Math 280 B2 Exam 2
NAME:
Solution
Problem 1 Let   x 2  y 2  z 2 .
a) (7 pt) Calculate grad 1 .
solution)
1
1
2
2
2 2
  x  y  z 
1

x
1

y
1

z
1
grad 
3
 xx 2  y 2  z 2   2   x3

3
y
 yx 2  y 2  z 2   2   3

3
 zx 2  y 2  z 2   2   z3

y
  x3 i  3 j  z3 k .



b) (5 pt) Calculate curl grad 1 .
solution)
Using the property curl grad  0, curl grad
1

 0
1
Problem 2(10 pt) Evaluate  
R xy
2
x 2  y 2 dxdy, where R xy is the region in the first quadrant
bounded by x 2  y 2  1, x 2  y  9, xy  2, and xy  4 by using change of variables.
solution)
Let u  x 2  y 2 and v  xy.
Then,
1  u  9, 2  v  4,
2x 2y
u, v

x, y
x, y

u, v
 R
x 2  y 2 dxdy 
xy
y
x
1

u,v
x,y
4
9
 2x 2  2y 2  2x 2  y 2 ,
1
,
2x 2  y 2 
4
9
 2  1 x 2  y 2  2x 2 1 y 2  dudv   2  1
1 dudv  8.
2
2
1
1
Problem 3(10 pt) Evaluate  x 3 cos x 2 dx by differentiating  x sin tx 2 dx with respect to t.
0
0
solution)
1
1
1
 0 x 3 cos tx 2 dx   0 t x sin tx 2 dx  dtd  0 x sin tx 2 dx
x1
 d  1 cos t  1
 d  1 cos tx 2
2t
2t
2t
dt
dt
x0
 12 cos t  1 sin t  12 .
2t
2t
2t
Put t  1. Then,
1
 0 x 3 cos x 2 dx 
Problem 4(7 pt) Evaluate 
solution)
24,4 6
C, 3,2 3 
ds 
24,4 6
 C, 3,2
3
1 cos 1  1 sin 1  1 .
2
2
2
yds 
yds, where C is the curve y  2 x .
dx 2  dy 2 
24
3
1
1  x  2
2
dx 
1  x 1 dx,
2 x 1  x 1 dx
24
24
3
2 x  1 dx  2  2 x  1 2
3
3
3
3
 4 25 2  4 2  4 125  8  156.
3
3

3
3
Problem 5 Consider 
3,4
1,2
6xy 2  y 3 dx  6x 2 y  3xy 2 dy.
a)(4 pt) Show that the integral is independent.
solution)
P 
y
Q

x
So, P 
y
Thus, the integral is path independent.
12xy  3y 2 ,
12xy  3y 2 .
Q
.
x
b) (8 pt) Evaluate the integral. (hint: find F such that F  6xy 2  y 3 , 6x 2 y  3xy 2 . )
solution)
Note that if F  3x 2 y 2  xy 3 , then F  6xy 2  y 3 , 6x 2 y  3xy 2 .
Hence,
dF  6xy 2  y 3 dx  6x 2 y  3xy 2 dy.
3,4
 1,2 6xy 2  y 3 dx  6x 2 y  3xy 2 dy

3,4
 1,2 d3x 2 y 2  xy 3 
 3x 2 y 2  xy 3  3,4
 236.
1,2
4
Problem 6(9 pt) Evaluate 
x 2 ydxx 3 dy
C
2
x 2 y 2 
that P
y
around the square with vertices 1, 1, oriented
Q
counterclockwise. (hint: Note
 x . )
solution)
Q
 x , we can use the unit circle with center 0, 0 instead of the given curve.
Since P
y
x y 1
2

2
0
 
 
2
x 2 ydx  x 3 dy
2
x 2  y 2 
cos 2 t sin t sin tdt  cos 3 t cos tdt, x  cos t, y  x sin t.
2
0
2
0
cos 2 tsin 2 t  cos 2 tdt   
2
0
cos 2 tdt
1  cos 2t dt  .
2
5