MATH 251.504
Solutions for the Final Examination
Fall 2006
1. (a) true, (b) true, (c) true, (d) false, (e) false, (f) true, (g) true, (h) true.
2. (a) curl F = h0, y cos(yz), −z cos(yz)i and div F = − sin y + ez .
(b) curl G = h−1, −1, −2yi, and so curl F = h−2, 0, 0i and div F = 0.
R
3. We have F = ∇f where f (x, y, z) = x3 yz 2 + sin x + y, and so C F · dr = f (π, −2, 0) −
f (0, 1, 3) = −3 by the Fundamental Theorem for Line Integrals.
4. We have div F = 6z(x2 + y 2 + z 2 )2 and so by the Divergence Theorem the given surface
integral is equal to
ZZZ
2
2
2 2
Z
1
Z
π
2
Z
2π
(6ρ5 cos ϕ)ρ2 sin ϕ dθ dϕ dρ
0
0
0
Z 2π
Z 1
Z π
2 1
3
7
sin 2ϕ dϕ
dθ = π.
=6
ρ dρ
4
0
0 2
0
6z(x + y + z ) dV =
E
5. By Green’s Theorem the given line integral is equal to
Z 3 Z 3−x
Z 3
x dy dx =
(2x2 − 6x) dx = −9.
0
−3+x
0
6. The cylindrical part S1 of the surface can be parametrized by r(θ, z) = hcos θ, sin θ, zi
where 0 ≤ θ ≤ 2π and −3 ≤ z ≤ 3, and since |rθ × rz | = |hcos θ, sin θ, 0i| = 1 we have
ZZ
Z 2π Z 3
f (x, y, z) dS =
(2 + z 2 ) dz dθ = 60π.
−3
0
S1
The base S2 can be parametrized by r(r, θ) = hr cos θ, r sin θ, −3i where 0 ≤ r ≤ 1 and
0 ≤ θ ≤ 2π, and since |rr × rθ | = |hr cos θ, r sin θ, 0i| = r we have
Z 2π Z 1
ZZ
f (x, y, z) dS =
(2r2 + 9)r dr dθ = 10π.
0
S2
Thus
RR
S
f (x, y, z) dS =
RR
S1
0
f (x, y, z) dS +
1
RR
S2
f (x, y, z) dS = 70π.
7. The area is equal to
ZZ q
∂z 2
+
1 + ∂x
D
∂z 2
∂y
Z
1
Z
dA = 2
0
0
x
√
2
1 + x2 dy, dx = (23/2 − 1).
3
8. The line integral around each of the four smooth parts of the boundary of S is zero,
and so the given surface integral is equal to zero.
9. Using the parametrization r(t) = hcos t, sin ti for 0 ≤ t ≤ 2π (which gives the opposite
orientation) we have r0 (t) = h− sin t, cos ti and hence
Z
Z 2π
Z 2π
F · dr = −
hsin t, − cos ti · h− sin t, cos ti dt =
dt = 2π.
C
0
0
2