PHY302 VC Quiz 60 points - Open tutorial - 60 minutes Spring 2021
Note: Be sure to attempt all 6 of the required problems. Partial credit will be given. If you
cannot do a calculation but know what calculation you should be doing, describe it in words.
The first four problems in this quiz refer to the same vector field ~
V:
~
V= [2xz]ˆx+ [2yz]ˆy+x2+y2+ 3z2ˆz.
1. (15 points) Find, if any, the most general expression of a scalar field φ(x, y, z) such that
~
∇φ=~
V.
Solution:
φ(x, y, z) = ˆdx Vx=x2z+g(y, z)
φ(x, y, z) = ˆdy Vy=y2z+h(x, z)
φ(x, y, z) = ˆdz Vz=x2z+y2z+z3+f(x, y)
All three of these expressions for φ(x, y, z) have to match. We can see this matching occurs
for
g(y, z) = y2z+z3+c, h(x, z) = x2z+z3+c, f(x, y)=(x2+y2)2+c,
where cis any constant. Thus we find
φ= (x2+y2+z2)z+c.
2. (10 points) Find ~
∇ × ~
Vand justify your answer.
Solution: Since φexists (see previous problem), we can write ~
V=~
∇φ, so ~
∇ × ~
V=~
0.
We can also check explicitly:
(~
∇ × ~
V)·ˆx =∂yVz−∂zVy= 2y−2y= 0
(~
∇ × ~
V)·ˆy =−∂xVz+∂zVx=−2x+ 2x= 0
(~
∇ × ~
V)·ˆz =∂xVy−∂yVx= 0 −0=0
so
~
∇ × ~
V=~
0.
3. (10 points) Express ~
Vin spherical polar coordinates.
Solution:
Easiest method, using the result from Problem 1 above and the definition of the spherical
polar coordinates in VC51:
~
V=~
∇φ, φ = (x2+y2+z2)z+c=r2z+c=r3cos θ+c.
Then we can use VC (65), which writes the gradient for SP coordinates (note the scalar
function is φbut the azimuthal angle is ϕ:
~
V=~
∇φ=ˆr∂φ
∂r +ˆ
θ1
r
∂φ
∂θ +ˆϕ1
rsin θ
∂φ
∂ϕ
= 3r2cos θˆr−r2sin θˆ
θ.
If we don’t use the result from Problem 1, the same result just takes longer. We begin by
rewriting the cartesian components in terms of SP coordinates, just using the VC52 defini-
tions:
Vx= 2xz = 2r2sin θcos θcos ϕ
Vy= 2yz = 2r2sin θcos θsin ϕ
Vz=x2+y2+z2+ 2z2=r2+ 2r2cos2θ.
These rewritings aren’t enough though, because we need to know the SP components in terms
of SP coordinates. To find these components, we compute the dot product of ~
Vwith each of
the basis vectors ˆr,ˆ
θ, and ˆϕ, using the unit vectors in VC53. First, with ˆr:
ˆr·~
V= (sin θcos ϕˆx+ sin θsin ϕˆy+ cos θˆz)·~
V
= sin θcos ϕVx+ sin θsin ϕVy+ cos θVz
= 2r2sin2θcos θcos2ϕ+ 2r2sin2θcos θsin2ϕ+r2cos θ+ 2r2cos3θ
=r2cos θ+ 2r2cos θsin2θcos2ϕ+ sin2θsin2ϕ+ cos2θ
=r2cos θ+ 2r2cos θsin2θ+ cos2θ=r2cos θ+ 2r2cos θ
Vr= 3r2cos θ.
Next for ˆ
θ:
Vθ=ˆ
θ·~
V=Vxcos θcos ϕ+Vycos θsin ϕ−Vzsin θ
= 2r2sin θcos2θcos2ϕ+ 2r2sin θcos2θsin2ϕ−r2sin θ−2r2sin θcos2θ
= 2r2sin θcos2θcos2ϕ+ sin2ϕ−1−r2sin θ
Vθ=−r2sin θ.
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And last with ˆϕ:
Vϕ=ˆϕ·~
V=−Vxsin ϕ+Vycos ϕ
=−2r2sin θcos θcos ϕsin ϕ+ 2r2sin θcos θsin ϕcos ϕ
Vϕ= 0.
Putting these all together we have
~
V= (~
V·ˆr)ˆr+ (~
V·ˆ
θ)ˆ
θ+ (~
V·ˆϕ)ˆϕ= 3r2cos θˆr−r2sin θˆ
θ.
4. Bonus (5 extra points): Simplify the expression ijk∂iφ∂jVk.
Solution: We note
∂iφ= (~
∇φ)i, ijk∂jVk= (~
∇ × ~
V)i,
so we have
ijk∂iφ∂jVk=~
∇φ·(~
∇ × ~
V)
and from above, ~
∇ × ~
V= 0, so we get
ijk∂iφ∂jVk= 0.
The last three problems in this quiz refer to the same scalar field ψ:
ψ=xy(z+ 1).
5. (5 points) Calculate ∇ψ. We have
∇ψ=∂ψ
∂x ˆx+∂ψ
∂y ˆy+∂ψ
∂z ˆz
=y(z+ 1) ˆx+x(z+ 1) ˆy+xy ˆz.
6. (10 points) Calculate
¨S
ˆn· ∇ψdS,
where Sis the unit square in the x= 0 plane with one vertex at the origin. That is, the
vertices of the square are at (x, y, z) = (0,0,0),(0,1,0),(0,0,1),and (0,1,1).
Solution: First we find the unit normal ˆn. It is ˆx, since the square is in the x= 0 plane.
Actually we have a sign choice here; you could choose ˆn=−ˆx, and you should have received
full credit if you made this choice. I’ll write the solution here using ˆn=ˆxthough.
¨S
ˆn· ∇ψdS=ˆ1
0
dyˆ1
0
dz[ˆx· ∇ψ]
=ˆ1
0
dyˆ1
0
dz[y(z+ 1)] = ˆ1
0
dyy1
2z2+z
z=1
z=0
=ˆ1
0
dy3
2y
=3
4y2
y=1
y=0
=3
4.
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7. (10 points) Find, by any means you choose, the flux
‹S
ˆn· ∇ψdS,
where Sis now the surface of the unit cube with one vertex at the origin. That is, the
vertices of the cube are at (x, y, z) = (0,0,0),(0,1,0),(0,0,1),and (0,1,1), as well as
(1,0,0),(1,1,0),(1,0,1),and (1,1,1).
Solution:
The easiest method is to use Gauss’ theorem. We have
‹S
ˆn· ∇ψd~
S=˚V
∇·∇ψ dV,
for any closed surface Swhich encloses the volume V.
The divergence of ∇ψis
∇·∇ψ=∂y(z+ 1)
∂x +∂x(z+ 1)
∂y +∂xy
∂z = 0.
Since the divergence is zero everywhere, the integral over the divergence is zero everywhere,
so the flux integral is also zero. (Note you could also recognize ∇ · ∇ψ=∇2ψand just
calculate the Laplacian directly; you’d still get zero).
The computation can also be done by direct calculation but it is more cumbersome.
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