Physical Mathematics 2010: Problems 5 (week 6)
Vectors
1. Describe the following surfaces and lines
|r − a| = 1
|r − (r · n)n| = a ; |n| = 1
r · n = a ; |n| = 1
r × n = a ; |n| = 1, a · n = 0
Div, Grad Curl
2. Here you will need to use
∇ = e
i
∂
i
,
∇ · a = e
i
∂
i
e
j
a
j
= ∂
i
a
i
,
∇ × a =
ijk
e
i
∂
j
a
k
,
and
ijk
ilm
= δ
jl
δ
km
− δ
jm
δ
kl
Using cartesian components (Einstein index summation notation) show the following.
(a) ∇ × ∇f(x) = 0
(b) ∇ · (∇ × v(x)) = 0
(c) ∇(φ(x)ψ(x)) = φ(x)∇ψ(x) + ψ(x)∇φ(x)
(d) ∇ · (φ(x)v(x)) = (∇φ(x)) · v + φ(x)(∇ · v(x)).
(e) ∇ × (φ(x)v(x)) = (∇φ(x)) × v + φ(x)(∇ × v(x)).
(f) ∇ · r = 3
(g) ∇|r| = ˆr
(h) ∇ × r = 0
(i) Where φ(|r|) is a function only of |r| (e.g. a central potential)
∇φ(|r|) =
dφ(|r|)
d|r|
∇|r| =
dφ(r)
dr
ˆr
∇ · (φ(|r|)r) = 3φ(|r|) + r
dφ(|r|)
d|r|
(j) Use the above to show that away from the origin
∇ · (
ˆr
r
2
) = 0
3. Use Cartesian coordinates to evaluate the following, assuming r = (x, y, z), r = |r| 6= 0
and ˆr = r/r and m is a constant vector with m = |m|:
(a) ∇r; ∇r
2
; ∇(1/r); ∇(m · r); ∇(m · r/r
3
); ∇(m · ˆr).
(b) ∇ · r; ∇ · (r/r
3
); ∇ · ˆr
(c) ∇ × r; ∇ × ˆr
(d) ∇
2
(1/r); ∇
2
(m · r/r
3
); ∇
2
([3(m · r)
2
− m
2
r
2
]/r
5
)
[Hint: “Laplacian = div grad”]
[Electromagnetism, Semester 2, Problem Sheet 0]
ANSWER:
Most of these are given in the lecture notes.
Integral theorems
4. Evaluate the line integral
R
(x
2
− y
2
) dx − 2xy dy along each of the following paths
from (0, 0) to (1, 2):
(a) y = 2x
2
(b) x = t
2
, y = 2t
(c) y = 0 from x = 0 to x = 2; then along the straight line joining (2, 0) to (1, 2).
ANSWER:
All three give −11/3. The first integral becomes
R
1
0
dx(x
2
− 20x
4
), the second becomes
R
1
0
dt(2t
5
− 16t
3
) and the third is
R
1
0
dx(x
2
) +
R
2
0
dy(−2y).
5. Which, if either, of the following forces (given in Cartesian coordinates) are conservative
fields?
F
1
= (−y, x, z) , F
2
= (y, x, z)
In the conservative case(s), deduce a scalar field such that F = −∇φ. Calculate the
work done for each field in moving a particle around a circle x = cos t, y = sin t in the
(x, y) plane.
Calculate the work done in moving a particle in force field F
1
from (1, 0, 0) to (−1, 0, π)
along the following paths:
(a) the helix x = cos t, y = sin t, z = t,
(b) the straight line joining the two points
Do you expect your answers to be the same? Explain your answer.
ANSWER:
∇ × F
1
= (0, 0, 2) so not conservative; ∇ × F
2
= (0, 0, 0) so is conservative. You can
guess that φ = xy − z
2
/2 + constant in latter case.
Circle: dr = (dx, dy, dz) = (− sin t dt, cos t dt, 0) for t = 0 . . . 2π. For F
1
, work done
is
R
2π
0
dt [cos
2
t +sin
2
t = 1] = 2π. For F
2
, integrand is cos
2
t −sin
2
t = cos 2t and work
done is 0, as expected for closed path with conservative force.
Helix: dr = (dx, dy, dz) = (− sin t dt, cos t dt, 1) for t = 0 . . . π. Integrand is cos 2t + t
and result is π
2
/2. Parameterise straight line as r = (x, y, z) = (1 − 2λ, 0, πλ) for
λ = 0 . . . 1 so dr = (dx, dy, dz) = (−2dλ, 0, πdλ). Work done is
R
1
0
dλ π
2
λ = π
2
/2.
Its a coincidence that the answers are the same: non-conservative forces have path-
dependent answers.
6. Given the vector A = (x
2
− y
2
, 2xy, 0)
(a) Find ∇ × A.
(b) Evaluate
R
dS · (∇ × A) over a rectangle in the (x, y) plane bounded by the lines
x = 0, x = a, y = 0, y = b.
(c) Evaluate
H
A·dr around the boundary of the rectangle and verify Stokes’ theorem
for this case.
ANSWER:
∇ × A = (0, 0, 4y). For the surface integral dS =
ˆ
kdxdy, so answer is 2ab
2
.
For the line integral, split into 4 parts all with dz = 0: (0, 0) → (a, 0), with dy = 0 and
y = 0, giving a
3
/3; (a, 0) → (a, b), with dx = 0 and x = a, giving ab
2
; (a, b) → (0, b),
with dy = 0 and y = b, giving −a
3
/3 + ab
2
(remember integrating “backwards” in x);
(0, b) → (0, 0), with dx = 0 and x = 0, giving 0. Total for line integral is same as
surface integral, as per Stokes’ Theorem.
7. Using a vector field V = r, verify the divergence theorem for a cylinder of radius a,
centred on the z-axis and extending from z = 0 to z = h.
(a) Calculate the divergence and show the volume integral is 3πa
2
h (preferably with-
out integrating anything).
ANSWER:
∇ · V = 3, so volume integral is thrice the volume of cylinder.
(b) Show that ˆn =
ˆ
k on the top surface, and that the surface integral gives πa
2
h
(again, avoid doing integrals if you can).
ANSWER:
dS · V = zdA = hdA so surface integral is h times area of circular top.
(c) Show that the surface integral is zero on the other flat surface (where ˆn = −
ˆ
k).
ANSWER:
dS · V = −zdA = 0 as z = 0.
(d) On the curved surface, show the answer is a
R
dS = a.2πah, and hence verify the
divergence theorem.
ANSWER:
Normal to surface is ˆn = (x, y, 0)/
p
x
2
+ y
2
so ˆn · V =
p
x
2
+ y
2
= a. Surface
element is dS = adφdz, so surface integral is a.a.2π.h.
Add all surface integrals to get same result.
8. Calculate
R
dS · r over the whole surface of a cylinder bounded by x
2
+ y
2
= 1, z = 0
and z = 3. r = (x, y, z).
ANSWER:
Trick question: see above.
9. The electric field from a point charge of charge q centred at the origin is E =
qr/(4πε
0
r
3
). Find the electric flux across a spherical surface r = a.
ANSWER:
dS = ˆrdS and ˆr · r = r. On the surface of a sphere, dS = a
2
dΩ where
R
dΩ =
R
π
0
dθ
R
2π
0
dφ sin θ = 4π.
Flux is
R
E · dS =
q
4πε
0
R
dΩ =
q
ε
0
.
10. Evaluate the line integrals
(a)
R
C
(x, y, z) · dl
(b)
R
C
(y, z, d) · dl
for the paths
(a) The straight line joining (0, 0, 0) to (1, 1, 1)
(b) The straight curve defined parametrically by (t, 0, t − t
2
) where 0 ≤ t ≤ 1.
11. Integrate F = (2x + y)ˆx + (3y − x)ˆy along the curve C defined by y = x
3
and z = 0,
from (1,1,0) to (2,8,0).
Z
CF · dx
ANSWER:
35
12. F = (2x + y
2
)i + (3y − 4x)j. Consider the triangle vertices (0,0), (2,0), (2,1). Evaluate
I
C
F · dr
I
A
∇ × F · ˆndA
13. Prove the F = (y
2
cos x + z
3
)ˆx + (2y sin x − 3)ˆy + (3xz
2
+ 2)ˆz is a conservative forces
Find the scalar potential for F
