MATH 251.504
Practice Problems for the Final Examination
Fall 2006
1. Use the Divergence Theorem to evaluate
RR
S
F · dS where
F(x, y, z) = 2xz i + (2ey + xz 3 ) j + (−z 2 + ey cos x) k
and S is the surface of the solid E that lies in the region x ≥ 0 and is bounded by the
parabolic cylinder y = 1 − x2 and the planes y = 0, z = 0, and z = x.
2. Find curl F and div F.
(a) F(x, y, z) = exyz i + z 3 j + cos(xy) k.
(b) F(x, y, z) = (y − z) i + (z − x) j + (y − z) k.
(c) F(x, y, z) = 4 i + (x − z)2 j + 4 k.
3. Determine whether or not F is conservative, and if it is conservative find a function f
such that F = ∇f .
2
2
(a) F(x, y) = 2xyex i + (ex + cos y) j.
(b) F(x, y, z) = (x + y sin z) i + x sin z j + (5 + xy cos z) k.
(c) F(x, y, z) = yz i +
4. Evaluate
R
C
2y
j + xy k.
+1
y2
y 2 z 2 ds where C is the line segment from (0, 3, 0) to (−1, 2, 1).
R
5. Evaluate C F · dr where F(x, y, z) = ln x i + yz j + 3 k and C is given by r(t) =
3 i + t j + t2 k for −2 ≤ t ≤ 2.
R
6. Use Green’s Theorem to evaluate C (3x2 y + 12xy 2 ) dx + (x3 + sin y) dy where C is the
boundary of the trapezoid with vertices (0, 0), (2, 0), (0, 2), and (2, 4) with clockwise
orientation.
1
7. Use the Fundamental Theorem for Line Integrals to evaluate
R
C
F · dr where
F(x, y, z) = y 4 z i + (4xy 3 z + ey z) j + (ey + xy 4 ) k
and C is given by r(t) = cos t i + sin t j + 2t j for 0 ≤ t ≤ 7π.
RR
8. Use Stokes’ Theorem to evaluate S curl F · dS where
F(x, y, z) =
x2 + y 2
i + (y 2 − x2 z) j + ez k
z
and S is the part of the sphere x2 + y 2 + z 2 = 10 that lies below the plane z = −1 with
upward orientation.
Solutions
1. Using the Divergence Theorem,
ZZ
ZZZ
Z
F · dS =
1
Z
1−x2
x
Z
2ey dz dy dx
div F dV =
S
0
E
Z
1
Z
=
0
Z
=
1−x2
h
0
0
Z
1
Z
1−x2
y
dy dx =
z=0
0
1
2ze
y
iz=x
2xe dy dx =
0
2
Z
2
0
(2xe1−x − 2x) dx = −e1−x − x2
0
i1
1
h
2xey
iy=1−x2
y=0
= e − 2.
0
0
2. (a) curl F = (−x sin(xy) − 3z 2 ) i + (xyexyz + y sin(xy)) j − xzexyz k; div F = yzexyz .
(b) curl F = −j − 2k; div F = −1.
(c) curl F = 2(x − z) i + 2(x − z) k; div F = 0.
2
dx
2
2
2
∂
∂
3. (a) Since ∂y
(2xyex ) = 2xex = ∂x
(ex + cos y) and F is defined on all of R2 , F is
2
conservative, i.e., F = ∇f for some function f . Since fx (x, y) = 2xyex we must have
2
2
f (x, y) = yex + g(y) for some single-variable function g. Then fy (x, y) = ex + g 0 (y), so
2
that g 0 (y) = cos y, in which case we may take g(y) = sin y so that f (x, y) = yex + sin y
(note that f is unique only up to addition of a constant term).
(b) Since curl F = 0 and F is defined on all of R3 , F is conservative, i.e., F = ∇f for
some function f . Since fx (x, y, z) = x + y sin z we have f (x, y, z) = 12 x2 + xy sin z +
∂
g(y, z) and so
g(y, z) for some two-variable function g. Then fy (x, y, z) = x sin z + ∂y
∂
g(y, z) = 0, which means that g(y, z) = h(z) for some single-variable function h.
∂y
Then fz (x, y, z) = xy cos z + h0 (z) and thus h0 (z) = 5, so that we may take h(y) = 5z
and hence f (x, y, z) = 12 x2 + xy sin z + 5z (note that f is unique only up to addition of
a constant term).
(c) Since curl F = hx, 0, −zi, which is not zero everywhere, F is not conservative.
0
4. Parametrize
√ C by r(t) = h−t, 3 − t, ti for 0 ≤ t ≤ 1. Then r (t) = h−1, −1, 1i so that
0
|r (t)| = 3 and hence
√
Z
Z 1
√
√ 3 3 4 1 5 i1 17 3
2 2
2 2
y z ds =
(3 − t) t 3 dt = 3 3t − t + t
=
.
2
5
10
0
C
0
5. We have r0 (t) = h0, 1, 2ti and hence
Z
Z 2
Z 2
i2
1
3
F · dr =
hln 3, t , 3i · h0, 1, 2ti dt =
(t3 + 6t) dt = t4 + 3t2
= 0.
4
−2
−2
−2
C
6. Using Green’s Theorem,
Z
Z 2 Z x+2
Z 2
iy=x+2
2
2
2
3
(3x y + 12xy ) dx + (x + sin y) dy =
dx
−24xy dy dx =
−12xy
y=0
0
C
0
0
Z 2
(−12x3 − 48x2 − 48x) dx
=
0
i2
4
3
2
= −272.
= −3x − 16x − 24x
0
3
7. Since F = ∇f where f (x, y, z) = xy 4 z + ey z (as can be found by a procedure similar to
that in the solution of 3(b)), we have, by the Fundamental Theorem for Line Integrals,
Z
F · dr = f (r(7π)) − f (r(0)) = f (−1, 0, 14π) − f (1, 0, 0) = 14π
C
8. The boundary C of S is a circle which we parametrize by r(t) = h3 cos t, 3 sin t, −1i for
0 ≤ t ≤ 2π. Then r0 (t) = h−3 sin t, 3 cos t, 0i. Thus, by Stokes’ Theorem,
ZZ
Z
Z 2π
curl F · dS =
F · dr =
h−9, 9, 1/ei · h−3 sin t, 3 cos t, 0i dt
0
S
C
Z 2π
i2π
=
27(sin t + cos t) dt = 27(− cos t + sin t)
= 0.
0
0
4