Name
Sec
MATH 251H
Exam 2
Spring 2011
Section 200
Solutions
1-12
/60
14
/15
13
/15
15
/15
Total
/105
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
2
1
2
3
4
y2
a.
b.
c.
d.
e.
2
y
∫0 ∫0
y
∫ 0 ∫ 0 xy dx dy.
1. Compute
Correct Choice
xy dx dy =
2
∫0
x2y
2
y
dy = 1
2
x=0
4
1 y
3
y
dy
=
∫0
2 4
2
2
=2
y=0
2. Find the area of one loop of the rose r = sin(3θ).
a.
b.
c.
d.
e.
π
12
π
6
π
4
π
3
π
2
Correct Choice
sin(3θ) = 0 at θ = 0 and 3θ = π or θ = π/3
A=
π/3
sin(3θ)
∫0 ∫0
= 1
2
3. Compute
π/3
∫0
π/3
sin(3θ)
π/3
dθ = 1 ∫ sin 2 (3θ) dθ
2 0
r=0
sin(6θ) π/3
1 − cos(6θ)
dθ = 1 θ −
= π
2
4
6
12
θ=0
r dr dθ =
r2
2
∫0
∫∫∫ x 2 + y 2 dV
over the region between the cones z =
x2 + y2
and z = 4 − x 2 + y 2 .
a. 8π
3
16π
b.
3
c. 32π
3
d. 16π
5
32π
e.
Correct Choice
5
In cylindrical coordinates, the cones are z = r and z = 4 − r which intersect at r = 2.
2π
2
4−r
∫0 ∫0 ∫r
r 2 r dz dr dθ = 2π ∫
2
0
4−r
r3z
2
5
dr = 2π ∫ r 3 (4 − 2r) dr = 2π r 4 − 2 r
0
5
z=r
2
r=0
= 2π 16 − 64
5
= 32π
5
1
4. Find the mass of the hemisphere x 2 + y 2 + z 2 ≤ 4 with y ≥ 0 if the density is δ = y.
a. π
2
π
2π
4π
8
b.
c.
d.
e.
M=
Correct Choice
π
π
2
π
2
π
∫∫∫ δ dV = ∫ 0 ∫ 0 ∫ 0 ρ sin ϕ sin θρ 2 sin ϕ dρ dθ dϕ = ∫ 0 ρ 3 dρ ∫ 0 sin θ dθ ∫ 0 sin 2 ϕ dϕ
ρ4
4
=
2
π
− cos θ
1 ϕ − sin 2ϕ
2
2
0
0
π
0
= 4(2) π
2
= 4π
5. Find the center of mass of the hemisphere x 2 + y 2 + z 2 ≤ 4 with y ≥ 0 if the density is δ = y.
0, 64π , 0
15
16
0,
,0
15
2
0, π , 0
12
0, 15 , 0
16
0, 122 , 0
π
a.
b.
c.
d.
e.
M xz =
Correct Choice
π
π
2
π
2
π
∫∫∫ yδ dV = ∫ 0 ∫ 0 ∫ 0 ρ 2 sin 2 ϕ sin 2 θρ 2 sin ϕ dρ dθ dϕ = ∫ 0 ρ 4 dρ ∫ 0 sin 2 θ dθ ∫ 0 sin 3 ϕ dϕ
ρ5 2
5 0
16π
=
1−
5
ȳ = M xz = 64π
M
15
=
6. Compute
π π
5
1 θ − sin 2θ
(1 − cos 2 ϕ) sin ϕ dϕ = 2
∫
5
2
2
0 0
1 + 1 − 1 = 64π
3
3
15
1 = 16
x̄ = z̄ = 0 by symmetry.
4π
15
∮ F⃗ ⋅ ds⃗
π
2
− cos ϕ +
⃗ = (−16x 2 y, 9xy 2 ) counterclockwise around the ellipse
for F
HINTS: The ellipse may be parametrized by ⃗r(θ) = (3 cos θ, 4 sin θ).
Since sin(2θ) = 2 sin θ cos θ, we have 4 sin 2 θ cos 2 θ = sin 2 (2θ).
a.
b.
c.
d.
e.
−864π
−288π
144π
288π
864π
cos 3 ϕ
3
π
0
2
x 2 + y = 1.
9
16
Correct Choice
⃗ (r⃗(θ)) = (−16 ⋅ 9 cos 2 θ ⋅ 4 sin θ, 9 ⋅ 3 cos θ ⋅ 16 sin 2 θ)
⃗v = (−3 sin θ, 4 cos θ)
F
⃗ ⋅ ⃗v = 64 ⋅ 27 cos 2 θ sin 2 θ + 27 ⋅ 64 cos 2 θ sin 2 θ = 27 ⋅ 64 ⋅ 2 cos 2 θ sin 2 θ = 27 ⋅ 32 sin 2 (2θ)
F
1 − cos(4θ)
= 27 ⋅ 16(1 − cos(4θ))
2
2π
∮ F⃗ ⋅ ds⃗ = ∮ F⃗ ⋅ ⃗v dθ = ∫ 0 27 ⋅ 16(1 − cos(4θ)) dθ = 27 ⋅ 16 θ − sin44θ
= 27 ⋅ 32
2π
0
= 864π
2
3
3
π, π
is a critical point of the function f(x, y) = sin(x) cos(y) −
x+
y
3 6
4
4
Use the Second Derivative Test to classify this critical point.
7. The point
a. Local Maximum
Correct Choice
b. Local Minimum
c. Inflection Point
d. Saddle Point
e. Test Fails
3
4
3
f y = − sin(x) sin(y) +
4
f x = cos(x) cos(y) −
f xx = − sin(x) cos(y)
f yy = − sin(x) cos(y)
f xy = − cos(x) sin(y)
D = f xx f yy − f 2xy
fx π , π
3 6
= cos π
3
fy π , π
3 6
f xx π , π
3 6
π
f yy
,π
3 6
f xy π , π
3 6
D π, π
3 6
= − sin
= − sin
= − sin
= − cos
= −3
4
cos π
6
3
=0
4
π sin π + 3 = 0
3
6
4
π cos π = − 3 < 0
3
6
4
π cos π = − 3
3
6
4
π sin π = − 1
4
3
6
2
−3 − −1
= 1 >0
4
4
2
⃗ (x, y) =
8. Which of the following is the plot of the vector field F
−
1
x + y2
2
Local Maximum
|x|, |y|
?
a.
b.
c.
d.
e.
(d) Correct Choice
All arrows must be up and right. So (c) or (d). On the x-axis, y = 0 and so the arrows are horizontal
there. So (d).
3
⃗ (r, θ) = (r cos θ, r sin θ, 2r cos θ, 2r sin θ) for
9. In ℝ 4 , consider the parameric 2-surface (x, y, z, w) = R
0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π. Compute
∫∫(xz dy dz − y 2 dz dw).
⃗
R
a.
b.
c.
d.
e.
32π
16π
−16π
−32π
−64π
Correct Choice
sin θ
dy dz =
∂(y, z)
dr dθ =
∂(r, θ)
dz dw =
∂(z, w)
dr dθ =
∂(r, θ)
xz = 2r 2 cos 2 θ
2 cos θ
r cos θ −2r sin θ
2 cos θ
2 sin θ
−2r sin θ 2r cos θ
dr dθ = (−2r sin 2 θ − 2r cos 2 θ) dr dθ = −2r dr dθ
dr dθ = (4r cos 2 θ + 4r sin 2 θ) dr dθ = 4r dr dθ
− y 2 = −r 2 sin 2 θ
2π
2
∫∫ R⃗ (xz dy dz + y 2 dz dw) = ∫ ∫(−2r 2 cos 2 θ2r dr dθ − r 2 sin 2 θ4r dr dθ) = −4 ∫ 0 ∫ 0 (r 3 ) dr dθ = −8π
⃗ (u, t) =
10. Find the equation of the plane tangent to the parameric surface R
⃗ (2, 0) where u = 2 and t = 0.
point P = R
⃗ at u = 2 and t = 0.
Hint: Evaluate the normal N
ue t , ue −t , 2 u
r4
4
2
0
= −32π
at the
a. x + y − 2 z = −4 2
b. x + y − 2 z = 0
Correct Choice
c. x + y − 2 z = 16 2
d. 2 x − 2 y + 2z = −4 2
e. 2 x − 2 y + 2z = 0
î
̂
k̂
⃗ = ⃗e u × ⃗e t = î 2 ue −t − ̂ − 2 ue t + k̂ (−u − u) =
⃗e u = (e t
e −t
2) N
⃗e t = (ue t −ue −t
0)
⃗ (2, 0) = 2, 2, 2 2
⃗ =
P=R
N
2 ue −t , 2 ue t , −2u = 2 2 , 2 2 , −4
⃗⋅X = N
⃗⋅P
N
2 2 x + 2 2 y − 4z = 2 2 (2) + 2 2 (2) − 4 2 2 = 0
2 ue −t , 2 ue t , −2u
4
⃗ = (xy tan z, yz cos x, xz sin y), then ∇
⃗ =
⃗ ⋅∇
⃗ ×F
11. If F
a.
b.
c.
d.
e.
−2z cos y − 2x(tan 2 z + 1)
−2z cos y + 2x(tan 2 z + 1)
2z cos y − 2x(tan 2 z + 1)
2z cos y + 2x(tan 2 z + 1)
0
Correct Choice
⃗ = 0 for any twice differentiable vector field.
⃗ ⋅∇
⃗ ×F
∇
⃗ = (2xz − 3y, 8yz − 3x, x 2 + 4y 2 + 2z) for which f(0, 0, 0) = 0.
12. Let f be the scalar potential for F
Then f(1, 1, 1) =
a.
b.
c.
d.
e.
1
2
3
4
5
⃗
⃗f = F
∇
(1)
Correct Choice
or
(1) ∂ x f = 2xz − 3y
 f = x 2 z − 3xy + g(y, z)
(2) ∂ y f = 8yz − 3x
(3) ∂ z f = x 2 + 4y 2 + 2z
 (4) ∂ y f = −3x + ∂ y g
∂ y g = 8yz  g = 4y 2 z + h(z)  f = x 2 z − 3xy + 4y 2 z + h(z)
dh(z)
 (5) ∂ z f = x 2 + 4y 2 +
dz
dh(z)
(3) and (5) 
= 2z  h = z 2 + C  f = x 2 z − 3xy + 4y 2 z + z 2 + C
dz
To have f(0, 0, 0) = 0  C = 0. So f(1, 1, 1) = 1 − 3 + 4 + 1 = 3.
(2) and (4)

5
Work Out: (Points indicated. Part credit possible. Show all work.)
13. (15 points) The plane x + 2y + 4z = 8 intersects the 1st octant (x > 0, y > 0, z > 0) in a triangle.
Find the point on this triangle at which the function f = xy 2 z 3 is a maximum.
Method of Eliminating a Variable:
x = 8 − 2y − 4z
f = (8 − 2y − 4z)y 2 z 3 = 8y 2 z 3 − 2y 3 z 3 − 4y 2 z 4
3
2 3
f y = 16yz − 6y z − 8yz 4 = 0

2yz 3 (8 − 3y − 4z) = 0
f z = 24y 2 z 2 − 6y 3 z 2 − 16y 2 z 3 = 0

2y 2 z 2 (12 − 3y − 8z) = 0
Since f = xy 2 z 3 is positive in the 1st octant, we know the solution cannot have y = 0 or z = 0.
So we solve 8 − 3y − 4z = 0 and 12 − 3y − 8z = 0.
Solving each for 3y and equating gives 3y = 8 − 4z = 12 − 8z, or 4z = 4. So z = 1.
Then 3y = 4. So y = 4 , and x = 8 − 2 4 − 4(1) = 4 .
3
3
3
4
4
, ,1 .
The point is
3 3
Method of Lagrange Multipliers:
⃗ f = (y 2 z 3 , 2xyz 3 , 3xy 2 z 2 )
⃗ g = (1, 2, 4)
f = xy 2 z 3
∇
g = x + 2y + 4z
∇
Lagrange equations: y 2 z 3 = λ, 2xyz 3 = 2λ, 3xy 2 z 2 = 4λ
Substitute the first eq into the other two: 2xyz 3 = 2y 2 z 3 , 3xy 2 z 2 = 4y 2 z 3 .
Since f = xy 2 z 3 is positive in the 1st octant, we know the solution cannot have x = 0 or y = 0 or
z = 0. So we cancel: x = y, 3x = 4z, So y = x and z = 3 x.
4
Substitute into the constraint: x + 2y + 4z = 8 x + 2x + 3x = 8 6x = 8 x = 4 y = 4 z = 1.
3
3
4 , 4 ,1 .
The point is
3 3
14. (15 points) Compute
z=
x +y
2
2
⃗ dV for F
⃗ = (xy 2 , yz 2 , zx 2 ) over the solid above the cone
⃗ ⋅F
∫∫∫ ∇
below the sphere x 2 + y 2 + z 2 = 4.
⃗ = y 2 + z 2 + x 2 = ρ 2 in spherical coordinates and dV = ρ 2 sin ϕ dρ dθ dϕ
⃗ ⋅F
∇
The cone is ρ cos ϕ = ρ 2 sin 2 ϕ cos 2 θ + ρ 2 sin 2 ϕ sin 2 θ = ρ sin ϕ. So tan ϕ = 1 or ϕ = π/4.
⃗ dV =
⃗ ⋅F
∫∫∫ ∇
π/4
2π
2
∫0 ∫0 ∫0
ρ 2 ρ 2 sin ϕ dρ dθ dϕ =
ρ5
5
2
0
(2π) − cos ϕ
π/4
0
= 64π
5
1− 1
2
6
∫∫ E ∇⃗ × F⃗ ⋅ k̂ dx dy
15. (15 points) Compute
⃗ = (−16x 2 y, 9xy 2 , 0) over the interior of the
for F
2
x 2 + y = 1.
9
16
⃗ ⋅ k̂ in rectangular coordinates.
⃗ ×F
HINTS: First compute ∇
Then compute the integral in elliptic coordinates x = 3u cos θ, y = 4u sin θ.
ellipse
⃗ =
⃗ ×F
∇
î
̂
∂x
∂y
k̂
∂ z = î(0) − ̂ (0) + k̂ (9y 2 − −16x 2 )
−16x 2 y 9xy 2
0
⃗ ⋅ k̂ = 9y 2 + 16x 2 = 9(4u sin θ) 2 + 16(3u cos θ) 2 = 9 ⋅ 16u 2 = 144u 2
⃗ ×F
∇
∂(x, y)
=
∂(u, θ)
∂x
∂u
∂x
∂θ
∫∫ ∇⃗ × F⃗ ⋅ k̂ dx dy =
E
∂y
∂u
∂y
∂θ
2π
∫0
=
3 cos θ
4 sin θ
−3u sin θ 4u cos θ
= 12u
4
∫ 0 144u 2 12u du dθ = 2π144 ⋅ 12 u4
1
dx dy = J du dθ = 12u du dθ
1
0
= 864π
7