Math 311
Exam 3
Section 503
Solutions
Spring 2002
P. Yasskin
Multiple Choice (8 points each.)
1. If F = xy, yz, xz then ⃗
∇ ⋅ F⃗ =
a. y − z + x
b. −y, z, −x
c. x + y + z
CORRECT
d. −y, −z, −x
e. −x + y − z
⃗∇ ⋅ F
⃗ = ∂ x xy  + ∂ y yz  + ∂ z xz  = y + z + x
∇ × F⃗ =
2. If F = xy, yz, xz then ⃗
a. y − z + x
b. −y, z, −x
c. x + y + z
d. −y, −z, −x
CORRECT
e. −x + y − z

⃗∇ × F
⃗=
i

j

k
∂x ∂y ∂z



= i 0 − y  − j z − 0  + k 0 − x  = −y, −z, −x 
xy yz xz
3. If fx, y, z = x sinyz  − y cosxz  + z tanxy  then ⃗
∇ × ⃗∇f =
a. z sinyz  i + z cosxz  j + xy sec 2 xy  k
b. sinyz  i
− cosxz  j + tanxy  k
c. cosyz  i + sinxz  j + sec 2 xy  k
d. 0
CORRECT
e. Does not exist.
⃗∇ × ⃗∇f == 0
for any twice differentiable function f.
4. Compute the line integral ∫ y dx − x dy counterclockwise around the semicircle x 2 + y 2 = 4 from
to −2, 0 .
(HINT: Parametrize the curve.)
a. −4π
CORRECT
b. −2π
c. π
d. 2π
e. 4π
2, 0 
⃗rθ  = 2 cos θ, 2 sin θ 
⃗vθ  = −2 sin θ, 2 cos θ 
⃗x, y  = y, −x 
⃗⃗rθ  = 2 sin θ, −2 cos θ 
F
F
⃗ ⋅ ds⃗ = ∫ π F
⃗⃗rθ  ⋅ ⃗vθ  dθ = ∫ π −4sin 2 θ + cos 2 θ  dθ = −4π
∫ y dx − x dy = ∫ F
0
0
1
⃗ ⋅ ds⃗ for the vector field F
⃗=
5. Compute the line integral ∫ F
⃗rt = e cost 2  , e sint 2 
a.
b.
c.
d.
e.
−2
for 0
1, 1
along the curve
x y
(HINT: Find a potential f so that F = ⃗
∇f.)
≤t≤ π.
CORRECT
0
2
e
1
π
F=⃗
∇f
1, 1
x y
df df
,
dx dy
= e, 1 
=
A = ⃗r0 = e cos0  , e sin0  
f = ln x + ln y
B = ⃗r π  = e cosπ  , e sinπ   = e −1 , 1 
⃗ ⋅ ds⃗ = ∫ ⃗∇f ⋅ ds⃗ = fB  − fA  = fe −1 , 1  − fe, 1  = ln e −1 + ln 1  − ln e + ln 1  = −2
∫F
B
A
6. Compute ∮5x + 3y  dx + x − 2y  dy counterclockwise around the edge of the rectangle 1
(HINT: Use Green’s Theorem.)
3≤y≤6.
a. 36
b. 24
c. 12
d. −24
CORRECT
e. −36
Q = x − 2y 
∂ Q − ∂P
∮ Pdx + Q dy = ∫∫
∂x ∂y
≤ x ≤ 5,
P = 5x + 3y 
∂R
6
5
3
1
dx dy = ∫ ∫ 1 − 3  dx dy = −25 − 1 6 − 3  = −24
R
⃗ ⋅ dS⃗ for the vector field F
⃗=
7. Compute ∫∫ F
∂C
cylinder C = x, y, z  ∣ x 2 + y 2
a. 360π
b. 180π
CORRECT
c. 90π
d. 60π
e. 30π
zx 3 , zy 3 , z 2 x 2 + y 2 
≤ 4, 0 ≤ z ≤ 3
with outward normal. (HINT: Use Gauss’ Theorem.)
⃗∇ ⋅ F
⃗ = 3zx 2 + 3zy 2 + 2zx 2 + y 2  = 5zx 2 + y 2  = 5zr 2
⃗ ⋅ dS⃗ = ∫∫∫ ⃗∇ ⋅ F
⃗ dV = ∫ 3 ∫ 2π ∫ 2 5zr 2 r dr dθ dz = 5 z 2
∫∫ F
∂C
C
0
0
0
over the total surface of the solid
2
3
0
2π 
r4
4
2
0
= 5 9 2π4  = 180π
2
2
8.
(20 points) Compute ∫∫ x 2 y dx dy over the
y
20
R
diamond shaped region R bounded by
y = 1x ,
y = 2x ,
y = 22 ,
y = 42
x
x
For full credit you must use curvilinear coordinates.
Half credit for rectangular coordinates.
10
0
1.25
2.5
3.75
x
u = xy
v=x y
2

Boundaries are
x2y
v
u = xy = x
2 2
u2 = x y = y
v
x2y

v=2 v=4
x = uv
2
y = uv
Solve for x and y:

1
∂x
− v2
u
2
∂v =
u
= − v2 − u2 − 1u
2
∂y
u
2u − u
v
v
∂v
v2
∂x, y 
Jacobian:
J=
= 1v
Integrand:
x2y = v
∂u, v 
4 2
∫∫ x 2 y dx dy = ∫ ∫ v 1v du dv = 4 − 2 2 − 1  = 2
∂x, y 
=
∂u, v 
∂x
∂u
∂y
∂u
u=1 u=2
2
2u
v
= 1v − 2v = − 1v
1
R
3
9. (24 points) Stokes’ Theorem states that if S is a nice surface in R 3 and ∂S is its boundary curve traversed
⃗ is a nice vector field on S then
counterclockwise as seen from the tip of the normal to S and F
⃗ ⋅ dS⃗ = ∮ F
⃗ ⋅ ds⃗
∫∫ ⃗∇ × F
∂S
S
Verify Stokes’ Theorem if F = −yx , xy , x + y
z = 2 with normal pointing up and in.
2
2
2
2

and S is the part of the cone z =
x2 + y2
below
∇ × F⃗. (HINT: Use rectangular coordinates.)
9a. (4 points) Compute ⃗



∂x
− yx 2
∂y
∂z
xy 2
x2 + y2
i
⃗∇ × F
⃗=
j
k
∫∫
9b. (10 points) Compute



= i 2y − 0  − j 2x − 0  + k y 2 − −x 2  = 2y, −2x, x 2 + y 2 
⃗∇ × F
⃗ ⋅ dS⃗.
S
(HINT: Here is the parametrization of the cone and the steps you should use. Remember to check the orientation
of the surface.)
⃗ r, θ = r cos θ, r sin θ, r
R
⃗ r = cos θ, sin θ, 1 
R
⃗θ =
R

−r sin θ, r cos θ, 0 



⃗ = i −r cos θ  − j r sin θ  + k r  =
N
⃗∇ × F
⃗
⃗∇ × F
⃗
∫∫
⃗ r, θ
R
⋅

−r cos θ, −r sin θ, r 
Points up and in.
= 2r sin θ, −2r cos θ, r 2 
⃗ = −2r 2 sin θ cos θ + 2r 2 cos θ sin θ + r 3 = r 3
N
⃗∇ × F
⃗ ⋅ dS⃗ = ∫ 2π ∫ 2 r 3 dr dθ = 2π r 4
4
0
0
S
⃗ ⋅ ds⃗ .
9c. (10 points) Compute ∮ F
2
0
= 8π
Recall F =

−yx 2 , xy 2 , x 2 + y 2  .
∂S
(HINT: Parametrize of the boundary circle. Remember to check the orientation of the curve.)
(CHECK the answers to 9b and 9c agree.)
⃗rθ  = 2 cos θ, 2 sin θ, 2 
⃗vθ  = −2 sin θ, 2 cos θ, 0 
⃗⃗rθ  =
F

Counterclockwise from +z-axis.
−2 sin θ ⋅ 4 cos 2 θ, 2 cos θ ⋅ 4 sin 2 θ, 4 cos 2 θ + 4 sin 2 θ  = −8 sin θ cos 2 θ, 8 cos θ sin 2 θ, 4 
⃗ ⋅ ⃗v = 16 sin 2 θ cos 2 θ + 16 cos 2 θ sin 2 θ = 32 sin 2 θ cos 2 θ
F
⃗ ⋅ ds⃗ = ∫
∮ F
∂S
2π
0
32 sin 2 θ cos 2 θ dθ = ∫
2π
0
8 sin 2 2θ  dθ = 8 ⋅ 1 2π − 0  = 8π
2
4
10. (10 points Extra Credit) Compute the line integral ∮
boundary of the plus sign shown below.
Be sure to justify any theorem you use.
y
x + y2
2
dx −
x
dy counterclockwise around the
x2 + y2
(Hint: The answer is not zero.)
2
1.75
1.5
1.25
1
0.75
0.5
0.25
0
-2-1.75
-1.5
-1.25
-1-0.75
-0.5
-0.25
00.25
0.5
0.75
1 1.25
1.5
1.75
2
-0.25
-0.5
-0.75
-1
-1.25
-1.5
-1.75
-2
P=
y
x + y2
2
∂xQ − ∂yP =
−x
x2 + y2
2
2
x + y −1  − −x 2x
Q=
x
2
+ y2 2
P and Q are defined everywhere except the origin.
−
x
2
+ y 2 1  − y 2y
−x 2 − y 2 + 2x 2 − x 2 − y 2 + 2y 2 = 0
=
2
2 2
2
2 2
x + y 
x + y 
By Green’s Theorem, the path may be continuously deformed without changing the value of the integral provided
the region between the curves does not contain the origin. So integrate counterclockwise around the unit circle.
x = cos θ,
y = sin θ,
dx = − sin θ dθ,
dy = cos θ dθ
2
2
Note: x + y = 1
So
P = y = sin θ
Q = −x = − cos θ
2π
y
x
dx − 2
dy = ∮ sin θ − sin θ  dθ − cos θ cos θ dθ = − ∫ dθ = −2π
∮ 2
2
2
x +y
0
x +y
5