Math 6C – Chapter 14 Quiz – SOLUTIONS
1. Calculate the line integral,
2 y
, over the elliptical path
x
F dr , with F ( x, y)
parametrized by
cos t
from t = 0 to t = .
2sin t
r (t )
Since
sin t
and
2 cos t
r '(t )
4sin t
, we get:
cos t
F (r (t ))
4sin t
cos t
0
F dr
sin t
2
2
dt 4sin t 2cos t
2 cos t
0
dt
2 2sin t cos t dt 2 sin 2 t sin 2 t cos 2 t dt
0
2
0
2
2 sin t 1dt 2 sin t dt 2 dt
0
2
0
2
0
1 cos(2t )
2
dt 2 1 cos(2t ) dt 2
2
0
0
2 3
2. Calculate the line integral,
2t
from t = 0 to t = 1.
2
t
r (t )
2
Since r '(t ) and
2t
y
, over the path parametrized by
xy
F dr , with F ( x, y)
t2
F (r (t )) 3 , we get:
2t
1
F
1
2
2
4
dr
dt 2t 4t
2t
0
1
t3
t5
2 4 22
2 4
5
3 5 15
3
0
t2
3
2t
0
3. Calculate
1
1
dt 2 t dt 4 t dt
2
4
0
0
F dr with F (x, y) g(x, y) , where g(x, y) 2x xy , over the path
2
t 2
parametrized by r (t ) from t = 0 to t = 1.
t 3
Since
F ( x, y) g ( x, y) , we have:
F dr g (r (1)) g (r (0)) g (1,1) g (0,0)
1 0 1
4. Suppose that
2 xy
with
2
x
g ( x, y)
g (0,0) 0 .
Then
g ( x, y) ?
2 xy g x
2
x g y
g
g x 2xy g 2xy dx x2 y c( y)
g y x2 x2 c '( y) c( y) C
Then
5. Calculate
g ( x, y) x2 y c , so g(0,0) 0 c 0 , and so
g ( x, y) x2 y .
2 xy
, over the path
2
x
F dr with F ( x, y) g ( x, y)
1,1 along the circle
from the point
x2 y 2 2 .
F dr g (r (1)) g (r (0)) g
1 0 1
2, 0 g (1,1)
2, 0 to
6.
Suppose that
y2 2x
with
2 xy
g ( x, y)
g (0,0) 0 .
Then
g ( x, y) ?
y2 2x g x
2 xy g y
g
g x y 2 2x g
y
2
2x dx xy 2 x2 c( y)
g y 2xy 2xy c '( y) c( y) C
Then
7. Calculate
g ( x, y) xy 2 x2 c , so g(0,0) 0 c 0 , and so
g ( x, y) xy 2 x2 .
2 2x
, over the path
2 xy
y
F dr with F ( x, y) g ( x, y)
from the point ( 1, 2 ) to
from the point ( 0, 1 ) to
( 3, 1 ) along the line segment connecting these two points.
F dr g 3,1 g (1,2) 6 3 9
8. Calculate
(
y2 2x
, over the path
2 xy
F dr with F ( x, y) g ( x, y)
, –1 ) along the curve
y = cos x.
F dr g , 1 g(0,1)
xy
M ( x, y)
9. If F ( x, y)
x and
N ( x, y) y
(1,3), then
2
0 2
is the boundary of the square with vertices (1,1), (3,1), (3,3),
Mdx Ndy ?
By Green’s theorem we have that
Mdx Ndy
N
dx
R
M
dxdy
dy
3 3
1
1
x dxdy
x dxdy
y
y
1 1
R
Mdx Ndy
3 3
1 1
1
dxdy
y
3
2ln y x
10. Suppose
N M
5
x y
3
1 1
1
x dxdy
3
2
1
3 3
3
3
3
1
1
1
dy dx dy x dx
y
1
2ln3 8
1
and region R has an area of 2 and has boundary
that is a simple closed curve.
Mdx Ndy ?
Then
By Green’s theorem,
Mdx Ndy
11. Suppose
N
dx
R
F is a gradient field, and so
M
dxdy
dy
R 5dxdy 5R dxdy 5 2 10 .
F g , where g x, y, z x2 cos y sin( z y 2 ) . If
r t 5 cos t, 3 sin t, 2 cos t
12. If
is the circle
with
0 t 2 , then
is the ellipse parametrized by
F dr =
F dr 0 over any closed path such as .
x2 y 2 9 , then
y dx x dy = ?
2
2
y2 M
about the given circle of radius 3.
This integral is the circulation of F ( x, y)
x 2 N
By Green’s theorem,
N M
Mdx Ndy
2 x 2 y dxdy 2 x y dxdy .
dxdy
dx
dy
R
R
R
Changing to poplar coordinates, we get
2 3
0
2 3
r cos r sin r drd
0
2
13. If F ( x, y)
interior of
0
3
0
cos d r dr
r cos drd
2
0 0
2
0
2
2 3
0 0
r 2 sin drd
3
0
sin d r 2 dr 0
M
2 and is the unit circle centered at the origin. If R is the circular
x2 y 2 N
N M
, then
= ?
[ Hint: this time it’s easier to compute the line integral! ]
x y
x2 y 2
dA
R
By Green’s theorem,
N
dx
R
M
dxdy Mdx Ndy .
dy
We will use the unit-circle parametrization,
cos t
, from t = 0 to t =
sin
t
r (t )
2
1
and F (r (t )) , for all t, we get
1
Mdx Ndy sin t cos t dt 0 .
0
y2
14. Compute the flux of the field F ( x, y, z ) x 2 through the portion of the surface
z
above (or below) the rectangle in the xy-plane with corners at (0,0), (3,0), (3,2), (0,2).
The flux is given by the integral
described by
sin t
cos t
2 . Since r '(t )
S F nˆ d .
Let
z y 2 x2
G( x, y, z) x2 y 2 z , so the S is the surface
G( x, y, z) 0 . Then,
d
G
G kˆ
dxdy
F nˆ F ( x, y, y 2 x 2 )
2 x, 2 y,1
1
dxdy 4 x2 4 y 2 1 dxdy , and
2x
1
2 y
4 x2 4 y 2 1 1
y2
1
x2
4 x2 4 y 2 1 2 2
y x
2x
2 y
1
that is
2 xy 2 2 x2 y y 2 x2
.
F nˆ
4 x2 4 y 2 1
Thus,
2 3
S F nˆ d
0 0
2 3
2 xy 2 2 x2 y y 2 x2
4 x2 4 y 2 1 dxdy
4 x2 4 y 2 1
0
2 xy 2 2 x 2 y y 2 x 2 dxdy
2 3
2
2 3
0
2 3
2 3
0 0 xy dxdy 20 0 x ydxdy 0 0 y dxdy 0 0 x dxdy
2
2
0
2
3
2 y dy
2
2
2
3
2
2
3
2
3
0 xdx 20 ydy0 x dx 0 y dy 0 dx 0 dy 0 x dx
2
y3
2
3
2
2
0
3
2
0
0
y2
x2
2
2
2
2
2
3
2
0
0
y3
x3
3
3
x3
3 2
3
3
0
23 32
22 33 23
33
2
3 2 22
3 2
2 3 3
3
yz
15. Compute the divergence of the vector field F ( x, y ) xz .
xy
x yz
F ( x, y) xz 0 0 0
y
xy
z
0
yz
16. Compute the outward flux of F ( x, y ) xz through the surface x2 2 y 2 3z 2
xy
Since the divergence of this field is identically zero, the divergence theorem gives us:
0dV 0 .
F nˆ d FdV
V
S
V
yz
17. Compute the curl of the vector field F ( x, y ) xz .
xy
1 .
x yz
F ( x, y) xz
y
xy
z
y
z
x
,
xy
z
xz
x
,
xy
y
yz
yz
xz
x x, y y, z z 2 x, 0, 2 z
yz
18. Compute the circulation of F ( x, y ) xz around the square curve
xy
(2,2,1), (0,2,1).
with corners (0,0,1), (2,0,1),
F dr ?
By Stoke’s theorem,
F dr S F nˆ d , S being the interior of the given square.
0
parallel to the xy-plane, it follows that n̂ 0 , and that d
1
the results from problem #17, the above integral becomes
Since S is
dxdy . Since z=1 on this square, and using
S F nˆ d S (2x,0, 2 1) (0,0,1) d
2 2
2 dxdy 8
0 0
19. Suppose that
xyz
F g with g ( x, y, z)
. Then
F ?
1 x6 y8 z10
The curl of any gradient field is zero:
F 0
20. Compute the circulation of the field F ( x, y) featured in #19, around the square curve
(0,0,1), (2,0,1), (2,2,3), (0,2,3).
F dr ?
The circulation around any closed curve is zero, for any gradient field:
F dr 0
with corners
0
