Homework 3, Math 244, Spring 2017
Last name, first name, ID number :
Score:
1. (5 points) Use Green’s Theorem to find the counterclockwise circulation and outward flux for the
field F (x, y) = arctan( xy )i + ln(x2 + y 2 )j and the curve C defined as the boundary of the region R
described, in polar coordinates, by the inequalities 1 ≤ r ≤ 2, 0 ≤ θ ≤ π.
Solution: We have M (x, y) = arctan xy and N (x, y) = ln(x2 + y 2 ) so
∂M
∂N
∂N
∂M
y
x
2x
2y
,
,
,
.
=− 2
= 2
= 2
= 2
2
2
2
∂x
x +y
∂y
x +y
∂x
x +y
∂y
x + y2
Using the first version of Green’s Theorem, we get
ZZ
∂M
∂N
Flux =
+
dxdy
∂x
∂y
Z ZR
y
=
dxdy
2
2
R x +y
Z πZ 2
=
sin θdrdθ (in polar coordinates)
0
1
= 2.
Using the second version of Green’s Theorem, we get
ZZ
∂N
∂M
−
dxdy
Circulation =
∂x
∂y
Z ZR
x
=
dxdy
2
2
R x +y
Z πZ 2
=
cos θdrdθ (in polar coordinates)
0
1
= 0.
2. (5 points) Use a parametrization to express the area of the portion of the paraboloid z = x2 + y 2
between the palnes z = 1 and z = 4 as a double integral and evaluate the integral.
Solution: We can use the parametrization r(a, θ) = a cos(θ)i + a sin(θ)j + a2 k√for 1 ≤ a ≤ 2 and
0 ≤ θ ≤ 2π. Calculating, we find |ra × rθ | = | − 2a2 cos(θ)i − 2a2 sin(θ)j + ak| = a 4a2 + 1. Therefore
Z
2π
Z
2
p
a 4a2 + 1dadθ
0
1
3 1
1
= 2π. (4a2 + 1) 2 12
0
3
3
π
= (17 2 − 5 2 )
6
Area =
3. (5 points) Find the outward flux of the field F (x, y, z) = xzi + yzj + k accross the surface S cut
from the sphere x2 + y 2 + z 2 = 25 by the plane z = 3.
1
Solution: The surface is described by the equation f (x, y, z) = 25 for z ≥ 3 where f (x, y, z) =
x2 + y 2 + z 2 . Therefore the projection R of S on the xy-plane is the disk centered at (0, 0) with
radius 4, whose
p unit normal vector is p = k Calculating on the surface S, we find ∇f = 2x + 2y + 2z
and |∇f | = 2 x2 + y 2 + z 2 = 10 so the outward
p pointing unit normal vector at a point (x, y, z) on
∇f
x+y+z
S is n = |∇f
and
|∇f
·
p|
=
|2z|
=
2
=
25 − (x2 + y 2 ). Moreover, we find
5
|
x2 z + y 2 z + z
z(x2 + y 2 + 1)
F·n =
=
=
5
5
p
25 − (x2 + y 2 )(x2 + y 2 + 1)
.
5
Thus
ZZ
F · ndσ
ZZ p
25 − (x2 + y 2 )(x2 + y 2 + 1)
10
=
dxdy
. p
5
2 25 − (x2 + y 2 )
R
ZZ
=
(x2 + y 2 + 1)dxdy
R
Z 2π Z 4
=
(r2 + 1)rdrdθ (in polar coordinates)
Flux =
S
0
0
= 144π.
4. (5 points) Let n be the outer unit normal (normal away from the origin) of the parabolic shell
1
1
S : 4x2 +RRy + z 2 = 4, y ≥ 0 and let F (x, y, z) = (−z + 2+x
)i + arctan(y)j + (x + 4+z
)k. Find the
value of S ∇ × F · ndσ.
RR
H
Solution: Stokes Theorem says that S ∇ × F · ndσ = C F · dr where C is the boundary for the
base of S in the xz-plane which is the ellipse 4x2 + z 2 = 4. We can use the counterclockwise
parametrization r(t) = cos(t)i − 2 sin(t)k, 0 ≤ t ≤ 2π, for C so
dr = − sin(t)dti − 2 cos(t)dtk.
Along C, we have
F = 2 sin(t) +
1
1
i + cos(t) +
k.
2 + cos(t)
4 − 2 sin(t)
Thus
F · dr = −2 −
sin(t)
2 cos(t)
−
2 + cos(t) 4 + 2 sin(t)
and
I
Z
F · dr =
C
2π
−2 −
0
sin(t)
2 cos(t)
−
dt
2 + cos(t) 4 + 2 sin(t)
2π
= −2t + ln(2 + cos(t)) − ln(4 + 2 sin(t))
0
= −4π.
2
5. (5 points) Use Stokes Theorem to calculate the circulation of the field F = 2yi + 3xj − z 2 k around
the circle C : x2 + y 2 = 9 in the xy-plane, counterclockwise when viewed from above.
H
RR
Solution: Stokes Theorem says that C F · dr = S ∇ × F · ndσ where S is the surface enclosed by
C. Calculating, we find ∇ × F = k and n = k so ∇ × F · n = 1. Thus
ZZ
ZZ
dσ
∇ × F · ndσ =
S
Z
x2 +y ≤ 9
2π Z 3
adadθ
=
0
0
= 9π.
6. (5 points) Use the Divergence Theorem to find the outward flux of F = x3 i + y 3 j + z 3 k across the
boundary of the region D : x2 + y 2 + z 2 ≤ a2 .
Solution: Calculating, we find div.F (x, y, z) = 3(x2 + y 2 + z 2 ). Thus
ZZZ
Outward Flux =
Z
3(x2 + y 2 + z 2 )dV
Z a
ρ4 sin(φ)dρdφdθ spherical coordinates
D
2π Z π
=3
0
0
0
12πa5
=
5
3