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MATH 4013 - FALL 2003 - EXAM III - VERSION 2
Do all of your work in your blue book. Show all your work. Justify your answers. When you are
asked to graph something be sure to label and number your axes.
1. (12 points) Let c : [0, 1]→ R3 be given by c(t) = (1
2
t2, 4t, 3t). Express the following integrals
in the form
∫ b
a h(t) dt. Simplify your answers as much as possible without evaluating the
integrals.
(a)
∫
c
f ds, where f(x, y, z) = yz + 16x.∫
c
f ds =
∫ b
a
f(c(t)) ||c′(t)|| dt.
f(c(t)) = (4t)(3t) + 16(1
2
t2) = 12t2 + 8t2 = 20t2.
c′(t) = (t, 4, 3), ||c′(t)|| =
√
t2 + 25, a = 0, b = 1, so we have
∫ 1
0
20t2
√
t2 + 25 dt.
(b)
∫
c
F · ds, where F(x, y, z) = (y + z, x, 2x).∫
c
F · ds =
∫ b
a
F(c(t)) · c′(t) dt
F(c(t)) = (4t+ 3t, 1
2
t2, 2(1
2
t2)) = (7t, 1
2
t2, t2).
F(c(t)) ·c′(t) = (7t)(t)+(1
2
t2)(4)+(t2)(3) = 7t2 +2t2 +3t2 = 12t2, so we have
∫ 1
0
12t2 dt.
2. (12 points) Let C1 be the straight line segment going from (0, 9) to (0, 0). Let C2 be the part
of the graph of y = x2 going from (0, 0) to (3, 9). Let C be C1 followed by C2. Evaluate∫
C
3y dx+ 2(x2 + y) dy. Hint: Use x and/or y as a parameter.
Here is a picture of C:
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(0,0)
x
y
C
C
1
2
(0,9) (3,9)
On C1 we have x = 0 while y goes from 9 to 0. We use y as a parameter. Then dx = 0 dy,
and we have
∫ 0
9
3y 0 dy + 2(0 + y) dy =
∫ 0
9
2y dy = y2
∣∣∣0
4
= −81
On C2 we have y = x
2 while x goes from 0 to 3. We use x as a parameter. Then dy = 2x dx,
and we have
∫ 3
0
3x2 dx+ 2(x2 + x2) 2x dx =
∫ 3
0
3x2 + 8x3 dx = x3 + 2x4
∣∣∣3
0
= 27 + 162 = 189
Thus we have
∫
C
3y dx+ 2(x2 + y) dy = −81 + 189 = 108.
3. (12 points) Let S be the surface parametrized by Φ(u, v) = (u4, u − v, u) for (u, v) ∈ [0, 3] ×
[0, 3].
(a) Compute Tu, Tv, Tu×Tv, and ||Tu×Tv||.
Tu = (4u
3, 1, 1), Tv = (0,−1, 0).
Tu × Tv =
∣∣∣∣∣∣∣
i j k
4u3 1 1
0 −1 0
∣∣∣∣∣∣∣ = i
∣∣∣∣∣ 1 1−1 0
∣∣∣∣∣ − j
∣∣∣∣∣ 4u3 10 0
∣∣∣∣∣ + k
∣∣∣∣∣ 4u3 10 −1
∣∣∣∣∣ = i(1) − j(0) +
k(−4u3) = (1, 0,−4u3).
||Tu ×Tv|| =
√
(1)2 + (0)2 + (−4u3)2 =
√
+ + 16u6.
(b) Find the equation ax+ by + cz = d of the tangent plane to S at Φ(1, 2) = (1,−1, 1).
At this point we have u = 1 and v = 2, so Tu×Tv = (1, 0,−4). We use this as our normal
vector to get (1, 0,−4)·(x−1, y−−1, z−1) = 0, so (1)(x−1)+(0)(y+1)+(−4)(z−1) = 0,
hence x− 1− 4z + 4 = 0, therefore x− 4z + 3 = 0, thus x− 4z = −3.
4. (24 points) Let S be the surface in Problem 3. Express the following as double iterated
integrals. Simplify your answers as much as possible without evaluating the integrals.
(a) The area of S.
The area of S is
∫ ∫
D
||Tu ×Tv|| du dv =
∫ 3
0
∫ 3
0
√
1 + 16u6 du dv.
(b)
∫ ∫
S
f dS, where f(x, y, z) = xz.∫ ∫
S
f dS =
∫ ∫
D
f(Φ(u, v))||Tu ×Tv|| du dv
f(Φ(u, v)) = (u4)(u) = u5, so we have
∫ 3
0
∫ 3
0
u5
√
1 + 16u6 du dv.
(c)
∫ ∫
S
F · dS, where F(x, y, z) = (5xz3, y, x).∫ ∫
S
F · dS =
∫ ∫
D
F(Φ(u, v)) · (Tu ×Tv) du dv.
F(Φ(u, v)) = (5(u4)(u3), u− v, u4) = (5u7, u− v, u4).
F(Φ(u, v))·(Tu×Tv) = (5u7, u−v, u4)·(1, 0,−4u3) = (5u7)(1)+(u−v)(0)+(u4)(−4u3) =
5u7 + 0− 4u7 = u7, so we have
∫ 3
0
∫ 3
0
u7 du dv.
5. (12 points) Let S be the portion of the graph of z = 7 − x − 3y lying above the rectangle
D = [0, 3]× [0, 4]. Find
∫ ∫
S
x+ 3y + z dS. (Yes, here you must evaluate the integral.)
Recall that the graph of a function z = g(x, y) is parametrized by Φ(x, y) = (x, y, g(x, y)).
This gives Tx = (1, 0, gx) and Ty = (0, 1, gy), and so Tx ×Ty = (−gx,−gy, 1), and therefore
||Tx ×Ty|| =
√
g2x + g
2
y + 1.
In our case this gives ||Tx ×Ty|| =
√
(−1)2 + (−3)2 + 1 =
√
1 + 9 + 1 =
√
11.
Also, f(Φ(x, y)) = f(x, y, 7− x− 3y) = x+ 3y + (7− x− 3y) = 7.
Thus we have
∫ ∫
S
f dS =
∫ ∫
D
f(Φ(x, y)) ||Tx ×Ty|| dx dy =
∫ ∫
D
7
√
11 dx dy =
7
√
11 Area(D) = 7
√
11(3)(4) = 84
√
11.
6. (12 points) Let S be the surface {(x, y, z) |x2 + y2 + z2 = 36, x ≥ 0, y ≥ 0, z ≥ 0}, oriented by
the outward pointing normal (x, y, z)/6. Let F(x, y, z) = (x, y − z, y + z). Find
∫ ∫
S
F · dS.
(You must evaluate the integral. Hints: Do not use a parametrization. The area of a sphere
of radius R is 4πR2.)
Here is a picture of S:
(6,0,0)
y
z
x
(0,6,0)
(0,0,6)
We express the integral as
∫ ∫
S
F · n dS.
F ·n = (x, y− z, y+ z) · (x, y, z)/6 = (x2 + y2− zy+ yz+ z2)/6 = (x2 + y2 + z2)/6 = 36/6 = 6.
S is one-eighth of the sphere of radius 4, so its area is
1
8
(4π(6)2) =
1
8
(144π) = 18π.
Thus our integral is (6)(18π) = 108π.
7. (12 points) Use Green’s Theorem to compute∫
∂D
(7x3 − 12x2y) dx+ (5y3 + 12xy2) dy,
where D is the disk {(x, y) |x2 + y2 ≤ 4}. You will receive no credit for using other methods.
(You must evaluate the integral that Green’s Theorem gives you. Hint: Use polar coordinates.)
Here is a picture of D:
(2,0)(−2,0)
(0,−2)
(0,2)
D
y
x
Green’s Theorem says that
∫
∂D
P dx+Qdy =
∫ ∫
D
∂Q
∂x
− ∂P
∂y
dx dy.
Q = 5y3 + 12xy2, so
∂Q
∂x
= 12y2.
P = 7x3 − 12x2y, so ∂P
∂y
= −12x2.
∂Q
∂x
− ∂P
∂y
= 12y2 − (−12x2) = 12y2 + 12x2 = 12(x2 + y2).
Note that x2+y2 is NOT equal to 4. We are working now in the disk D = {(x, y) |x2+y2 ≤ 4},
NOT on the circle ∂D = {(x, y) |x2 + y2 = 4}.
We wish to evaluate
∫ ∫
D
12(x2 + y2) dx dy. In polar coordinates D is given by 0 ≤ r ≤ 2
and 0 ≤ θ ≤ 2π, and we have x2 + y2 = r2. (You should know this from understanding
the geometry of polar coordinates, not from laboriously working it out from x = r cos θ and
y = r sin θ.)
It is crucial to remember that when you change coordinates in an integral you MUST take
account of how this distorts area by using the change of variables formula. In the case of polar
coordinates this is given by
dx dy =
∣∣∣∣∣∂(x, y)∂(r, θ)
∣∣∣∣∣ dr dθ = r dr dθ.
Putting all this together gives∫ 2π
0
∫ 2
0
12r2 r dr dθ =
∫ 2π
0
∫ 2
0
12r3 dr dθ =
∫ 2π
0
3r4
∣∣∣2
0
dθ =
∫ 2π
0
48 dθ = 48(2π) = 96π.
8. (4 points) Compute
∫
c∇ f · ds, where f(x, y, z) = x5y3z and c is a path running from (1, 1, 1)
to (1, 2, 3).
By the fundamental theorem for line integrals
∫
c
∇ f · ds = f(1, 2, 3)− f(1, 1, 1) = 24− 1 = 23