Name
Sec
MATH 251/253
Exam 2
Sections 508/200,501,502
1-12
/60 14
/15
13
/15 15
/15
Spring 2008
Solutions
P. Yasskin
Total
Multiple Choice: (5 points each. No part credit.)
/105
1. Find the directional derivative of f = xyz at the point (x, y, z) = (1, 2, 3)
in the direction of the vector ⃗v = (3, 4, 12).
a. 47
13
b. 30
c. 30
13
d. 54
e. 54
13
Correct Choice
⃗ f = (yz, xz, xy)
∇
⃗f
D v̂ f = v̂ ⋅ ∇
(1,2,3)
v̂ = ⃗v = 1 (3, 4, 12)
13
|v⃗|
= 1 (3, 4, 12) ⋅ (6, 3, 2) = 54
13
13
⃗f
∇
(1,2,3)
= (6, 3, 2)
2. The point (2, 1) is a critical point of the function f = (x 3 − 3x 2 )(y 3 − 3y).
Use the 2 nd Derivative Test to classify (2, 1).
a. local minimum
b. local maximum
Correct Choice
c. saddle point
d. inflection point
e. The test fails.
f x = (3x 2 − 6x)(y 3 − 3y)
f xx = (6x − 6)(y 3 − 3y)
f y = (x 3 − 3x 2 )(3y 2 − 3)
f yy = (x 3 − 3x 2 )6y
f xy = (3x 2 − 6x)(3y 2 − 3)
f xx (2, 1) = (6)(−2) = −12 < 0
f yy (2, 1) = (−4)6 = −24 < 0
D = f xx f yy − (f xy ) 2 = 288 > 0
local maximum
f xy (2, 1) = 0
1
2
1
∫ 1 ∫ 1/y ye xy dx dy
3. Compute
a. e 2 − 2e
Correct Choice
b. e 2 − e
c. e 2 − 2e − 1
d. e 2 − e − 1
e. e 2 − 2
2
1
2
∫ 1 ∫ 1/y ye xy dx dy = ∫ 1
e xy
1
dy =
x=1/y
2
∫ 1 [e y − e] dy =
2
e y − ey
y=1
= e 2 − 2e
4. The region of integration of the integral in the previous problem is:
y
2
y
a.
2
b.
1
1
0
0
0
y
1
2
x
2
y
c.
0
1
0
1
2
x
2
d.
1
1
0
0
0
y
1
2
x
2
x
2
e.
CorrectChoice
1
0
0
1
2
x
2
5. Find the mass of the quarter circle x 2 + y 2 ≤ 9 for x ≥ 0 and y ≥ 0
if the density is ρ =
x2 + y2 .
a. 9 π
4
9π
Correct Choice
2
27 π
4
81 π
4
243 π
2
b.
c.
d.
e.
M=
∫ ∫ ρ dA =
π/2
∫0
3
∫ 0 r r dr dθ = π2 r3
3
3
0
= 9π
2
6. Find the center of mass of the quarter circle x 2 + y 2 ≤ 9 for x ≥ 0 and y ≥ 0
if the density is ρ =
a.
b.
c.
d.
4π , 4π
9
9
2π , 2π
9
9
π, π
9 9
9 , 9
2π 2π
x2 + y2 .
Correct Choice
e. (9, 9)
My =
∫ ∫ xρ dA =
π/2
∫0
4
∫ 0 r cos θ r r dr dθ = r4
3
3
0
3
sin θ
π/2
0
= 81
4
π/2
r
− cos θ
= 81
4 0
4
0
You can skip one of these since x̄ = ȳ by symmetry.
Mx =
π/2
3
∫ ∫ yρ dA = ∫ 0 ∫ 0 r sin θ r r dr dθ =
4
My
= 81 2 = 9
M
4 9π
2π
ȳ = M x = 81 2 = 9
M
4 9π
2π
x̄ =
3
7. Which of the following is the polar graph of the polar curve r = 1 − sin θ ?
a.
b.
c.
d.
⇑
CorrectChoice
⇑
e.
(e) is the rectangular graph.
(c) is the polar graph.
8. The function f(x, y) = x sin 2y − y cos 2x satisfies which differential equation?
⃗ ⋅∇
⃗ f = −4f
a. ∇
Correct Choice
⃗ ⋅∇
⃗ f = −2f
b. ∇
⃗ ⋅∇
⃗f = 0
c. ∇
⃗ ⋅∇
⃗ f = 2f
d. ∇
⃗ ⋅∇
⃗ f = 4f
e. ∇
⃗ f = (sin 2y + 2y sin 2x, 2x cos 2y − cos 2x)
∇
⃗ ⋅∇
⃗ f = 4y cos 2x − 4x sin 2y = −4(x sin 2y − y cos 2x) = −4f
∇
4
9. Find the volume above the cone z = 2 x 2 + y 2
a. 40 π
below the paraboloid z = 8 − x 2 − y 2 .
Correct Choice
3
b. 16π
c. 56 π
3
d. 80 π
3
e. 32π
In cylindrical coordinates, the equations are z = 2r and z = 8 − r 2 .
They intersect when 2r = 8 − r 2 or r 2 + 2r − 8 = 0 or (r − 2)(r + 4) = 0 or r = 2.
V=
∫∫∫ 1 dV =
2π
2
8−r 2
∫ 0 ∫ 0 ∫ 2r
r dz dr dθ =
2π
2
∫0 ∫0
rz
8−r 2
2r
dr dθ =
2π
2
∫ 0 ∫ 0 r(8 − r 2 − 2r) dr dθ
You could have started from the last integral as polar coordinates.
4
3
V = 2π 4r 2 − r − 2r
4
3
2
0
= 2π 16 − 4 − 16
3
= 40 π
3
10. Find the average value of the function f(x, y, z) = z 2
on the hemisphere x 2 + y 2 + z 2 ≤ 4 for z ≥ 0.
HINT: f ave = 1
V
∫∫∫ f dV
a. 0
b. 4
5
c. 8
5
d. π
4
π
e.
2
Correct Choice
The volume is V = 1 ⋅ 4 π2 3 = 16 π
2 3
3
∫∫∫ f dV =
f ave
2π
π/2
2
∫0 ∫0 ∫0
(ρ cos ϕ) 2 ρ 2 sin ϕ dρ dϕ dθ = 2π
− cos 3 ϕ
3
π/2
0
ρ5
5
2
0
= 2π 1
3
32
5
= 64 π
15
= 3 ⋅ 64π = 4
5
16π
15
5
11. Compute
∫ F⃗ ⋅ ds⃗
counterclockwise around the circle x 2 + y 2 = 4 with z = 4
⃗ = (−yz, xz, z 2 ).
for the vector field F
a. 2π
b. 4π
c. 8π
d. 16π
Correct Choice
e. 32π
⃗r(θ) = (2 cos θ, 2 sin θ, 4)
⃗ (r⃗(θ)) = (−8 sin θ, 8 cos θ, 16)
F
⃗v = (−2 sin θ, 2 cos θ, 0)
∫ F⃗ ⋅ ds⃗ = ∫ F⃗ ⋅ ⃗v dθ = ∫ 0 (16 sin 2 θ + 16 cos 2 θ) dθ = ∫ 0
2π
2π
16 dθ = 32π
1 dS on the parametric surface R
⃗ (u, v) = (u 2 + v 2 , u 2 − v 2 , 2uv)
x
for 1 ≤ u ≤ 3 and 1 ≤ v ≤ 4.
12. Compute
∫∫
a. 6 2
b. 12 2
c. 24 2
Correct Choice
d. 64 2
e. 272 2
î
̂
k̂
⃗e u = (2u 2u 2v)
⃗e v = (2v −2v 2u)
⃗ =
N
=
⃗ = (4u 2 + 4v 2 , 4v 2 − 4u 2 , −8uv)
N
(4u 2 + 4v 2 ) 2 + (4v 2 − 4u 2 ) 2 + (−8uv) 2 =
32u 4 + 64u 2 v 2 + 32v 4
32(u 4 + 2u 2 v 2 + v 4 ) = 4 2 (u 2 + v 2 )
1 =
1
x
u2 + v2
4 3
∫∫ 1x dS = ∫ 1 ∫ 1 u 2 1+ v 2 4 2 (u 2 + v 2 ) du dv = 4 2
4
3
∫ 1 ∫ 1 1 du dv = 24
2
6
Work Out: (15 points each. Part credit possible. Show all work.)
13.
A plate has the shape of the region
y
between the curves
y = 1 + 1 ex,
y = 2 + 1 ex,
2
2
1
x
y = 5 − 1 ex
y = 3− e ,
2
2
where x and y are measured in centimeters.
4
2
If the mass density is ρ = ye x gm/cm 2 ,
-1
find the total mass of the plate.
1
x
HINT: Use the curvilinear coordinates
and
v = y + 1 ex
u = y − 1 ex
2
2
y = 1u+ 1v
v − u = ex
2
2
The boundaries are:
u = 1,
u = 2,
u + v = 2y
−1 , 1
v−u 2
⃗e u =
=
1 ,1
v−u 2
⃗e v =
M=
∂(x, y)
∂(u, v)
=
5
2
∫∫ ρ dA = ∫∫ ρ J du dv = ∫ 3 ∫ 1
= 1
2
= 1
2
v = 3,
⃗ (u, v) = ln(v − u), 1 u + 1 v
R
2
2
v=5
∂(x, y)
=
∂(u, v)
−1
v−u
1
v−u
1
2
1
2
−1
1
−
= −1
2(v − u)
2(v − u)
v−u
−1
= 1
v−u
v−u
ρ = ye x = 1 u + 1 v (v − u)
2
2
The mass is
J=
x = ln(v − u)
since v > u always.
1 u + 1 v (v − u) 1
du dv = 1
2
2
2
(v − u)
u 2 + vu 2 dv = 1 5 3 + v dv = 1
∫3 2
2 ∫3 2
2
u=1
15 + 25 − 1 9 + 9 = 11
2
2
2 2
2
2
5
3 v + v2
2
2
5
2
∫ 3 ∫ 1 (u + v) du dv
5
v=3
7
14. Find the point in the first octant on the graph of z =
8
x2y
closest to the origin.
Minimize the square of the distance f = D 2 = x 2 + y 2 + z 2 subject to the constraint z =
f = x 2 + y 2 + 64
x4y2
f x = 2x − 256
=0
x5y2
So y =
4
64
x =
So x = 2
15. Compute
f y = 2y − 128
= 0 or x 6 y 2 = 128
x4y3
8
x
y=
∫∫ ∇⃗ × F⃗ ⋅ dS⃗
2
x 4 y 4 = 64
x 6 82 = 128 or x 4 = 16
x
and
8
=
2
8 .
x2y
z=
8
=
22 2
2
over the paraboloid z = x 2 + y 2 with z ≤ 4 oriented down and out,
⃗ = (−yz, xz, z 2 ).
for the vector field F
⃗ (r, θ) = (r cos θ, r sin θ, r 2 ).
HINT: The paraboloid may be parametrized by R
k̂
̂
î
⃗e r = (cos θ
sin θ
2r)
⃗e θ = (−r sin θ r cos θ 0)
Reverse
⃗ =
⃗ ×F
∇
⃗ = (−2r 2 cos θ, −2r 2 sin θ, r) This is up and in.
N
⃗ = (2r 2 cos θ, 2r 2 sin θ, −r)
N
î
∂
∂x
̂
∂
∂y
k̂
∂
∂z
−yz
xz
z2
⃗ R
⃗ (r, θ)
⃗ ×F
∇
= î(−x) − ̂ (y) + k̂ (z + z) = (−x, −y, 2z)
= (−r cos θ, −r sin θ, , 2r 2 )
⃗⋅N
⃗ = −2r 3 cos 2 θ − 2r 3 sin 2 θ − 2r 3 = −4r 3
⃗ ×F
∇
∫∫ ∇⃗ × F⃗ ⋅ dS⃗ =
∫∫ ∇⃗ × F⃗ ⋅ N⃗ dr dθ =
2π
2
∫0 ∫0
−4r 3 dr dθ = −2π r 4
2
0
= −32π
8