Name
Sec
MATH 251 Honors
Sections 200
Final
Spring 2010
Solutions
2,4-13
/55
16
/10
14
/21
17
/10
15
/10 Total
/106
P. Yasskin
Multiple Choice: (5 points each. No part credit.)
1.
Honors students, skip this question. Do not bubble anything on the scantron.
2.
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
Honors students, skip this question. Do not bubble anything on the scantron.
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
17. (10 points) The equation xy = wz defines a 3-dimensional surface in ℝ 4 .
It may be parametrized by
⃗ (u, v, θ) = (u cos θ, u sin θ, v cos θ, v sin θ).
(w, x, y, z) = R
Consider the portion S of this 3-surface inside the 3-sphere w 2 + x 2 + y 2 + z 2 ≤ 9.
HINT: What does the 3-sphere equation say about the parameters u, v, and θ?
a.
Find the 3-volume of the 3-surface S.
⃗,
HINT: Successively find ⃗e u , ⃗e v , ⃗e θ , N
k̂
̂
î
⃗ , V. Be very careful with signs.
N
̂l
⃗e u = ( cos θ
sin θ
0
0 )
⃗e v = ( 0
0
cos θ
sin θ )
⃗e θ = (−u sin θ u cos θ −v sin θ v cos θ)
⃗ =î
N
cos θ
sin θ
0
0
0
cos θ
sin θ
− ̂
u cos θ −v sin θ v cos θ
0
0
0
cos θ
sin θ
−u sin θ −v sin θ v cos θ
cos θ
+ k̂
sin θ
0
0
sin θ
0
cos θ
− ̂l
−u sin θ u cos θ v cos θ
0
sin θ
0
0
cos θ
−u sin θ u cos θ −v sin θ
Expand the î and ̂ determinants on the first row and the k̂ and ̂l determinants on the
second row:
⃗ = î sin θ
N
cos θ
sin θ
−v sin θ v cos θ
cos θ
− ̂ cos θ
sin θ
−v sin θ v cos θ
− k̂ sin θ
cos θ
sin θ
−u sin θ u cos θ
+ ̂l cos θ
cos θ
sin θ
−u sin θ u cos θ
= (v sin θ, −v cos θ, −u sin θ, u cos θ)
⃗ =
N
V=
∫∫∫
v 2 sin 2 θ + v 2 cos 2 θ + u 2 sin 2 θ + u 2 cos 2 θ ==
⃗ du dv dθ =
N
∫∫∫
v2 + u2
v 2 + u 2 du dv dθ
The limits on u and v are determined the 3-sphere equation:
w 2 + x 2 + y 2 + z 2 ≤ 9  u 2 cos 2 θ + u 2 sin 2 θ + v 2 cos 2 θ + v 2 sin 2 θ = u 2 + v 2 ≤ 9
So u and v are constrained to a circle of radius 3 and it is useful to switch to
polar coordinates:
u = r cos ϕ
v = r sin ϕ
du dv = r dr dϕ
with
0≤r≤3
0 ≤ ϕ ≤ 2π
v2 + u2 = r
9
TRICKY PART:
.......................................................................................................
It is tempting to let the range of θ values be 0 ≤ θ ≤ 2π, but this actually double
covers the surface.
To see this, notice that the point with (u, v, θ) = (a, b, γ) has rectangular coordinates
(w, x, y, z) = (a cos γ, a sin γ, b cos γ, b sin γ)
while the point with (u, v, θ) = (−a, −b, γ + π) has rectangular coordinates
(w, x, y, z) = (−a cos(γ + π), −a sin(γ + π), −b cos(γ + π), −b sin(γ + π)) = (a cos γ, a sin γ, b cos γ, b sin γ)
which is exactly the same point. So either
i.
u and v cover only half of the circle and 0 ≤ θ ≤ 2π,
ii.
u and v cover the whole circle and 0 ≤ θ ≤ π only.
OR
We will use the second option.
NO LOSS OF CREDIT IF YOU MISS THIS TRICKY PART.
..................................
ONE POINT EXTRA CREDIT IF YOU GET THIS TRICKY PART.
.........................
So:
V=
b.
π
2π
3
∫0 ∫0 ∫0
3
r 2 r dr dϕ dθ = (π)(2π) r
3
3
0
= 18π 2
⃗ = (z, x − y, −x, w + z) through the 3-surface S oriented by your
Find the flux of F
⃗.
N
⃗ R
⃗ (u, v, θ) , F
⃗⋅N
⃗,
HINT: Successively find F
∫∫∫ F⃗ ⋅ dV⃗.
⃗ (u, v, θ) = (u cos θ, u sin θ, v cos θ, v sin θ)
Recall: The surface is (w, x, y, z) = R
⃗ R
⃗ (u, v, θ)
F
= (z, x − y, −x, w + z) = (v sin θ, u sin θ − v cos θ, −u sin θ, u cos θ + v sin θ)
From part (a):
⃗ = (v sin θ, −v cos θ, −u sin θ, u cos θ)
N
⃗⋅N
⃗ = v 2 sin 2 θ − v cos θ(u sin θ − v cos θ) + u 2 sin 2 θ + u cos θ(u cos θ + v sin θ)
F
= v 2 sin 2 θ + v 2 cos 2 θ + u 2 sin 2 θ + u 2 cos 2 θ − v cos θ(u sin θ) + u cos θ(v sin θ) = v 2 + u 2 = r 2
∫∫∫ F⃗ ⋅ dV⃗ =
∫∫∫ F⃗ ⋅ N⃗ du dv dθ =
π
2π
3
∫ 0 ∫ 0 ∫ 0 r 2 r dr dϕ dθ = (π)(2π)
r4
4
3
0
= 81 π 2
2
10