MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 4.
Problem 1.
Find the constant a such that
F = (2x + 2xy 3 ) i + (ax2 y 2 + 2) j
is a gradient field and find a potential function using both the algebraic and geometric methods.
Solution: The curl is
∂M
∂N
−
= 2axy 2 − 6xy 2
∂x
∂y
which is 0 only if a = 3.
We will use the algebraic method to find a potential f . We have
fx = 2x + 2xy 3 , fy = 3x2 y 2 + 2.
Integrating the first equation we find
Z
f (x, y) = 2x + 2xy 3 dx = x2 + x2 y 3 + g(y).
Substitute this result into the second equality:
3x2 y 2 + 2 = fy = 3x2 y 2 + g 0 (y)
Z
0
g (y) = 2 ⇒ g(y) = 2 dy = 2y + C
f (x, y) = x2 + x2 y 3 + 2y + C .
For the geometric method, we use a horizontal path from (0, 0) to (p, 0) and a vertical
path from (p, 0) to (p, q):
p q
Z p
Z q
3
2 2
2
2 3
f (p, q)−f (0, 0) =
2x+2x0 dx+
3p y +2 dy = x + p y +2y = p2 +p2 q 3 +2q
0
0
0
0
which is the same result as that obtained from the algebraic method. The constant C
appearing in the algebraic method is equal to f (0, 0).
Problem 2.
Verify that the field
G = xy 2 i + y 3 j
is not conservative. Try to find a potential function using the algebraic method. What goes
wrong? Try to find a potential function using the geometric method. What goes wrong?
Solution: The curl is
∂M
∂N
−
= −2xy
∂x
∂y
which is not zero, so the field is not conservative. Let’s try the algebraic method:
1
2
MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 4.
Z
g(x, y) =
1
xy 2 dx = x2 y 2 + h(y)
2
∂g
= x2 y + h0 (y) ⇒ h0 (y) = y 3 − x2 y.
∂y
This is a problem: h0 should be independent of x, but the expression on the right-hand
side is not.
Let’s try the geometric method, with a horizontal path from (0, 0) to (p, 0) and a
vertical path from (p, 0) to (p, q):
4 q
Z q
Z p
y
q4
y 3 dy =
= .
x02 dx +
g(p, q) =
4 0
4
0
0
y3 =
But this is not a potential function for G because
∂g
∂x
= 0 6= xy 2 .
Page 399, problem 21.
(a) Determine where the vector field
F=
x + xy 2
x2 + 1
i
−
j
y2
y3
is conservative.
Solution: The curl is
∂N
∂M
2x 2x
−
= − 3 + 3 = 0.
∂x
∂y
y
y
So the field is conservative wherever it is defined; that is, where y 6= 0 .
(b) Determine a scalar potential for F.
Solution: By the algebraic method, we first integrate with respect to x:
Z
x + xy 2
x2 + x2 y 2
f (x, y) =
dx
=
+ g(y),
y2
2y 2
−x2 − 1
x2
=
f
=
−
+ g 0 (y)
y
y3
y3
Z
1
1
1
0
g (y) = − 3 ⇒ g(y) = − 3 = 2 + C
y
y
2y
So f (x, y) =
x2 + x2 y 2 + 1
+ C.
2y 2
(c) Find the work done by F in moving a particle along the parabolic curve y = 1 + x − x2
from (0, 1) to (1, 1).
Solution: Since y is never zero on the curve, we can safely use the Fundamental
Theorem of Calculus: The work is
Z
3 1
F · ds = f (1, 1) − f (0, 1) = − = 1 .
2 2
MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 4.
3
Problem 3.
Consider the field
F = rn (x i + y j).
Calculate the curl of F. For each value of n for which curl F = 0, find a potential f such that
F = ∇f .
1
Solution: First note that r = (x2 + y 2 ) 2 . The curl is
n
n
n
∂ 2
∂ 2
n
(x + y 2 ) 2 y −
(x + y 2 ) 2 x = (x2 + y 2 ) 2 −1 (2xy − 2yx) = 0 .
∂x
∂y
2
Let us find a potential by the algebraic method:
n+2
Z
(x2 + y 2 ) 2
2
2 n
f (x, y) = (x + y ) 2 x dx =
+ g(y).
n+2
n
(x2 + y 2 ) 2 y =
n
∂f
= (x2 + y 2 ) 2 y + g 0 (y) ⇒ g(y) = C.
∂y
So
n+2
(x2 + y 2 ) 2
rn+2
f (x, y) =
=
.
n+2
n+2
A different discussion is necessary when n = −2, since in this case the integration
above has a different answer:
f = ln r .
Page 388, problem 4.
Verify Green’s theorem for the vector field F = 2y i + x j and the semicircular region D
determined by x2 + y 2 ≤ a2 , y ≥ 0.
Solution: The curl of F is 1 − 2 = −1. Using polar coordinates:
Z πZ a
Z π 2 a
Z Z
Z π
r
a2
πa2
curl(F) dA =
(−1)r dr dθ =
−
− dθ = −
dθ =
.
2 0
2
2
D
0
0
0
0
We parameterize the the lower boundary of the semicircle with (x, y) = (0, t), −a ≤
t ≤ a, (dx, dy) = (0, dt). We parameterize the upper boundary with (x, y) = (a cos t, a sin t),
0 ≤ t ≤ π, (dx, dy) = (−a sin t dt, a cos t dt). Then
I
Z
a
F· ds =
∂D
Z π
Z
2(0) dt+
(2a sin t)(−a sin t)+(a cos t)(a cos t) dt = 0+a2
−a
0
π
0
π
π
1
3
πa2
3
2
2
.
=a
1 − (1 − cos(2t)) dt = a − t + sin(2t) = −
2
2
4
2
0
0
RR
H
So
D curl(F) dA = ∂D F · ds, as Green’s theorem asserts.
Z
1−3 sin2 t dt
4
MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 4.
Page 388, problem 6.
Let F = 3xy i + 2x2 j and suppose C is the counterclockwise curve enclosing both the square
with vertices (0, 0), (0, −2), (2, −2), (2, 0) and the upper semicircle of radius 1 and center
(1, 0). Evaluate
I
F · ds
C
both directly and also by means of Green’s theorem.
Solution: We parameterize the left side with (x, y) = (0, −t), 0 ≤ t ≤ 2, the bottom
with (x, y) = (t, −2), 0 ≤ t ≤ 2, the right side with (x, y) = (2, t), −2 ≤ t ≤ 0, and the
top with (x, y) = (cos t + 1, sin t), 0 ≤ t ≤ π, (dx, dt) = (− sin t dt, cos t dt).
I
Z
F · ds = −
C
Z
0
Z
2(0)2 dt +
Z
2
3x(−2) dt
0
π
2(2) dt +
3(cos t + 1)(sin t)(− sin t) + 2(cos t + 1)2 (cos t) dt
−2
0
Z π
= 0 − 12 + 16 +
−3 cos t sin2 t + (−3 sin2 t + 4 cos2 t) + 2(cos3 t) + 2 cos t dt
0
Z π
=4+
−3 cos t sin2 t + (−3 + 7 cos2 t) + 2 cos t(1 − sin2 t) + 2 cos t dt
0
Z π
=4+
−5 cos t sin2 t + −3 + 4 cos t + 7 cos2 t dt
0
Z π
7
=4+
−5 cos t sin2 t + −3 + 4 cos t + (1 + cos(2t)) dt
2
0
Z π
1
7
−5 cos t sin2 t + + 4 cos t + cos(2t)) dt
=4+
2
2
0
π
5
t
7
π
3
= 4 + − sin t + + 4 sin t + sin(2t)) = 4 + .
3
2
4
2
0
To use Green’s theorem, we first find the curl:
∂N
∂M
−
= 4x − 3x = x.
∂x
∂y
+
2
0
2
The integral is computed by observing the connection with average values:
RR
x dxdy
xavg = R
.
area R
Since xavg = 1 and R has area 4 + π2 , we find that
ZZ
π
curl F dA = 4 + .
2
R
Problem 8, Page 389
Use Green’s theorem to find the work done by the vector field F = (4y − 3x)i + (x − 4y)j on
a particle as it moves counterclockwise around the ellipse given by x2 + 4y 2 = 4
MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 4.
Solution:
5
ZZ
I
1 − 4 dxdy = −3A
F · ds =
D
C
where A = π(1)(2) = 2π is the area of the ellipse.
Problem 19, Page 389
Let C be any simple closed curve in the plane. Show that
I
3x2 y dx + x3 dy = 0
Solution: By Green’s theorem,
I
ZZ
ZZ
∂ 3
∂
2
3
2
3x y dx + x dy =
(x ) −
(3x y) dxdy =
3x2 − 3x2 dxdy = 0
∂y
intC ∂x
intC
Problem 20, Page 389
Show that
I
−y 3 dx + (x3 + 2x + y)dy
is positive for any closed curve C to which Green’s theorem applies.
I
Solution: By Green’s theorem
ZZ
ZZ
3
3
2
2
−y dx + (x + 2x + y)dy =
(3x + 2) − (−3y ) dxdy =
3(x2 + y 2 ) + 2 dxdy
R
But
3(x2
+
y2)
R
+ 2 is positive on R, thus so is the integral.
Problem 21, Page 390
Let C be the boundary of any rectangular region, D in R2 . Show that
I
(x2 y 3 − 3y)dx + x3 y 2 dy
is translation invariant, i.e. it depends only on the area of the region.
Solution:
I
2 3
3 2
ZZ
2 2
(x y − 3y)dx + x y dy =
C
2 2
ZZ
3x y − (3x y − 3) dydx = 3
D
dxdy = 3 area(D).
D
Problem 24, page 390.
Let f (x, y) be a C 2 harmonic function, i.e.
∂2f
∂2f
+
=0
∂x2
∂x2
and let C be the boundary of a region to which Green’s theorem applies. Show that
Z
∂f
∂f
dx −
dy = 0.
∂y
∂x
C
6
MATH 52, FALL 2007 - SOLUTIONS TO HOMEWORK 4.
Solution:
I
Z
ZZ
ZZ
∂f
∂2f
∂2f
∂f
F · ds =
0 dxdy = 0.
− 2−
dxdy =
dx −
dy =
∂x
∂x
∂x2
C ∂y
C
D
D
Problem 5, Handout.
For what simple closed curve C does the line integral
Z
(x2 y + y 3 − y)dx + (3x + 2y 2 x)dy
C
have the largest value. What is that value?
Solution: By Green’s theorem
Z
ZZ
2
3
2
(x y + y − y)dx + (3x + 2y x)dy =
(3 + 2y 2 ) − (x2 + 3y 2 − 1) dxdy
C
intC
ZZ
=
4 − x2 − y 2 dxdy
D
The integrand is positive everywhere inside the disk of radius 2, zero on the circle
of radius 2, and negative outside the disk. So to maximize the integral, C should be
the oriented boundary of the disk, i.e. the circle of radius 2 oriented in the counterclockwise direction.
The maximum value is
Z 2π Z 2
Z
r4 2
4 − x2 − y 2 =
(4 − r2 )r dr dθ = 2π 4r −
= 8π.
4 r=0
0
0
D