Solutions: MATH 251, Exam 3
Jens Forsgård
April 22, 2016
1. Let F be the vector field F = hyz, xz, xyi.
(a) Compute the divergence of F .
3 pts.
∇ · F = 0 + 0 + 0 = 0.
(b) Compute the curl of F .
3 pts.
∇ × F = hx − x, −(y − y), z − zi = h0, 0, 0i
(c) Find a function f (x, y, z) such that F = ∇f .
4 pts.
R
R
R
We have that f = yzdx = xzdy = xydz, which gives f = xyz + A(y, z) = xyz + B(x, z) =
xyz + C(x, y). We can choose A(y, z) = B(x, z) = C(x, y) = 0, which gives f = xyz.
2
2. Compute the work done by the vector field F = hxey , x + yi on a particle moving from (0, 0) to (1, 2)
along a straight line.
6 pts.
We can parametrize the line segment by r(t) = ht, 2ti for 0 ≤ t ≤ 1. This gives dr = hdt, 2dti. Plugging
this into the integral yields
Z
Z
F · dr =
C
1
"
4t2
te
0
2
e4t
+ (t + 2t)2 dt =
+ 3t2
8
#1
=
0
e4
1
+3− .
8
8
3. Let C be the curve from (0, 1) to (5, 1) pictured below. Compute the curve integral
F (x, y) = hx + y, x + yi.
R
C
F · dr, where
7 pts.
We have that Q0x − Py0 = 1 − 1 = 0, hence F is a conservative vector field. By the same method as in
Problem 1 we compute the potential function to be f (x, y) = (x + y)2 /2. It follows, by the fundamental
theorem of curve integrals, that
Z
1
F · dr = f (5, 1) − f (0, 1) = 18 − .
2
C
4. Let C be the unit circle Rwith positive orientation. Let F = hxy + arctan(ex ), x2 + y 2 + earctan y i.
Compute the curve integral C F · dr.
7 pts.
We have that Q0 x − P 0 y = 2x − x = x. Since C is closed we can use Green’s theorem;
Z
ZZ
Z 1 Z 2π
r2 cos(θ) dθdr = 0.
F · dr =
x dxdy =
C
0
D
1
0
5. Let S be the surface which is the part of the graph of the function z = 1 − x2 − y 2 which lies
above the plane z = −1. Orient S by choosing the normal vector pointing upwards (i.e., with positive
z-coordinate). Let F = hzx, zy, 2i.
(a) Find a parametrization of S (and do not forget to specify the domain D for the parameters). 3 pts.
Since the surface is the graph of the function f (x, y) = 1 − x2 − y 2 , we can parametrize by
r(u, v) = hu, v, 1 − u2 − v 2 i.
That z ≥ −1 gives 1 − u2 − v 2 ≥ −1 which is equivalent to u2 + v 2 ≤ 2.
(b) Compute the flux of F along S.
7 pts.
We have that
ru0 = h1, 0, −2ui and rv0 = h0, 1, −2vi,
and obtain that
ru0 × rv0 = h2u, 2v, 1i,
which has positive z-coordinate. Let D = {u2 + v 2 ≤ 2}. We find that
ZZ
ZZ
F · dS =
h(1 − u2 − v 2 )u, (1 − u2 − v 2 )v, 2i · h2u, 2v, 1idudv
S
D
ZZ
=2
(1 − u2 − v 2 )(u2 + v 2 ) + 1 dudv
D
√
2
Z
Z
=2
0
2π
r2 − r4 + 1 r dθdr
0
r6
r2
r
−
+
= 4π
4
6
2
4 8 2
= 4π
− +
4 6 2
8π
.
=
3
4
√2
0
2