Math 227 Problem Set VII Solutions
RR
1. Evaluate, for each of the following, the flux S F · n̂ dS where n̂ is the outward normal to the surface
S.
(a) F = (x2 + y 2 + z 2 )n (xı̂ı + y ̂ + z k̂) and the surface S is the sphere x2 + y 2 + z 2 = a2 .
(b) F = xı̂ı + y ̂ + z k̂ and S is the surface of the rectangular box 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c.
p
(c) F = y ı̂ı + z k̂ and S is the surface of the solid cone 0 ≤ z ≤ 1 − x2 + y 2 .
Solution. (a) The surface is g(x, y, z) = x2 + y 2 + z 2 − a2 = 0. So, on the surface of the sphere,
∇g
|∇
∇g|
n̂ =
xı̂ı+y̂
+zk̂
=√ 2 2 2
=⇒
x +y +z
=⇒
n+1−1/2
n+1−1/2
F · n̂ = x2 + y 2 + z 2
= a2
= a2n+1
ZZ
ZZ
F · n̂ dS = a2n+1
dS = 4πa2n+3
S
S
since the surface area of a sphere of radius a is 4πa2 .
(b)
Z
Z
0ı̂ı + y ̂ + z k̂ · (−ı̂ı) dy dz = 0
F · n̂ dS =
x=0
x=0
Z face
Z face
x=a
face
F · n̂ dS =
x=a
face
aı̂ı + y ̂ + z k̂ · ı̂ı dy dz
=a
Z
x=a
face
dy dz = abc
Similarly,
Z
y=0
face
F · n̂ dS =
Z
z=0
face
Z
F · n̂ dS = 0
y=b
face
F · n̂ dS =
Z
z=c
face
F · n̂ dS = abc
The total flux is 3abc .
(c) The flux through the base is
Z
base
F · n̂ dS =
ZZ
(yı̂ı) · (−k̂) dx dy = 0
Parametrize the top part of the cone by r(r, θ) = r cos θı̂ı + r sin θ̂ + (1 − r)k̂ with 0 ≤ θ ≤ 2π and
0 ≤ r ≤ 1. Then
Tr = cos θı̂ı + sin θ̂ − k̂
=⇒
Tθ = −r sin θı̂ı + r cos θ̂
n̂dS = Tr × Tθ dr dθ = cos θı̂ı + sin θ̂ − k̂ × − r sin θı̂ı + r cos θ̂ dr dθ
= − r cos θı̂ı − r sin θ̂ + rk̂ dr dθ
We have the orientation correct because the k̂ component of n̂ is positive. The flux through the top is
Z
top
Z
1
F · n̂ dS =
1
=
Z
Z
2π
dr
2π
dr
Z
0
0
0
0
dθ r sin θı̂ı + (1 − r)k̂ · − r cos θı̂ı − r sin θ̂ + rk̂
dθ r2 sin θ cos θ + r(1 − r) =
1
1
3
×0+
2π
2
−
2π
3
=
π
3
2. When a fluid of constant density flows through a circular pipe, its velocity is v = U (1 − r2 /a2 )k̂ where
r and z are cylindrical coordinates, a is the radius of the pipe and U is the velocity at the centre of the
pipe. Find the rate at which mass is transported down the pipe. If the pipe remains circular at each
cross section, but the radius changes slowly along the pipe, show that U a2 will remain constant.
a
z
Solution. The rate of transport of mass across a face is
Z
face
ρv · n̂dS =
Z
face
ρU (1 − r2 /a2 )dS =
2π
Z
dθ
0
Z
0
a
dr rρU (1 − r2 /a2 ) = 2πρU
h
r2
2
−
ia
r4
2
4a 0
=
π
2
2 ρU a
By conservation of mass, and incompressibility, the mass crossing all faces must be the same. So U a2
must be constant, even if U and a vary from face to face.
3. Prove the mean value theorem for flux integrals: If F is a continuous vector field on a connected surface
S, then
ZZ
F · n̂ dS = [F(x0 ) · n̂(x0 )] A(S)
S
for some point x0 on S. Here A(S) is the surface area of S.
Hint. First prove that, for any continuous real valued function f and any positive function g,
ZZ
ZZ
f g dS = f (x0 )
g dS
S
S
for some point x0 on S.
Solution. Define
fm = inf f (x)
x∈S
fM = sup f (x)
x∈S
Then fm ≤ f (x) ≤ fM for all x on S. Hence
ZZ
S
fm g dS ≤
ZZ
S
f g dS ≤
ZZ
fM g dS
S
=⇒
fm
RR
f g dS
≤ fM
≤ RRS
g dS
S
Because f is continuous, it takes every value between fm and fM . So there is some x0 on S with
RR
f g dS
f (x0 ) = RRS
g dS
S
Now just apply this with f (x) = F(x) · n̂(x) and g(x) = 1. With this g,
2
RR
S
g dS = A(S).
4. (Hand in only parts (b) and (c) of this problem.)
Evaluate ∇ · F and ∇ × F for each of the following vector fields.
(b) F = xy 2ı̂ı − yz 2̂ + zx2 k̂
(d) F = θ̂θ (the polar basis vector in 2d)
(a) F = xı̂ı + y ̂ + z k̂
(c) F = r̂ (the polar basis vector in 2d)
Solution.
(a) ∇ · (xı̂ı + y ̂ + z k̂) = 3
(b) ∇ · (xy 2ı̂ı − yz 2̂ + zx2 k̂) = y 2 − z 2 + x2
(c) ∇ · √ 2x 2 ı̂ı + √ 2y 2 ̂ = √ 21 2
x +y
x +y
x +y
x
√
ı̂
ı
+
̂

=
0
(d) ∇ · √ −y
2
2
2
2
x +y
∇ × (xı̂ı + y ̂ + z k̂) = 0
∇ × (xy 2ı̂ı − yz 2̂ + zx2 k̂) = 2yzı̂ı − 2xz̂ − 2xy k̂
∇ × √ 2x 2 ı̂ı + √ 2y 2 ̂ = 0
x +y
x +y
x
√
∇ × √ −y
ı̂
ı
+
̂ = √ 2k̂ 2
2
2
2
2
x +y
x +y
x +y
x +y
5. Does ∇ × F have to be perpendicular to F?
Solution. No. The field in part (b) of the last question provides a counterexample.
6. (Hand in only parts (a) and (c) of this problem.)
Verify the vector identities
(a) ∇ · (f F) = f∇
∇ · F + F · ∇f
(b) ∇ · (F × G) = G · (∇
∇ × F) − F · (∇
∇ × G)
(c) ∇ × (F × G) = F(∇
∇ · G) − (∇
∇ · F)G + (G · ∇ )F − (F · ∇)G
(d) ∇ · f (∇
∇g × ∇h) = ∇ f · (∇
∇g × ∇h)
(e) ∇ · (∇
∇ × F) = 0
(f) ∇ × (∇
∇f ) = 0
(g) ∇ × (∇
∇ × F) = ∇(∇
∇ · F) − ∇2 F
P3
Solution. Write ı̂ı = ı̂ı2 , ̂ = ı̂ı2 , k̂ = ı̂ı3 and x = x1 , y = x2 , z = x3 so that ∇ = n=1 ı̂ın ∂∂xn .
(a)
3
3
3
P
P
P
∂f
∂Fn
∂
(f
F
)
=
f
+
∇ · F + F · ∇f
∇ · (f F) =
n
∂xn
∂xn
∂xn Fn = f∇
n=1
n=1
n=1
(b)
∇ · (F × G) =
3
P
n=1
ı̂ın ·
∂
∂xn (F
× G) =
3
P
n=1
∂F
∂xn
ı̂ın ·
3
P
F×
×G +
n=1
∂G
∂xn
· ı̂ın
Applying a · (b × c) = (a × b) · c and (a × b) · c = a · (b × c),
∇ · (F × G) =
3
P
n=1
ı̂ın ×
∂F
∂xn
·G+
3
P
n=1
F·
∂G
∂xn
= (∇
∇ × F) · G − F · (∇
∇ × G)
3
P
∇ × F) · G −
× ı̂ın = (∇
F · ı̂ın ×
n=1
∂G
∂xn
(c)
∇ × (F × G) =
3
P
n=1
ı̂ın ×
∂
∂xn (F
× G) =
3
P
n=1
ı̂ın ×
∂F
∂xn
3
P
×G +
ı̂ın × F ×
n=1
Applying a × (b × c) = (c · a)b − (b · a)c,
∇ × (F × G) =
3 h
P
∂F
Gn ∂x
−
n
n=1
i
∂Fn
∂xn G
+
3 h
P
∂Gn
n=1
∂xn
∂G
F − Fn ∂x
n
= (G · ∇)F − (∇
∇ · F)G + (∇
∇ · G)F − (F · ∇)G
3
i
∂G
∂xn
(d) By part (a), followed by part (b)
∇ · f (∇
∇g × ∇f ) = f∇
∇ · (∇
∇g × ∇ h) + ∇f · (∇
∇g × ∇ h)
= f (∇
∇ × ∇g) · ∇h − ∇g · (∇
∇ × ∇ f ) + ∇ f · (∇
∇g × ∇h)
But ∇ × ∇ g = ∇ × ∇h = 0, so ∇ · f (∇
∇g × ∇ f ) = ∇f · (∇
∇g × ∇h).
(e)
∂F2
∂F1
∂F1
∂F3
∂F2
3
∇ × F = ∂F
ı̂
ı
−
̂

+
−
−
−
∂y
∂z
∂x
∂z
∂x
∂y k̂
∂ ∂F3
∂F1
∂F1
∂F3
∂F2
∂
∂
2
−
+
−
−
∇ · ∇ × F = ∂x ∂y − ∂F
∂z
∂y
∂x
∂z
∂z
∂x
∂y
=
∂ 2 F3
∂x∂y
−
∂ 2 F3
∂y∂x
−
∂ 2 F2
∂x∂z
+
∂ 2 F2
∂z∂x
+
∂ 2 F1
∂y∂z
−
∂ 2 F1
∂z∂y
=0
(f)
∂f
∇f = ∂f
 + ∂f
∂x ı̂ı + ∂y ̂
∂z k̂
∂ ∂f
∂ ∂f
∂ ∂f
ı̂
ı
−
−
∇ × ∇f = ∂y
∂z
∂z ∂y
∂x ∂z −
∂ ∂f
∂z ∂x
∂ ∂f
̂ + ∂x
∂y −
∂ ∂f
∂y ∂x
k̂ = 0
(g)
∇ × (∇
∇ × F) =
3
X
ℓ,m,n=1
ı̂ıℓ × ı̂ım × ı̂ın
∂ Fn
∂xℓ ∂xm
Using a × (b × c) = (c · a)b − (b · a)c, we have ı̂ıℓ × ı̂ım × ı̂ın = δℓ,nı̂ım − δℓ,mı̂ın (where δm,n is one when
m = n and zero when m 6= n) and hence
∇ × (∇
∇ × F) =
3
X
ı̂ım
∂Fn
∂
∂xm ∂xn
m,n=1
−
3
X
m,n=1
ı̂ın
∂ 2 Fn
∂x2m
2
= ∇(∇
∇ · F) − ∇ F
7. Let r(t) be the position of a particle of mass m at time t, v(t) its velocity and a(t) its acceleration.
Suppose that F is a given vector field and ma(t) = F r(t) .
(a) Show that
d
dt mr(t) × v(t) = r(t) × F r(t)
(i.e. “rate of change of angular momentum=torque”).
(b) In the case of planetary motion (about a sun at the origin) F r(t) is parallel to r(t). What
conclusion do you draw?
(c) Show that a planet moving about a sun under the influence of gravity remains in a fixed plane.
Solution. (a)
d
dt
mr(t) × v(t) = mv(t) × v(t) + mr(t) × a(t) = mr(t) × a(t) = r(t) × F r(t)
d
(b) If F r(t) is parallel to r(t) then r(t) × F r(t) = 0. So dt
mr(t) × v(t) = 0. That is, mr(t) × v(t)
is independent of time. This is called conservation of angular momentum.
(c) Let M = mr(t) × v(t). By part b, this vector is a constant. Furthermore M is always perpendicular
to r(t). Thus r(t) remains perpendicular to a fixed vector. That is, r(t) remains in a fixed plane, the
plane through the origin (the sun) perpendicular to M.
4
8. Let D be a subset of IR3 with the property that each point of D can be joined to the origin by a straight
line segment in D. Suppose that ∇ · F = 0 in D. Let r(t) = txı̂ı + ty̂ + tz k̂ be a parametrization of the
line segment from the origin to (x, y, z) in D. Define
Z 1
G(x, y, z) =
tF r(t) × dr
dt (t) dt
0
Show that ∇ × G = F throughout D.
Solution 1. Since r(t) is linear in t,
G(x, y, z) =
so that
Z
∇×G=
Z
1
0
F r(t) × r(t) dt
1
∇ × F r(t) × r(t) dt =
0
Z
1
0
∇ × [H r(t) ] dt
where H(r) = F(r) × r. By the chain rule,
= t ∂H
∂x (tx, ty, tz)
∂
∂x [H(tx, ty, tz)]
So
∂
∂y [H(tx, ty, tz)]
= t ∂H
∂y (tx, ty, tz)
∇ × [H r(t) ] = t∇
∇ × H(r)
∂
∂z [H(tx, ty, tz)]
= t ∂H
∂z (tx, ty, tz)
r=(tx,ty,tz)
By the vector identity ∇ × (F × G) = F(∇
∇ · G) − (∇
∇ · F)G + (G · ∇)F − (F · ∇)G,
∂F
∂F
∇ × H(r) = ∇ × F × r = F(∇
∇ · r) − (∇
∇ · F)r + (r · ∇ )F − (F · ∇)r = 2F + x ∂F
∂x + y ∂y + z ∂z
since ∇ · r = 3, (F · ∇ )r = F and, by hypothesis, ∇ · F = 0. Subbing in
Z 1h
i
2 ∂F
2 ∂F
2tF(tx, ty, tz) + t2 x ∂F
(tx,
ty,
tz)
+
t
y
(tx,
ty,
tz)
+
t
z
(tx,
ty,
tz)
dt
∇×G=
∂x
∂y
∂z
0
=
Z
1
0
d
dt
h
i
h
it=1
t2 F(tx, ty, tz) dt = t2 F(tx, ty, tz)
= F(x, y, z)
t=0
2
− ∂G
∂z = F1 . The other components are similar.
Z 1
t F1 tx, ty, tz y − F2 tx, ty, tz x dt
G3 (x, y, z) =
Solution 2. We shall show that
∂G3
∂y
0
G2 (x, y, z) =
Z
1
0
so
∂G3
∂y
∂G2
∂z
⇒
∂G3
∂y
−
∂G2
∂z
1
=
Z
1
=
Z
1
=
Z
0
0
t F1 tx, ty, tz +
t
0
t F3 tx, ty, tz x − F1 tx, ty, tz z dt
∂F3
∂z
∂F1
∂y
tx, ty, tz tx −
tx, ty, tz ty −
∂F1
∂z
∂F2
∂y
tx, ty, tz tx dt
tx, ty, tz tz − F1 tx, ty, tz dt
h
2 ∂F1
1
2tF1 tx, ty, tz + t2 y ∂F
∂y tx, ty, tz + t z ∂z tx, ty, tz
i
2 ∂F3
2
− t2 x ∂F
tx,
ty,
tz
x
−
t
x
tx,
ty,
tz
x dt
∂y
∂z
Since, by hypothesis, ∇ · F = 0,
Z 1h
i
∂G3
2 ∂F1
∂G2
2 ∂F1
2 ∂F1
dt
tx,
ty,
tz
+
t
x
−
=
2tF
tx,
ty,
tz
+
t
y
tx,
ty,
tz
+
t
z
tx,
ty,
tz
1
∂y
∂z
∂x
∂y
∂z
0
=
Z
0
1
d
dt
h
i
h
it=1
t2 F1 (tx, ty, tz) dt = t2 F1 (tx, ty, tz)
= F1 (x, y, z)
t=0
5