S. F. Ellermeyer
MATH 2203 — Final Exam Solutions
May 3, 2004
Name
Instructions. This exam contains ten problems, but only eight of them will be graded.
You may choose any eight to do. Please write DON’T GRADE on the two that you don’t
want me to grade. In writing your solution to each problem, include sufficient detail and use
correct notation. (For instance, don’t forget to write “=” when you mean to say that two
things are equal.) Your method of solving the problem must be clear to the reader
(me). If I have to struggle to understand what you have written, then you might not get full
credit or even any credit for your solution even if you get a correct answer.
1. Find the surface area of the surface with parametric equations
x = uv
y =u+v
z =u−v
2
u + v2 ≤ 1.
Solution: The surface area of this surface, S, is
ZZ
ZZ
1 dS =
|ru × rv | dA
S
D
where D is the unit disk.
Since
we have
¯
¯ i j k
¯
ru × rv = ¯¯ v 1 1
¯ u 1 −1
¯
¯
¯
¯ = −2i + (u + v) j+ (v − u) k,
¯
¯
q
(−2)2 + (u + v)2 + (v − u)2
p
= 4 + 2 (u2 + v 2 )
√ √
= 2 2 + u2 + v2
|ru × rv | =
so the surface area is
ZZ
√ ZZ √
2 + u2 + v2 dA
|ru × rv | dA = 2
D
D
√ Z 2π Z 1 √
2 + r2 r dr dθ
= 2
0
¶
µ 0
√
8
π.
= 2 6−
3
1
2. Draw a sketch of the vector field
F (x, y) = xi − yj.
Make sure to include all four quadrants in the xy plane in your picture so that the
overall nature of the vector field is evident.
Solution:
4
y
2
-4
0
-2
2
x
-2
-4
3. Evaluate the line integral
Z
yz dy + xy dz
C
where C is the curve
Solution:
√
x= t
y=t
z = t2
0 ≤ t ≤ 1.
Z
Z 1³
´
√
2
yz dy + xy dz =
tt (1) + tt (2t) dt
0
C
Z 1
¡3
¢
=
t + 2t5/2 dt
0
23
= .
28
2
4
4. The figure below shows the vector field
­
®
F (x, y) = 2xy, x2
and three curves that start at (1, 2) and end at (3, 2).
Explain why
Z
F· dr
C
has the same value for all three of these curves and find this common value. (Hint:
Use the Fundamental Theorem for Line Integrals.)
Solution: Since
∂Q
= 2x
∂x
∂P
= 2x,
∂y
we see that
∂Q ∂P
=
.
∂x
∂y
Hence, the vector field F is conservative. This explains why line integrals of F are
independent of path. To find a function f such that ∇f = F, we observe that f must
satisfy
∂f
= 2xy
∂x
3
and hence that
f (x, y) = x2 y + h (y) .
Also, f must satisfy both
∂f
= x2 + h0 (y)
∂y
and
∂f
= x2
∂y
which means that h0 (y) = 0 and hence that h (y) = C. Choosing C = 0, we see that
f (x, y) = x2 y
satisfies ∇f = F. By the Fundamental Theorem for Line Integrals, we obtain (for any
path C starting at (1, 2) and ending at (3, 2))
Z
Z
F· dr= ∇f · dr
C
C
=f (3, 2) − f (1, 2)
= 18 − 2
= 16.
5. Use Green’s Theorem to evaluate the line integral
Z
x2 y 2 dx + 4xy 3 dy
C
along the positively oriented triangle, C, with vertices at (0, 0), (1, 3), and (0, 3). D
Solution: The triangle is pictured below.
4
By Green’s Theorem,
¶
ZZ µ
Z
∂Q ∂P
2 2
3
x y dx + 4xy dy =
−
dA
∂x
∂y
C
D
ZZ
¢
¡ 3
=
4y − 2x2 y dA
Z 1DZ 3
¢
¡ 3
=
4y − 2x2 y dy dx
0
3x
318
=
.
5
6. Let f be a scalar field and let F be a vector field. Decide whether or not each of the
following expressions is meaningful. It the expression is meaningful, then state whether
it is a scalar field or a vector field.
(a) grad (div (f ))
1. not meaningful
2. meaningful. It is a scalar field.
3. meaningful. It is a vector field.
(b) curl (grad (f ))
1. not meaningful
2. meaningful. It is a scalar field.
3. meaningful. It is a vector field.
(c) curl (curl (F))
1. not meaningful
2. meaningful. It is a scalar field.
3. meaningful. It is a vector field.
(d) curl (f )
1. not meaningful
2. meaningful. It is a scalar field.
3. meaningful. It is a vector field.
(e) div (grad (f ))
1. not meaningful
2. meaningful. It is a scalar field.
3. meaningful. It is a vector field.
7. Evaluate the surface integral
ZZ
S
5
y dS
where S is the surface
¢
2 ¡ 3/2
x + y 3/2
3
0≤x≤1
0 ≤ y ≤ 1.
z=
Solution:
ZZ
y dS =
S
ZZ
q
y (zx )2 + (zy )2 + 1 dA
Z ZD p
=
y x + y + 1 dA
D
Z 1Z 1 p
=
y x + y + 1 dx dy
0
0
Z
¯x=1
2 1
¯
=
y (x + y + 1)3/2 ¯ dy
3 0
x=0
Z 1³
´
2
=
y (y + 2)3/2 − y (y + 1)3/2 dy
3 0
8
12 √
16 √
=
.
3+
2−
35
105
105
Note: The integral is a little tricky to compute. Break it into two integrals and then
use a rather tricky substitution. For example, to compute
Z 1
y (y + 2)3/2 dy,
0
use the substitution
u=y+2
du = dy.
With this substitution, the integral can be written as
Z 3
Z 1
3/2
y (y + 2) dy =
(u − 2) u3/2 du
0
2
which is not hard to do!
8. Use Stokes’ Theorem to evaluate
ZZ
S
curl (F) · dS
where
F (x, y, z) = yzi + xzj + xyk
and S is the part of the paraboloid
z = 9 − x2 − y 2
6
that lies above the plane z = 5 and is oriented upward.
Solution: Since the curve of intersection of the paraboloid and the plane is the circle
x2 + y 2 = 4,
we see that the positively oriented boundary curve of this piece of the paraboloid is
x = 2 cos (t)
y = 2 sin (t)
z=5
0 ≤ t ≤ 2π.
By Stokes’ Theorem,
I
ZZ
curl (F) · dS= F· dr
S
ZC 2π
F (r (t)) · r0 (t) dt
=
Z0 2π
=
(yzi + xzj + xyk) · (−yi + xj) dt
0
Z 2π
¢
¡ 2
=
−y z + x2 z dt
0
Z 2π
¡ 2
¢
=5
x − y 2 dt
Z0 2π
¢
¡ 2
= 20
cos (t) − sin2 (t) dt
Z0 2π
= 20
cos (2t) dt
0
= 0.
9. A vector field, F, and a counterclockwise—oriented curve, C, are pictured. Is
Z
F· dr
C
positive, negative, or zero? (You must provide a correct explanation of your answer in
order to receive credit of course.)
7
Answer: The line integral is positive because
Z
Z
F· dr = F · T ds
C
C
where T is the unit tangent vector to the curve. Since all of the angles formed by
vectors of F and and unit tangent vectors T are acute, then F · T >0 at each point on
the curve, which means that the line integral is positive.
10. Let F be the vector field
¡
¢
F (x, y, z) = xyzi − x2 y 2 z 2 j − x2 + y 3 + z 4 k.
Show by direct calculation that div (curl (F)) = 0.
Solution:
Thus
¯
¯
¯ i
¯
j
k
¯ ∂
¯
∂
∂
¯
¯
curl (F) = ¯ ∂x
∂y
∂z
¯
¯ xyz −x2 y 2 z 2 −x2 − y 3 − z 4 ¯
¡
¢
¡
¢
= −3y 2 + 2x2 y 2 z i + (2x + xy) j + −2xy 2 z 2 − xz k
div (curl (F))
¢
¢
∂ ¡
∂
∂ ¡
−3y 2 + 2x2 y 2 z +
(2x + xy) +
−2xy 2 z 2 − xz
=
∂x
∂y
∂z
2
2
= 4xy z + x − 4xy z − x
= 0.
8