Name
1-8
/40 11
/15
Spring 2007
9
/ 5 12
/15
P. Yasskin
10
/15 13
/15
ID
MATH 253
Exam 2
Sections 501-503
Solutions
Multiple Choice: (5 points each. No part credit.)
/105
Total
1. Find the equation of the line perpendicular to the graph of
xyz − x 2 − y 2 − z 2 = −8 at the point (1, 2, 3).
Where does this line intersect the xz-plane?
a. (−7, 0, −5)
b. (−7, 0, 11)
c. (9, 0, −5)
Correct Choice
d. (9, 0, 11)
e. (5, 0, −1)
⃗ F = ⟨yz − 2x, xz − 2y, xy − 2z⟩.
Then ∇
⃗ =∇
⃗ F = ⟨4, −1, −4⟩.
Then the normal at P = (1, 2, 3) is N
Let F = xyz − x 2 − y 2 − z 2 .
P
This is also the tangent vector to the perpendicular line. So ⃗v = ⟨4, −1, −4⟩.
So the perpendicular line is X = P + tv⃗ = (1, 2, 3) + t⟨4, −1, −4⟩ or
x = 1 + 4t
y = 2−t
z = 3 − 4t
This intersects the xz-plane when y = 0 or t = 2.
So x = 1 + 4t = 9 and z = 3 − 4t = −5.
2. The point (x, y) = (1, 2) is a critical point of the function f(x, y) = (x 2 + y 2 − 4) − 4x − 8y.
2
Use the Second Derivative Test to classify it as a
a. local maximum
b. local minimum
Correct Choice
c. inflection point
d. saddle point
e. Test Fails
f x = 2(x 2 + y 2 − 4)2x − 4
f y = 2(x 2 + y 2 − 4)2y − 8
f x (1, 2) = f y (1, 2) = 0
2
2
2
2
f xx = 2(2x)2x + 2(x + y − 4)2
f yy = 2(2y)2y + 2(x + y − 4)2
f xy = 2(2y)2x
f xx (1, 2) = 8 + 4 = 12 > 0
f yy (1, 2) = 32 + 4 = 36 > 0
f xy (1, 2) = 16
D = f xx f yy − f xy 2 = 12 ⋅ 36 − 16 2 = 176 > 0
local minimum
1
3.
y
Find the center of mass of the triangle
with vertices (0, 0), (1, 1) and (−1, 1)
1.0
0.5
if the mass density is ρ = y.
-1
0, 1
3
0, 1
2
0, 2
3
0, 3
4
4
0,
5
a.
b.
c.
d.
e.
1
x
Correct Choice
x̄ = 0 by symmetry,
M=
0
1
1
∫∫ ρ dA = ∫ 0 ∫ −y y dx dy = ∫ 0
y-mom = M x =
x = −y
x=y
1
3
y
yx
dy = 2 ∫ y 2 dy = 2
= 2
x=−y
0
3
3 y=0
Use a y-integral.
y
∫∫ yρ dA =
1
y
∫ 0 ∫ −y
0≤y≤1
1
y
y 2 dx dy =
1
∫0
y2x
1
y
x=−y
dy = 2 ∫ y 3 dy = 2
0
1
y4
4
y=0
= 1
2
y-mom
ȳ =
= Mx = 1 3 = 3
M
M
2 2
4
4.
Compute
y
∫∫(x 2 + y 2 ) dA over the region bounded
1
by the polar curve r = θ and the x-axis.
-3
5
a. π
20
π4
16
π4
12
π3
9
π3
6
b.
c.
d.
e.
π
θ
∫0 ∫0
-2
-1
0
x
Correct Choice
r 2 r dr dθ =
π
∫0
r4
4
θ
dθ = 1
4
0
π
∫ 0 θ 4 dθ = 14
θ5
5
π
0
5
= π
20
2
5. Find the mass of the 1/8 of the solid sphere x 2 + y 2 + z 2 ≤ 16 in the first octant
if the mass density is δ = z.
a. 64π
Correct Choice
b. 16π
c. 8π
d. 4π
e. π
M=
π/2
π/2
4
∫∫∫ δ dV = ∫∫∫ z dV = ∫ 0 ∫ 0 ∫ 0 ρ(cos ϕ) ρ 2 (sin ϕ) dρ dϕ dθ
=
π/2
∫0
dθ ∫
π/2
0
4
cos ϕ sin ϕ dϕ ∫ ρ 3 dρ = π
0
2
π/2
sin 2 ϕ
2
4
ρ4
4
ϕ=0
ρ=0
6. Find the volume of the solid between the cone z = 2 x 2 + y 2
= π 1 4 3 = 16π
2 2
and the paraboloid z = 8 − x 2 − y 2 .
HINT: Find the radius where the cone and paraboloid intersect.
a. 30π
b. 80π
c.
d.
e.
z=
V=
3
40π
Correct Choice
3
20π
3
10π
3
2r = 8 − r 2
r 2 + 2r − 8 = 0
2π
2
8−r 2
∫∫∫ 1 dV = ∫ 0 ∫ 0 ∫ 2r
(r − 2)(r + 4) = 0
r dz dr dθ = 2π ∫
2
0
rz
8−r 2
z=2r
4
2
= 2π ∫ (8r − r 3 − 2r 2 ) dr = 2π 4r 2 − r − 2r
0
4
3
r = 2 since r ≥ 0.
2
dr = 2π ∫ r(8 − r 2 − 2r) dr
0
3
2
r=0
= 2π 16 − 4 − 16
3
= 40π
3
∫∫ S F⃗ ⋅ dS⃗
⃗ = (z, z, x + y) over the surface S which is parametrized
for the vector field F
⃗ (u, v) = (u + v, u − v, uv) for 0 ≤ u ≤ 2 and 0 ≤ v ≤ 3 and oriented along N
⃗ = ⃗e u × ⃗e v .
by R
7. Compute
a. −60
b. −12
c. 0
d. 12
Correct Choice
e. 60
⃗ (u, v) = (u + v, u − v, uv)
R
î
̂
k̂
⃗e u =
1,
1,
v
⃗e v =
1, − 1,
u
⃗ = ⃗e u × ⃗e v = î(u + v) − ̂ (u − v) + k̂ (−1 − 1) = (u + v, v − u, −2)
N
⃗ = (z, z, x + y)
⃗ R
⃗ (u, v) = (uv, uv, 2u)
⃗⋅N
⃗ = uv(u + v) + uv(v − u) + 2u(−2) = u(2v 2 − 4)
F
F
F
∫∫ S F⃗ ⋅ dS⃗ = ∫ ∫ F⃗
⃗ (u, v) ⋅ N
⃗ du dv =
R
3
2
∫ 0 ∫ 0 u(2v 2 − 4) du dv =
u2
2
2
0
2v 3 − 4v
3
3
0
= (2)(18 − 12) = 12
3
∫∫ C F⃗ ⋅ dS⃗
⃗ = (xz, yz, z 2 ) over the cylindrical surface x 2 + y 2 = 9 for
for the vector field F
0 ≤ z ≤ 2 oriented outward.
8. Compute
a. 18π
b. 62 π
3
c. 21π
d. 124 π
3
e. 36π
Correct Choice
̂
î
⃗ (θ, z) = (3 cos θ, 3 sin θ, z)
R
k̂
⃗e θ = −3 sin θ, 3 cos θ,
0
⃗e z =
1
0,
0,
⃗ = ⃗e θ × ⃗e z = î(3 cos θ) − ̂ (−3 sin θ) + k̂ (0) = (3 cos θ, 3 sin θ, 0)
N
⃗ = (xz, yz, z 2 )
⃗ R
⃗ (θ, z) = (3z cos θ, 3z sin θ, z 2 )
⃗⋅N
⃗ = 9z cos 2 θ + 9z sin 2 θ = 9z
F
F
F
∫∫ C
⃗=
⃗ ⋅ dS
F
∫ ∫ F⃗ R⃗(θ, z) ⋅ N⃗ dθ dz =
2
2π
∫0 ∫0
2
9z dθ dz = 9(2π) z
2
2
0
= 36π
Work Out: (Points indicated. Part credit possible. Show all work.)
9.
(5 points) At the right is the contour
plot of a function f(x, y). If you start
at the dot at (5, 6) and move so that
your velocity is always in the direction
⃗ f, the gradient of f, roughly
of ∇
sketch your path on the plot.
NOTE : The numbers on the right are
the values of f on each level curve.
The curve starts at (5, 6) goes down and curves to the right towards higher values of the function
f, always perpendicular to each level curve. It should not go up.
4
10. (15 points) An aquarium in the shape of a rectangular solid has a base made of marble which costs
6 cents per square inch, a back and sides made of mirrored glass which costs 2 cents per square
inch and a front made of clear glass which costs 1 cent per square inch. There is no top. If the
volume of the aquarium is 9000 cubic inches, what are the dimensions of the cheapest such
aquarium?
Let x be the length side to side, y be the width front to back, and z be the height.
C = 6xy + 2(xz + 2yz) + 1xz = 6xy + 3xz + 4yz
The cost is
V = xyz = 9000
The volume constraint is
which we solve for
z = 9000
xy
C = 6xy + 27000
+ 36000
y
x
So the cost becomes
We want to minimize C. So we set the partials equal to zero:
C x = 6y − 36000
=0
C y = 6x − 27000
=0
2
x
y2
4500
4500x 4
y = 6000
x = 27000
=
=
= 1 x4
2
2
2
6000
⋅
6000
8000
6y
x
6000
2
x
Cancel an x and solve for x then y and z:
3
9000
1= x ,
x = 20
y = 6000
= 6000 = 15
z = 9000
xy = 300 = 30
8000
400
x2
x = 20
So the dimensions are:
11.
(15 points) Compute
y = 15
z = 30
y
∫∫ xy dA over the "diamond"
6
5
shaped region bounded by the curves
4
y2 − x2 = 4
y 2 − x 2 = 16
3
y − 2x = 1
y − 2x = 4
2
2
2
2
2
1
HINT: Let u = y 2 − x 2 and v = y 2 − 2x 2 .
u − v = y 2 − x 2 − y 2 + 2x 2 = x 2
J=
∂(x, y)
∂(u, v)
=
=
xy =
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
0
0
2u − v = 2y 2 − 2x 2 − y 2 + 2x 2 = y 2
=
1
2 u−v
−1
2 u−v
x=
2
u−v
4
y=
x
2u − v
2
2 2u − v
−1
2 2u − v
1
−1
2
−1
−
2 u − v 2 2u − v
2
u−v
2 2u − v
=
1
4 u − v 2u − v
u − v 2u − v
4
16
∫∫ xy dA = ∫ 1 ∫ 4
4 ≤ u ≤ 16
1≤v≤4
4 16
1
u − v 2u − v
du dv = ∫ ∫ 1 du dv = 1 (16 − 4)(4 − 1) = 9
1
4 4
4
4 u − v 2u − v
5
12. (15 points) Find the average temperature on the hemisphere surface x 2 + y 2 + z 2 = 9, with z ≥ 0,
if the temperature is T = z.
f dS
NOTE : The average of a function f is f ave = ∫∫
.
∫∫ dS
HINT: Parametrize the hemisphere.
⃗ (ϕ, θ) = (3 sin ϕ cos θ, 3 sin ϕ sin θ, 3 cos ϕ)
R
k̂
̂
î
⃗e ϕ = (3 cos ϕ cos θ, 3 cos ϕ sin θ, −3 sin ϕ)
⃗e θ = (−3 sin ϕ sin θ, 3 sin ϕ cos θ, 0 )
⃗ = ⃗e ϕ × ⃗e θ = î(9 sin 2 ϕ cos θ) − ̂ (−9 sin 2 ϕ sin θ) + k̂ (9 sin ϕ cos ϕ cos 2 θ + 9 sin ϕ cos ϕ sin 2 θ)
N
= (9 sin 2 ϕ cos θ, 9 sin 2 ϕ sin θ, 9 sin ϕ cos ϕ)
⃗ =
N
81 sin 4 ϕ cos 2 θ + 81 sin 4 ϕ sin 2 θ + 81 sin 2 ϕ cos 2 ϕ = 9 sin ϕ
2π
π/2
⃗ dϕ dθ =
N
2π
π/2
(Area of hemisphere = 1 (4πR 2 ).)
2
π/2
2
2π π/2
⃗ dϕ dθ = 2π π/2 3 cos ϕ 9 sin ϕ dϕ dθ = 27 ⋅ 2π sin ϕ
= 27π
∫∫ T dS = ∫ 0 ∫ 0 z N
∫0 ∫0
2
ϕ=0
∫∫ dS =
∫0 ∫0
∫0 ∫0
9 sin ϕ dϕ dθ = 18π
T dS
T ave = ∫∫
= 27π = 3
18π
2
dS
∫∫
13. (15 points) Sketch the region of integration and then compute the integral
y
1.0
y
0.5
1
1
∫0 ∫x
2
x 3 cos(y 3 ) dy dx.
1.0
0.5
0.0
0.0
0.0
0.5
1.0
0.0
x
0.5
1.0
x
Reverse the order of integration:
1
1
∫0 ∫x
x 3 cos(y 3 ) dy dx =
2
Substitute:
1
1
∫0 ∫x
2
u = y3
1
∫0 ∫0
y
x 3 cos(y 3 ) dx dy =
du = 3y 2 dy
x 3 cos(y 3 ) dy dx = 1
12
∫ cos(u) du =
1
∫0
x 4 cos(y 3 )
4
y
x=0
dy =
1
∫0
y2
cos(y 3 ) dy
4
y 2 dy = 1 du
3
1 sin(u) = 1 sin(y 3 )
12
12
1
0
= 1 sin 1
12
6