Name
Sec
MATH 251
Final
Sections 511
Spring 2010
Solutions
1-13
/65
15
/10
14
/21
16
/10
Total
/106
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
1.
Find the angle between the vectors ⃗
u = (2, 2, 1) and ⃗v = (1, 2, 2).
a.
arccos(8/9)
b.
arccos(8/3)
c.
arccos
d.
arccos 3/ 8
e.
arccos(3/8)
Correct Choice
8 /9
⃗
u ⋅ ⃗v = 2 + 4 + 2 = 8
2.
⃗| =
|u
4+4+1 = 3
|v⃗| =
1+4+4 = 3
u ⋅ ⃗v = 8
cos θ = ⃗
9
⃗ ||v⃗|
|u
At the point (x, y, z) where the line ⃗r(t) = (1 − t, t, 2 − 2t) intersects the plane x − 2y + 3z = 16,
we have x + y + z =
a.
−2
b.
2
c.
3
d.
5
e.
16
Correct Choice
Plug the line into the plane and solve for t:
(1 − t) − 2(t) + 3(2 − 2t) = 16
− 9t + 7 = 16
t = −1
Plug back into the line:
(x, y, z) = (1 − t, t, 2 − 2t) = (2, −1, 4)
3.
So
x+y+z = 5
If a jet flies around the world from West to East, directly above the equator,
in what direction does its unit binormal B̂ point?
a.
Down (toward the center of the earth)
b.
Up (away from the center of the earth)
c.
North
d.
South
e.
West
Correct Choice
T̂ points East, N̂ points Down and by the right hand rule B̂ points North.
1
4.
Find the z-intercept of the plane tangent to the surface
a.
6
b.
1
6
c.
5
d.
−5
e.
−6
Correct Choice
xy
⃗ =
⃗ F = y , x , − xy
∇
N
z2
z
z z
1x+ 1y− 1z = 12+
⃗⋅X = N
⃗⋅P
N
2
3
6
2
z = 3x + 2y − 6
z-intercept is c = −6.
F=
5.
xy
= 1 at the point (2, 3, 6).
z
⃗F
∇
1 , 1 ,− 1
2 3 6
13− 16 = 1
1z = 1x+ 1y−1
3
6
6
2
3
(2,3,6)
=
The temperature in an ideal gas is given by T = κ P
ρ where κ is a constant,
P is the pressure and ρ is the density. At a certain point Q = (3, 2, 1), we have
P(Q) = 8
⃗ P(Q) = (4, −2, −4)
∇
ρ(Q) = 2
⃗ ρ(Q) = (−1, 4, 2)
∇
⃗ T(Q) =
So at the point Q, the temperature is T(Q) = 4κ and its gradient is ∇
a.
κ(−8. 5, 6, 9)
b.
κ(4, −9, −6)
c.
κ(3, 2, −2)
d.
κ 1 ,2
2
κ − 1 ,2
2
e.
Correct Choice
By chain rule: (Think about each component separately.)
κP ⃗
κ
κ8
⃗ T = ∂T ∇
⃗ P + ∂T ∇
⃗ρ = κ ∇
⃗
∇
ρ P − ρ 2 ∇ρ = 2 (4, −2, −4) − 2 2 (−1, 4, 2)
∂P
∂ρ
= κ(2, −1, −2) + κ(2, −8, −4) = κ(4, −9, −6)
2
6.
If the temperature in a room is T = xyz 2 , find the rate of change of the temperature
as seen by a fly who is located at (3, 2, 1) and has velocity (1, 2, 3).
a.
32
b.
36
c.
44
d.
48
e.
52
Correct Choice
dT = ⃗v ⋅ ∇
⃗T
dt
7.
Find the volume below z = xy above the region between the curves y = 3x and y = x 2 .
81
2
81
4
81
8
243
2
243
8
a.
b.
c.
d.
e.
3x = x 2
V=
8.
= (1, 2, 3) ⋅ (yz 2 , xz 2 , 2xyz)| (3,2,1) = (1, 2, 3) ⋅ (2, 3, 12) = 44
(3,2,1)
Correct Choice

3
x = 0, 3
3x
∫0 ∫x
xy dy dx =
2
Compute
3
∫0
xy 2
2
∫∫ e −x −y dx dy
2
3x
dx =
y=x 2
3
∫0
x9x 2 − xx 4
2
2
dx =
9x 4 − x 6
8
12
3
x=0
6
= 3
4
1 − 1
3
2
= 243
8
over the disk enclosed in the circle x 2 + y 2 = 4.
2
C
a.
π (1 − e −4 )
2
b.
π(1 − e −4 )
c.
π e −4
2
d.
πe −4
e.
2πe −4
Correct Choice
2π
2
∫∫ e −x −y dx dy = ∫ 0 ∫ 0 e −r r dr dθ = 2π
2
2
2
− 1 e −r
2
2
2
0
= π(1 − e −4 )
3
9.
Find the mass of 2 loops of the helical ramp
parametrized by
⃗ (r, θ) = (r cos θ, r sin θ, 4θ) for r ≤ 3
R
if the density is ρ =
x2 + y2 .
a. 40π
b. 120π
c. 200π
d.
500 π
3
e.
244 π
3
Correct Choice
̂
î
⃗e r = ( cos θ,
sin θ,
⃗e θ = (−r sin θ, r cos θ
A=
∫∫ ρ dS = ∫∫ ρ
⃗ = ⃗e r × ⃗e θ = î(4 sin θ) − ̂ (4 cos θ) + k̂ (r cos 2 θ + r sin 2 θ)
N
k̂
= (4 sin θ, −4 cos θ, r)
0)
⃗ =
N
4)
⃗ dr dθ =
N
4π
3
∫0 ∫0 r
16 sin 2 θ + 16 cos 2 θ + 1 =
16 + r 2 dr dθ = 4π (16 + r 2 ) 3/2
3
3
0
16 + r 2
ρ=r
= 4π (125 − 64) = 244 π
3
3
⃗ = (y, −x, 2) through the helical ramp of problem 9 oriented up.
10. Find the flux of F
a.
4π
b.
8π
3
c.
216π
d.
108π
e.
1024 π
3
Correct Choice
⃗ = (y, −x, 1) = (r sin θ, −r cos θ, 2)
F
⃗⋅N
⃗ = 4r sin 2 θ + 4r cos 2 θ + 2r = 6r
F
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 6r dr dθ = 4π[3r 2 ] 0 = 108π
4π
11. Compute
(3,2)
∫ (2,1) F⃗ ⋅ ds⃗
3
3
⃗ = (2xy, x 2 ) along the curve ⃗r(t) = ((2 + t 2 )e sin πt , (1 + t 2 )e sin 2πt ).
for F
HINT: Find a scalar potential.
a.
12
b.
14
c.
22
d.
2
e.
Correct Choice
15 − 4 2
⃗ = (2xy, x 2 ) = ∇
⃗ f for f = x 2 y
F
(3,2)
(3,2)
∫ (2,1) F⃗ ⋅ ds⃗ = ∫ (2,1) ∇⃗ f ⋅ ds⃗ = f(3, 2) − f(2, 1) = 3 2 2 − 2 2 1 = 14
4
12.
∮(2x sin y − 5y) dx + (x 2 cos y − 4x) dy
Compute
y
counterclockwise around the cross shown.
3
2
HINT: Use Green’s Theorem.
a. −45
1
b. −10
c. 5
0
Correct Choice
0
1
2
3
x
d. 10
e. 45
Green’s Theorem says
∮ P dx + Q dy = ∫∫
∂R
R
∂Q
− ∂P
∂x
∂y
dx dy
Here P = 2x sin y − 5y and Q = x 2 cos y − 4x.
So ∂ x Q − ∂ y P = (2x cos y − 4) − (2x cos y − 5) = 1.
So
∫∫(∂ x Q − ∂ y P) dx dy = ∫∫ 1 dx dy =
R
13.
area = 5
R
∫∫ ∇⃗ × F⃗ ⋅ dS⃗
Compute
⃗ = (−yz, xz, xyz)
for F
S
over the quartic surface z = (x 2 + y 2 ) 2 for z ≤ 16
oriented down and out. The surface may be parametrized by
⃗ (r, θ) = (r cos θ, r sin θ, r 4 )
R
HINT: Use Stokes’ Theorem.
a. −128π
Correct Choice
b. −64π
c. −32π
d. 32π
e. 64π
Stokes’ Theorem says
∫∫ ∇⃗ × F⃗ ⋅ dS⃗ = ∮ F⃗ ⋅ ds⃗.
∂R
S
The boundary is the circle x 2 + y 2 = 4 with z = 16 which may be parametrized by
⃗r(θ) = (2 cos θ, 2 sin θ, 16). The velocity is ⃗v = (−2 sin θ, 2 cos θ, 0).
Since the surface is oriented down and out, the circle must be traversed clockwise.
So reverse the velocity: ⃗v = (2 sin θ, −2 cos θ, 0).
⃗ = (−yz, xz, xyz)
⃗ (r⃗(θ)) = (−32 sin θ, 32 sin θ, 64 cos θ sin θ)
F
F
⃗ ⋅ ⃗v = −64 sin 2 θ − 64 cos 2 θ = −64
F
∮ F⃗ ⋅ ds⃗ = ∫ 0
2π
∂R
⃗ ⋅ ⃗v dθ = − 2π 64 dθ = −128π
F
∫
0
5
Work Out: (Points indicated. Part credit possible. Show all work.)
14.
(21 points) Verify Gauss’ Theorem
∫∫∫ ∇⃗ ⋅ F⃗ dV = ∫∫ F⃗ ⋅ dS⃗
∂V
V
⃗ = (4xz 3 , 4yz 3 , z 4 ) and the solid V
for the vector field F
above the cone C given by z =
x2 + y2
or parametrized by R(r, θ) = (r cos θ, r sin θ, r),
below the disk D given by x 2 + y 2 ≤ 9 and z = 3.
Be sure to check and explain the orientations.
Use the following steps:
a.
(4 pts) Compute the volume integral by successively finding:
⃗ , dV,
⃗ dV
⃗ ⋅F
⃗ ⋅F
∇
∇
∫∫∫
V
⃗ = 4z 3 + 4z 3 + 4z 3 = 12z 3
⃗ ⋅F
∇
dV = r dr dθ dz
∫∫∫ ∇⃗ ⋅ F⃗ dV = ∫ 0 ∫ 0 ∫ r 12z 3 r dz dr dθ = 2π ∫ 0 [3z 4 ] z=r r dr = 2π ∫ 0 (3 5 − 3r 4 ) r dr
2π
3
3
3
3
3
V
3
2
6
= 2π 3 5 r − 3 r
2
6
b.
r=0
7
6
= 2π 3 − 3
2
2
= π3 6 (3 − 1) = 1458π
(8 pts) Compute the surface integral over the disk by parametrizing the disk and
successively finding:
⃗
⃗ (r, θ), ⃗e r , ⃗e θ , N
⃗, F
⃗ R
⃗ (r, θ) ,
⃗ ⋅ dS
R
F
∫∫
D
⃗ (r, θ) = (r cos θ, r sin θ, 3)
R
î
̂
k̂
⃗e r = (cos θ,
sin θ, 0)
⃗e θ = (−r sin θ, r cos θ, 0)
⃗ = ⃗e r × ⃗e θ = î(0) − ̂ (0) + k̂ (r cos 2 θ + r sin 2 θ) = (0, 0, r)
N
⃗ to point up which it does.
We need N
⃗ = (4xz 3 , 4yz 3 , z 4 )
F
⃗ R
⃗ (r, θ)
F
= (4r cos θ 27, 4r sin θ 27, 81)
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 81r dr dθ = 2π
2π
D
D
3
2
81 r
2
3
0
= 729π
6
⃗ = (4xz 3 , 4yz 3 , z 4 ) and C is the cone parametrized by
Recall:
F
⃗
R(r, θ) = (r cos θ, r sin θ, r).
c.
(7 pts) Compute the surface integral over the cone C by successively finding:
⃗
⃗, F
⃗ R
⃗ (r, θ) ,
⃗ ⋅ dS
⃗e r , ⃗e θ , N
F
∫∫
C
⃗ (r, θ) = (r cos θ, r sin θ, r)
R
î
k̂
̂
⃗e r = (cos θ,
sin θ, 1)
⃗e θ = (−r sin θ, r cos θ, 0)
⃗ = ⃗e r × ⃗e θ = î(−r cos θ) − ̂ (r sin θ) + k̂ (r cos 2 θ + r sin 2 θ) = (−r cos θ, −r sin θ, r)
N
⃗ to point down (out of the volume). Reverse N
⃗ = (r cos θ, r sin θ, −r)
We need N
⃗ = (4xz 3 , 4yz 3 , z 4 )
F
⃗ R
⃗ (r, θ)
F
= (4r cos θ r 3 , 4r sin θ r 3 , r 4 )
⃗⋅N
⃗ = 4r 5 cos 2 θ + 4r 5 sin 2 θ − r 5 = 3r 5
F
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ N⃗ dr dθ = ∫ 0 ∫ 0 3r 5 dr dθ = 2π
2π
P
d.
3
P
(2 pts) Combine
∫∫ F⃗ ⋅ dS⃗
and
∫∫ F⃗ ⋅ dS⃗
C
D
6
3r
6
to get
3
0
6
= 2π 3
2
= 729π
∫∫ F⃗ ⋅ dS⃗
∂V
∫∫ F⃗ ⋅ dS⃗ = ∫∫ F⃗ ⋅ dS⃗ + ∫∫ F⃗ ⋅ dS⃗ = 729π + 729π = 1458π
∂V
D
C
which agrees with part (a).
7
15. (10 points) Find the average value of the function f(x, y, z) = x 2 + y 2 + z 2
within the solid cylinder x 2 + y 2 ≤ 9 for 0 ≤ z ≤ 4
4 2π 3
∫∫∫ f dV
1 dV = ∫ ∫ ∫ r dr dθ dz = 36π = πR 2 H
∫∫∫
0 0
0
∫∫∫ 1 dV
3
2
4 2π 3
4 r4
4
2
2 )r
2 r
(r
f
dV
=
+
z
dr
dθ
dz
=
2π
+
z
dz
=
2π
∫
∫
∫
∫
∫
∫∫∫
0 0
0
0
0
4
2 r=0
f ave =
3
= 2π 81 z + 9 z
4
2 3
4
z=0
81 + 9 z 2 dz
4
2
= 2π(81 + 96) = 354π
f ave = 354π = 59
36π
6
16. (10 points) Find the value(s) of R so that the ellipsoid
is tangent to the plane
1 x + 4 y + 2z = 36.
2
3
2
x2 + y + z2 = R2
42
32
22
HINT: Their normal vectors must be parallel.
2
2
2
y
Let f = x 2 + 2 + z 2 and g = 1 x + 4 y + 2z.
2
3
4
3
2
⃗ f = 2x , 2y , 2z
⃗g = 1 , 4 , 2
⃗ f = λ∇
⃗g
∇
∇
∇
2 3
42 32 22
2y
2x = λ 1
2z = λ2
= λ4
2
3
42
32
22
x = 4λ
y = 6λ
z = 4λ
36 = 1 x + 4 y + 2z = 1 ⋅ 4λ + 4 ⋅ 6λ + 2 ⋅ 4λ = 18λ
2
3
2
3
x=8
y = 12

λ=2
z=8
2
2
2
2
2
y2
R 2 = x 2 + 2 + z 2 = 8 2 + 122 + 8 2 = 4 + 16 + 16 = 36  R = 6
4
3
2
4
3
2
8