Name
ID
MATH 311
Exam 3
Fall 2001
Section 200
Solutions
P. Yasskin
1.
35 points
1
/35
2
/35
3
/30
⃗ ⋅ dS⃗ over the complete surface of the box
Compute ∫∫ F
∂C
0≤x≤2
⃗ = x 2 y 2 z 3 , xy 3 z 3 , xy 2 z 4 .
where F
0≤y≤3
0≤z≤4
⃗ = ∇ ⋅ x 2 y 2 z 3 , xy 3 z 3 , xy 2 z 4  = 2xy 2 z 3 + 3xy 2 z 3 + 4xy 2 z 3 = 9xy 2 z 3
∇⋅F
By Gauss’ Theorem
⃗ ⋅ FdV = ∫ 4 ∫ 3 ∫ 2 9xy 2 z 3 dx dy dz = 9 x 2
⃗ ⋅ dS⃗ = ∫∫∫ ∇
∫∫ F
2
0 0 0
∂C
C
2
0
y3
3
3
0
z4
4
4
0
= 9 ⋅ 2 ⋅ 9 ⋅ 64 = 10 368
1
2.
35 points Consider the cone C given by
z=
x2 + y2
for z ≤ 4
4
⃗ = −yz, xz, −xy .
and the vector field F
⃗×F
⃗ ⋅ dS⃗ with
We want to compute ∫∫ ∇
2
C
0
normal pointing up and into the cone.
a.
⃗×F
⃗.
5 Compute ∇
 

⃗ × −yz, xz, xy  =
∇
b.

i

j
k
∂x ∂y ∂z
−yz xz −xy
= −2x, 0, 2z 
10  Parametrize the cone using cylindrical coordinates r and θ as the parameters
⃗×F
⃗ ⋅ dS⃗.
and give the range of the parameters. Then explicitly compute ∫∫ ∇

⃗ r, θ  = r cos θ, r sin θ, r  with 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π
R



cos θ,
sin θ,
1

−r sin θ, r cos θ, 0

i
⃗r =
R
⃗θ =
R


j
C
k
⃗ =R
⃗r × R
⃗ θ = −r cos θ, −r sin θ, r  which points up and in.
N
⃗×F
⃗ = −2x, 0, 2z  = −2r cos θ, 0, 2r 
∇
⃗×F
⃗×F
⃗ ⋅ dS⃗ = ∫∫ ∇
⃗⋅N
⃗ dr dθ = ∫ 2π ∫ 4 2r 2 cos 2 θ + 2r 2  dr dθ
∫∫ ∇
C
0
C
3
4
3
0
= 2r
∫
2π
0

0
cos 2 θ + 1  dθ = 128 π + 2π  = 128π
3
2
RECALL: C is the cone z = x 2 + y 2
⃗ = −yz, xz, −xy .
the cone and F
c.

for z ≤ 4 with normal pointing up and into
⃗×F
⃗ ⋅ dS⃗. Be sure to name or quote
10  Describe 2 other ways to compute ∫∫ ∇
C
any Theorem you use and discuss the orientation of any curves or surfaces.
⃗×F
⃗ ⋅ dS⃗ = ∫ F
⃗ ⋅ ds⃗ where ∂C is the circle
i. By Stokes’ Theorem, ∫∫ ∇
∂C
C
x 2 + y 2 = 16 and z = 4 traversed counterclockwise.
ii. By Stokes’ Theorem again,
⃗×F
⃗ ⋅ ds⃗ = ∫∫ ∇
⃗ ⋅ dS⃗ where D is the disk
∫F
∂C
D
x 2 + y 2 ≤ 16 and z = 4 with normal pointing up.
d.

⃗×F
⃗ ⋅ dS⃗ by one of these two methods.
10  Recompute ∫∫ ∇
C
i. ⃗
rt  = 4 cos θ, 4 sin θ, 4 
⃗vt  = −4 sin θ, 4 cos θ, 0 
⃗ = −yz, xz, −xy  = −16 sin θ, 16 cos θ, −16 sin θ cos θ 
F
⃗ ⋅ ds⃗ = ∫ 2π F
⃗ ⋅ ⃗v dθ = ∫ 2π 64 sin 2 θ + 64 cos 2 θ dθ = 128π
∫F
0
∂C
0
OR
ii. We parametrize the disk D.
⃗ r, θ  = r cos θ, r sin θ, 4  with 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π
R



cos θ,
sin θ,
0

−r sin θ, r cos θ, 0

i
⃗r =
R
⃗θ =
R


j
k
⃗ =R
⃗r × R
⃗ θ = 0, 0, r  which points up.
N
⃗×F
⃗ = −2x, 0, 2z  = −2r cos θ, 0, 8 
∇
⃗×F
⃗×F
⃗ ⋅ dS⃗ = ∫∫ ∇
⃗⋅N
⃗ dr dθ = ∫ 2π ∫ 4 8r  dr dθ = 2π 4r 2
∫∫ ∇
D
D
0
0
4
0
= 128π
3
3.
A hypersurface S in R 4 with coordinates w, x, y, z , may be parametrized by
30 points

⃗ r, θ, ϕ  = r cos θ, r sin θ, r cos ϕ, r sin ϕ 
w, x, y, z  = R
for 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π and 0 ≤ ϕ ≤ 2π.
a.

15  Find the tangent vectors, the normal vector and length of the normal vector.




cos θ,
sin θ,
cos ϕ,
sin ϕ

0,
0

i
⃗r =
R

⃗θ =
R

⃗ϕ =
R


⃗ = i
N
j
k
−r sin θ, r cos θ,
0,
−r sin ϕ, r cos ϕ
0,
sin θ
cos ϕ
sin ϕ
r cos θ
0
0
0,
l
cos θ

− j −r sin θ
−r sin ϕ r cos ϕ
cos θ
0
sin θ

+ k −r sin θ r cos θ
0


0
cos ϕ
sin ϕ
0
0
−r sin ϕ r cos ϕ
sin ϕ
0
r cos ϕ
cos θ
sin θ

− l −r sin θ r cos θ
0
0
cos ϕ
0
−r sin ϕ

= i −r cos θ r cos 2 ϕ + r sin 2 ϕ − j r sin θ r cos 2 ϕ + r sin 2 ϕ


+ k r cos ϕ r cos 2 θ + r sin 2 θ − l −r sin ϕ r cos 2 θ + r sin 2 θ
= −r 2 cos θ, −r 2 sin θ, r 2 cos ϕ, r 2 sin ϕ 
⃗ =
N
b.

−r 2 cos θ  2 + −r 2 sin θ  2 + r 2 cos ϕ  2 + r 2 sin ϕ  2 = 2r 4 = r 2 2
5 Find the hyperarea of the hypersurface.
 
⃗ dr dθ dϕ = ∫ 2π ∫ 2π ∫ 3 r 2 2 dr dθ dϕ = 2π  2 2 r 3
A = ∫∫∫ N
3
0
0
0
S
3
0
= 36π 2 2
4
RECALL: S is the hypersurface parametrized by
⃗ r, θ, ϕ  = r cos θ, r sin θ, r cos ϕ, r sin ϕ 
w, x, y, z  = R
for 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π and 0 ≤ ϕ ≤ 2π.

c.
5 Compute P = ∫∫∫ 2w2 + 2x 2 dS over the hypersurface.
 
S
⃗ dr dθ dϕ = ∫ 2π ∫ 2π ∫ 3 2r 3 dr dθ dϕ = 2π  2 2 r 4
P = ∫∫∫ 2r 2 N
4
0
0
0
S
d.
3
0
= 162π 2
5 Compute Q = ∫∫∫wdy dx dz − 5z dwdx dy  over the hypersurface.
 
S
Q = ∫∫∫

r cos θ 
S
∂y, x, z 
∂ w, x, y 
− 5r sin ϕ  
∂r, θ, ϕ 
∂r, θ, ϕ 
∂ x, y, z 
∂y, x, z 
=− 
= −N 1 = r 2 cos θ
∂r, θ, ϕ 
∂r, θ, ϕ 
dr dθ dϕ
∂w, x, y 
= −N 4 = −r 2 sin ϕ
∂r, θ, ϕ 
Q = ∫∫∫r cos θ r 2 cos θ  − 5r sin ϕ −r 2 sin ϕ   dr dθ dϕ
S
2π
2π
3
4
3
0
0
0
4
0
= ∫ ∫ ∫ r 3 cos 2 θ + 5r 3 sin 2 ϕ dr dθ dϕ = 2π r
∫
2π
0
cos 2 θ dθ + 5 ∫
2π
0
sin 2 ϕ dϕ
= π 81 π + 5π  = 243π 2
2
5