Math 20E Second Midterm Solutions May 26, 2004
1. There may be other answers than these
(a) gradient, scalar OR gradient, potential OR curl, solenoidal
(b) 0
(c) harmonic
(d) gradient, curl OR irrotational function, solenoidal function OR
gradient of a scalar potential, curl of a vector potential
2. First solution: Use the divergence theorem and ∇ · ∇ × F= 0.
Second solution: Use Stokes’ Theorem. Since we are integrating over a closed surface,
there is no boundary and so the integral is zero.
3. Since Fis defined everywhere, we can take the domain to be all of 3-space and R0=0.
(a) Since Fis homogenous of degree 2, the supplementary homework tells us that
G=12
2 + 2(2xzi−z2k)×R= 3(yz2i−3xz2j+ 2xyzk).
Another way to solve it is to compute the integral in the text:
G=Z1
0
t(24xzt2i−12z2t2k)×Rdt =Z1
0
t(12yz2t2i−36xz2t2j+ 24xyzt2k)dt
= 3yz2i−9xz2j+ 6xyzk.
(b) We can add the gradient of any function to Gwithout changing its curl. We need
φsuch that (G+∇φ)·k= 0. The general solution to this is φ=−3xyz2+f(x, y).
For simplicity, I took f= 0, but you could make another choice. This gave me
G+∇φ=−12xz2j.
4. Looking at the given equations for D, we see that, in terms of uand v, it is the square
R={(u, v)| −1≤u≤1 and −1≤v≤1}.
We have ∂(u, v)
∂(x, y)=
1 1
1−1
=−2.
Since ∂(x,y)
∂(u,v)
∂(x,y)
∂(u,v)= 1, we have ∂(x, y)/∂(u, v) = 1/−2 = −1/2. Alternatively, you
could express x, y in terms of u, v and compute ∂(x, y)/∂(u, v ) directly. The answer
can be written in various ways:
ZZR
v2euv
−1
2
dA =1
2ZZR
v2euv du dv =1
2Z1
−1Z1
−1
v2euv du dv.
Aside: to evaluate the integral, use Rteatdt =1
ateat −1
a2eat +Cand write
2Z1
−1Z1
−1
v2euv du dv = 2 Z1
−1
veuv
u=1
u=−1dv = 2 Z1
−1
(vev−ve−v)dv
= 2(vev−ev+ve−v+e−v)
v=1
v=−1= 2(e−1+e−1)−2(−e−1−e−1) = 8/e.
