MA231 Vector Analysis
Example Sheet 2: Hints and partial solutions
2010, term 1
Stefan Adams
A1 (a) 4x(b) y(1 + x2y2z2)−1/2(c) xyez.
A2 (a) ˆ
N=x
Ron the surface Sso that v·ˆ
N=kxk2/R =Rand RSv·ˆ
NdS=RRSdS=R
(Surface area of S)= 4πR3. (b) Here v·ˆ
N= 0 so the flux is zero. (c) Here v·ˆ
N=
1
R(−x2+y2+z2). But by symmetry RSx2dS=RSy2dS=RSz2dS. So RSv·ˆ
NdS=
1
RRSz2dS=1
3RRSx2+y2+z2dS=R
3RSdS=4πR3
3. Note that in each case the flux was
calculated without ever having to start in on a parameterisation. Moral — look for symmetry
tricks. The divergences in the three cases are 3,0,1and it is easy to confirm that each flux
integral agrees with the corresponding volume integral given by the divergence theorem.
A3 The divergence of ∇fis 12(x2+y2+z2). (a) 12 (b) 48πR5/5(use polar coordinates).
A4 (a) (0,0,2) (b) (0,0,0) (c) (2x, −2y, 1).
A6 curl(v)=1so that
Area(Ω) = ZΩ
Curl(v) dA=Z∂Ω
v·ˆ
Tds=Z2π
0
(0, α cos t)·(−αsin t, β cos t) dt=παβ .
A7 One possibility is to parametrise Sby x(u, v)=(u, v, 2−u2−v2)for (u, v)∈Ω = {u2+v2≤
2}. Then inward (non-unit) normal vector ˆ
N(u, v) = (−2u, −2v, −1). By parameterising ∂S
by α(t)=(√2 cos t, √2 sin t, 0) for t∈[0,2π]the tangent vector and normal vector are then
suitably oriented for Stokes’ theorem. Both sides of Stokes’ identity give the value 2π, for
example RS∇ × v·ˆ
NdS=RΩ(2u+ 2v+ 1) dudv=RΩdudv=Area of Ω, (using symmetry
to reduce the integrand from 2u+ 2v+ 1 to 1).
B1 (a) A parametrisation is α(θ, z)=(rcos θ, r sin θ, z )with r= 4. Result is zero. Why?
Symmetry because of the given vector field (x-dependence and the symmetry of the surface,
flux in and out the volume cancel each other out). (b) Summing the single parts (as in the
lecture), result is 3
2.
B2 (a) A simple calculation gives div f= 1, integrating this over the disc with radius Rgives the
result πR2. (b) Divergence = 4. Flux = 4×Volume of the pyramid with base area 9 and
height 2 = 24.
B3 (a) Use the definitions of the cross product, of the curl and diligence. (b) Use the product rule
(c) Chain rule. (d) Use the function φ(r) = 1
4πr2R∂B (a,r)fand show that φis constant. Why
gives this the proof of the statement? Answer: take the limit r↓0.
B4 The expression for div(f v)follows from the product rule, the integration by parts formula can
be derived by applying Gauss’s theorem. Adding IBP for f , g and g, f gives Green’s identity.
C3 (a) Stokes theorem for the vector field v= (f, 0,0) becomes
ZC
fˆ
T1ds=ZS∂f
∂z N2−∂f
∂y N3dS.
Now repeat with v= (0, f, 0) and v= (0,0, f ). (b) Prove each co-ordinate of the identity
separately, each is a case of the divergence theorem.
C4 Since f1and f2vanish on ∂Ω, the boundary terms in Green’s identity vanish and we get
λ2RΩf1f2=−RΩf1∆f2=−RΩf2∆f1=λ1RΩf1f2. Since λ16=λ2this implies the result.
Marking scheme:
B 1: part (a).
B 2: part (b).
B 3: Part (c) and (d) with 3 points for part (d).
B 4: all parts.