a|
3
∂y ̂
−3
·
r−a
|r−a|3
[ı̂ı·(r−a)]2 +[̂
·(r−a)]2 +[k̂·(r−a)]2
|r−a|5
=
2
3
|r−a|3
2
|r−a|
− 3 |r−a|
5 = 0
for all r, except r = a, where E is not defined.
r−a
(b) The unit outward normal n̂ at r ∈ Sǫ is |r−a|
. Furthermore, every r ∈ Sǫ obeys |r − a| = ǫ. So
ZZ
Sǫ
E · n̂ dS =
ZZ
Sǫ
r−a
|r−a|3
·
r−a
|r−a|
ZZ
dS =
1
|r−a|2
Sǫ
dS =
ZZ
Sǫ
1
ǫ2
dS =
2
1
ǫ2 4πǫ
= 4π
(c) By the divergence theorem
ZZ
∂V
ZZZ
E · n̂ dS =
V
∇ · E dV = 0
since ∇ · E is well–defined and zero everywhere in V .
(d) The divergence theorem does not apply directly in this case, because ∇ · E is not defined at the point
a ∈ V . (I find it helpful to think of ∇ · E as being infinite at a.) But, if we choose ǫ sufficiently small,
then the ball
Bǫ = r ∈ IR3 |r − a| < ǫ
is a subset of V . Because ∇ · E is well–defined at every point of V \ Bǫ , we may apply the divergence
theorem to the solid V \ Bǫ . The surface of V \ Bǫ is the union of ∂V and Sǫ , but with the outward
r−a
, the negative of the outward normal, n̂, to Sǫ at the
normal to V \ Bǫ at a point r ∈ Sǫ being − |r−a|
same point. Thus
ZZZ
ZZ
∂(V \Bǫ )
so that
ZZ
∂(V \Bǫ )
and
E · n̂ dS =
E · n̂ dS =
ZZ
∂V
ZZ
∂V
E · n̂ dS =
V \Bǫ
∇ · E dV = 0
E · n̂ dS −
ZZ
Sǫ
ZZ
Sǫ
E · n̂ dS = 0
E · n̂ dS = 4π
by part (b).
7. Let V be a solid in IR3 with surface ∂V . Show that
ZZ
r · n̂ dS = 3 volume(V )
∂V
See if you can explain this result geometrically. Do not hand this part in.
Solution. By the divergence theorem
ZZZ
ZZZ
ZZ
3 dV = 3volume(V )
∇ · r dV =
r · n̂ dS =
∂V
V
V
The volume of the cone with vertex (0, 0, 0) and base a tiny piece of surface dS is 31 times the area of
the base times the height of the cone. The height of the cone is |n̂ · r|, where r is a point in dS. So the
volume of the cone is 13 |n̂ · r| dS.
4
First assume that (0, 0, 0) is in V and V is convex. Then
◦ n̂ · r > 0, and the volume is 31 n̂ · r dS.
◦ the cone is contained in V and
◦ V is the union of all the tiny conical pieces with dS running over ∂V .
So
ZZ
volume(V ) = 13
r · n̂ dS
∂V
To generalise to the case that V is not convex or (0, 0, 0) is not in V , write V as the difference between a
large convex solid and one or more smaller convex solids.
n̂
dS
r
(0, 0, 0)
5