Name
Sec
MATH 253
Exam 2
Sections 501-503,200
1-13
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/15
14
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/25
Fall 2008
Solutions
P. Yasskin
Total
Multiple Choice: (4 points each. No part credit.)
/107
18
1. The point 2, 3 is a critical point of the function f = x y + 12
x + y .
Apply the Second Derivative Test to classify the point 2, 3.
a. local minimum
correct choice
b. local maximum
c. inflection point
d. saddle point
e. Test Fails
f x = y − 122
x
f xx = 243
x
f y = x − 182
y
f x 2, 3 = 0
f yy = 363
y
f xx 2, 3 = 3 > 0
f xy = 1
D2, −3 = f xx f − f 2xy = 3 4
3
f y 2, 3 = 0
− 1 2 = 3 > 0
f yy 2, 3 = 36 = 4
27
3
f xy 2, 3 = 1
local minimum
2. Find the volume below the function z = xy above the region in the xy-plane
bounded by x = 4y and x = y 2 .
a. 80
b.
c.
d.
e.
V=
3
160
3
512
3
640
3
1024
3
∫0 ∫y
4
4y
correct choice
xy dx dy =
2
∫0
4
x2y
2
4y
dy =
x=y 2
∫ 0 8y
4
3
−
y5
2
dy =
y6
2y −
12
4
= 24 4 −
4
0
4 6
= 512
12
3
1
3. Find the mass of the solid below the surface z = x 2 above the triangle
with vertices 0, 0, 1, 0 and 1, 2, if the density is ρ = y.
a. 1
b.
c.
d.
e.
M=
5
2
5
4
5
6
5
8
5
correct choice
x2
∫∫∫ ρ dV = ∫ 0 ∫ 0 ∫ 0
1
=
2x
∫ 0 2x 4 dx =
1
y dz dy dx =
1
2x 5
5
0
∫ 0 ∫ 0 yz xz=0 dy dx = ∫ 0 ∫ 0
1
2x
1
2
∫0
x2
sin 2 4θ dθ = 1
2
∫0
2x
yx 2 dy dx =
1
y2
2
2x
dx
0
= 2
5
4. Find the area of one leaf of the rose r = sin4θ.
a. π
b. π
2
c. π
4
d. π
8
e. π
16
r=0
A=
correct choice
sin4θ = 0
when
π/4
∫∫ 1 dA = ∫ 0 ∫ 0
sin4θ
sin8θ
= 1 θ−
4
8
r dr dθ =
π/4
0
= 1
4
or
π/4
∫0
4θ = 0, π
r2
2
sin4θ
0
π − sin2π
4
8
θ = 0, π
4
or
dθ = 1
2
π/4
∫0
sin0
− 1 0−
4
8
π/4
1 − cos8θ
dθ
2
= π
16
2
5. Find the center of mass of the quarter circle x 2 + y 2 ≤ 9 in the first quadrant,
if the density is ρ =
9, 9
4 4
9, 9
2 2
2, 2
9 9
9 , 9
2π 2π
2π , 2π
9
9
a. x̄ , ȳ  =
b. x̄ , ȳ  =
c. x̄ , ȳ  =
d. x̄ , ȳ  =
e. x̄ , ȳ  =
M=
x2 + y2 .
π/2
correct choice
∫∫ ρ dA = ∫ 0 ∫ 0 r r dr dθ =
My =
3
π/2
π
2
r3
3
3
0
= 9π
2
x̄ = ȳ by symmetry
∫∫ xρ dA = ∫ 0 ∫ 0 r cosθr r dr dθ = sinθ π/2
0
6. Compute
∫ −2 ∫ 0
2
4−x 2
3
x 2 +y 2
∫0
e z dz dy dx.
r4
4
3
0
= 81
4
My
= 81 2 = 9
M
4 9π
2π
x̄ =
HINT: Convert to cylindrical coordinates.
a. πe 4
b. πe 4 − 1
c. π e 4 − 3
2
d. πe 4 − 4
e. π e 4 − 5
2
I=
π
r2
∫0 ∫0 ∫0
2
correct choice
e z r dz dr dθ =
2
2
= π 1 er − r
2
2
2
0
π
π
∫ 0 ∫ 0 e z  r0 r dr dθ = ∫ 0 ∫ 0
2
2
2
e r − 1 r dr dθ = π ∫
2
2
0
2
e r r − r dr
= π e 4 − 4 − 1 = π e 4 − 5
2
2
3
7. Find the average value of fx, y, z =
x2 + y2 + z2
within the sphere x 2 + y 2 + z 2 ≤ 4.
a. 3
b. 2
c. 3
correct choice
2
d. 3
4
e. 3
8
π
∫∫∫ 1 dV = ∫ 0 ∫ 0 ∫ 0 ρ 2 sin ϕ dρ dθ dϕ =
2π
2
4 πr 3 = 32π
3
3
4
π 2π 2
∫∫∫ f dV = ∫ 0 ∫ 0 ∫ 0 ρ ρ 2 sin ϕ dρ dθ dϕ = 2π− cos ϕ π0 ρ4
V=
f ave =
∫∫∫ f dV
2
= 16π
0
= 16π 3 = 3
32π
2
V
8. Find the mass of a wire in the shape of the curve ⃗
rt =
e t , 2 t, e −t
for 0 ≤ t ≤ 1
if the linear density is ρ = x .
2
3
a. e + e
b.
c.
d.
e.
3
e3 + e − 4
3
3
1 e2
2
1 e2 + 1
2
2
1 e2 − 1
2
2
correct choice
⃗v =
e t , 2 , −e −t
|v⃗| =
M=
∫ ρ ds = ∫ 0 e 2t e t + e −t  dt = ∫ 0 e 3t + e t  dt =
1
9. Compute the line integral
10.
e 2t + 2 + e −2t = e t + e −t
1
ρ = e 2t
e 3t + e t
3
1
0
3
= e +e− 4
3
3
∫P ⃗∇f ⋅ ds⃗
for the function f = xy along the parabola y = x 2 from the point −1, 1 to the point 1, 1.
NOTE: The parabola may be parametrized as ⃗rt = t, t 2 .
a. 0
b. 1
c. 2
correct choice
d. 3
e. 4
⃗v = 1, 2t
⃗
∇f = y, x = t 2 , t
∫P ⃗∇f ⋅ ds = ∫ −1 ⃗∇f ⋅ ⃗v dt = ∫ −1 t 2 + 2t 2  dt =
1
1
3
3t
3
1
−1
=2
4
∮ C F⃗ ⋅ ds⃗
11. Compute the line integral
⃗ = −x 2 y, xy 2  counterclockwise
for the vector field F
around the circle x 2 + y 2 = 9.
a. 324π
b. 162π
c. 81π
d. 81π
2
e. 81π
4
correct choice
The circle may be parametrized as ⃗rθ = 3 cos θ, 3 sin θ.
⃗ = −27 cos 2 θ sin θ, 27 cos θ sin 2 θ
F
⃗v = −3 sin θ, 3 cos θ
∮ C F⃗ ⋅ ds⃗ = ∫ 0
2π
= 162 ∫
12.
Compute
2π
0
⃗ ⋅ ⃗v dθ =
F
sin 2θ
2
2
∫ 0 81 cos 2 θ sin 2 θ + 81 cos 2 θ sin 2 θ dθ = 162 ∫ 0
2π
dθ = 81
2
∫∫∫ R ⃗∇ ⋅ F⃗ dV
2π
∫0
2π
sin 2 2θ dθ = 81
2
∫0
2π
cos 2 θ sin 2 θ dθ
1 − cos 4θ dθ = 81 θ − sin 4θ
2
4
4
2π
0
= 81π
2
for the vector field
⃗ = xz, yz, z 2  over the solid bounded by
F
the surfaces z = x 2 + y 2 and z = 4.
a. 256π
b.
c.
d.
e.
3
512π
3
256π
5
512π
5
1024π
5
correct choice
⃗ = z + z + 2z = 4z
In cylindrical coordinates, r 2 ≤ z = 4, dV = r dr dθ dz and ⃗
∇⋅F
∫∫∫ R ⃗∇ ⋅ F⃗ dV = ∫ 0 ∫ 0 ∫ r
2π
4z r dz dr dθ = 2π ∫ 2z 2  4r 2 r dr = 2π ∫ 32 − 2r 4 r dr
2
2
4
6
2
= 2π 16r 2 − 2 r
6
0
= 2π 64 − 64
3
2
2
0
0
= 256π
3
5
⃗ s, t =
13. Find the equation of the plane tangent to the surface R
st, 1 s 2 + 1 t 2 , 1 s 2 − 1 t 2
2
2
2
2
at the point where s, t = 3, 1.
a. 3x − 5y + 4z = 0
correct choice
b. 3x + 5y + 4z = 0
c. 3x − 5y + 4z = 50
d. 3x + 5y + 4z = 50
e. −6x − 10y − 8z = −50
⃗ 3, 1 =
P=R
⃗e s = t, s, s
3, 1 9 + 1 , 1 9 − 1 = 3, 5, 4
2
2 2
2
⃗e t = s, t, −t
⃗
u = ⃗e s 3, 1 = 1, 3, 3
î ̂ k̂
⃗
N=⃗
u × ⃗v = 1 3 3
⃗v = ⃗e t 3, 1 = 3, 1, −1
= î−3 − 3 − ̂ −1 − 9 + k̂ 1 − 9 = −6, 10, −8
3 1 −1
⃗⋅X = N
⃗⋅P
N
− 6x + 10y − 8z = −6 ⋅ 3 + 10 ⋅ 5 − 8 ⋅ 4 = 0
⃗ s, t =
14. Find the area of the surface R
3x − 5y + 4z = 0
st, 1 s 2 + 1 t 2 , 1 s 2 − 1 t 2
2
2
2
2
for 0 ≤ s ≤ 1 and 0 ≤ t ≤ 1.
HINT: Look for a perfect square.
2
3
b. 1
3
c. 28 2
45
d. 28
45
a.
e.
2 2
3
correct choice
î ̂ k̂
⃗ = ⃗e s × ⃗e t = t s s
N
= î−st − st − ̂ −t 2 − s 2  + k̂ t 2 − s 2  = −2st, s 2 + t 2 , t 2 − s 2 
s t −t
⃗ =
N
=
A=
2
2
2st 2 + s 2 + t 2  + t 2 − s 2  =
2s 4 + 4s 2 t 2 + 2t 4 =
∫∫ dS = ∫∫ N⃗
=
∫0
1
2
ds dt =
2 s 2 + t 2 
∫0 ∫0
1 + t 2 dt =
3
4s 2 t 2 + s 4 + 2s 2 t 2 + t 4 + t 4 − 2s 2 t 2 + s 4
1
1
2
1 t + t3
3
3
1
t=0
=
∫0
1
2 s 2 + t 2  ds dt =
2
2
s3 + t2s
3
1 + 1
3
3
=
1
dt
s=0
2 2
3
6
Work Out: (Points indicated. Part credit possible. Show all work.)
15.
∫∫ x 2 y dA over the
(15 points) Compute
y
12
10
"diamond" shaped region in the first
8
quadrant bounded by the curves
y = 1x
y = 3x
y = 12
x
u = xy
HINT: Let
6
y = 42
x
4
v = x2y
2
0
0
1
2
3
4
x
Solve for x and y: Eliminate y = ux . So v = x 2 ux = xu.
2
So x = uv . Then y = ux = u uv = uv
− v2
u
1
u
∂x, y
=
∂u, v
2u
v
= 1v − 2v = − 1v
2
− u2
v
J=
∂x, y
∂u, v
= − 1v
= 1v
x2y = v
Rewrite the integrand:
∫∫ x 2 y dA = ∫ 1 ∫ 1 v 1v du dv = 4 − 13 − 1 = 6
4
3
16. (15 points) Compute the surface integral
∫∫ H ⃗∇ × F⃗ ⋅ dS⃗
⃗ = −x 2 y, xy 2 , z 2 
for the vector field F
x 2 + y 2 + z 2 = 9 with z ≥ 0 and upward normal, parametrized as
over the hemisphere
⃗ θ, ϕ = 3 sin ϕ cos θ, 3 sin ϕ sin θ, 3 cos ϕ.
R
̂
î
⃗
N
k̂
= ⃗e r × ⃗e θ
= î−9 sin 2 ϕ cos θ − ̂ 9 sin 2 ϕ sin θ + k̂  
+ k̂ −9 sin ϕ cos ϕ sin 2 θ − 9 sin ϕ cos ϕ cos 2 θ
⃗e θ = −3 sin ϕ sin θ 3 sin ϕ cos θ
0
⃗e ϕ = 3 cos ϕ cos θ 3 cos ϕ sin θ −3 sin ϕ
= −9 sin 2 ϕ cos θ, −9 sin 2 ϕ sin θ, −9 sin ϕ cos ϕ
⃗ = 9 sin 2 ϕ cos θ, 9 sin 2 ϕ sin θ, 9 sin ϕ cos ϕ.
Normal on the hemisphere must point up. So reverse N
⃗
⃗ =
∇×F
î
̂
k̂
∂x
∂y
∂z
= î0 − ̂ 0 + k̂ y 2 − −x 2  = 0, 0, x 2 + y 2  = 0, 0, 9 sin 2 ϕ
−x 2 y xy 2 z 2
π/2
∫∫ H ⃗∇ × F⃗ ⋅ dS⃗ = ∫ 0 ∫ 0
2π
⃗
⃗⋅N
⃗ dθ dϕ =
∇×F
π/2
∫0 ∫0
2π
81 sin 3 ϕ cos ϕ dθ dϕ = 162π
sin 4 ϕ
4
π/2
0
= 81π
2
7
∫∫ S F⃗ ⋅ dS⃗
17. (25 points) Compute
⃗ = xz, yz, z 2  over the complete surface
for the vector field F
of the solid bounded by the surfaces z = x 2 + y 2 and z = 4 with outward normal.
(See the figure in #11.)
∫∫ P F⃗ ⋅ dS⃗
a. (11 pts) First compute
for the paraboloid z = x 2 + y 2 ≤ 4 which may be
⃗ r, θ = r cos θ, r sin θ, r 2 .
parametrized as R
̂
î
k̂
⃗e r = cos θ
sin θ 2r
⃗e θ = −r sin θ r cos θ 0
⃗
N
= ⃗e r × ⃗e θ = î−2r 2 cos θ − ̂ 2r 2 sin θ + k̂ r
= −2r 2 cos θ, −2r 2 sin θ, r
⃗ = 2r 2 cos θ, 2r 2 sin θ, −r
Normal on the paraboloid must point down. So reverse N
⃗ = xz, yz, z 2  = r 3 cos θ, r 3 sin θ, r 4 
F
⃗⋅N
⃗ = 2r 5 cos 2 θ + 2r 5 sin 2 θ − r 5 = r 5
F
∫∫ P F⃗ ⋅ dS⃗ = ∫ 0 ∫ 0 F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 r 5 dr dθ = 2π r6
2π
2
2π
b. (11 pts) Second compute
2
6
∫∫ D F⃗ ⋅ dS⃗
2
0
= 64π
3
for the disk x 2 + y 2 ≤ 4 with z = 4 by writing a
parametrization.
⃗ r, θ = r cos θ, r sin θ, 4
R
̂
î
k̂
⃗e r = cos θ
sin θ 0
⃗e θ = −r sin θ r cos θ 0
⃗
N
= ⃗e r × ⃗e θ = î0 − ̂ 0 + k̂ r
= 0, 0, r
⃗.
Normal on the disk must point up. So don’t reverse N
⃗ = xz, yz, z 2  = 4r cos θ, 4r sin θ, 16
F
⃗⋅N
⃗ = 16r
F
∫∫ D F⃗ ⋅ dS⃗ = ∫ 0 ∫ 0 F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 16r dr dθ = 2π8r 2 |20 = 64π
2π
2
2π
2
c. (3 pts) Combine the results from (a) and (b) to obtain
∫∫ S F⃗ ⋅ dS⃗ = ∫∫ P F⃗ ⋅ dS⃗ + ∫∫ D F⃗ ⋅ dS⃗ =
∫∫ S F⃗ ⋅ dS⃗.
64π + 64π = 256 π
3
3
8