MAT 254 – Fall Quarter 2002
Test 3 - Answers
NAME_______________________________________________________
Show work and write clearly.
1. (20 pts.) Find the length of the curve y  ln(sin x) ,
ANS: y  

 x
6

3
.
1
cos x  cot x .
sin x
Arc length, s 
 /3

/6
1  cot 2 xdx 
 /3

csc 2 xdx 
/6
 /3

/6
csc xdx . This integral can be found
b
using a table of integrals (  csc xdx  ln csc x  cot x  C ). For this test, you were permitted to
a
use Simpson’s rule with n = 20 to estimate the integral (0.7677).
2. (20 pts.) Find the area of the surface of revolution obtained by rotating the curve
x3
1
y 

, 1  x  2 about the x-axis.
6
2x
x2
1
 2.
ANS: y  
2
2x
Surface area, SA = 2  f ( x) 1   f ( x)  dx  2 
2
2
1
2
2
1
 x3
 x2
1 
1 

 1  

 2  dx
2x 
2x 
 6
 2
3
3
2 x
2 x
 x4 1
1 
1 
1  x4 1
1
 1  

 2  

  4 dx  2  

  4 dx
1
1
2x 
2 4x 
2x  4
2 4x
 6
 4
 6
 2 
2
1
 x3
1 



2 x 
 6
2
3
2 x
 x2
1 
1  x 2
1 




dx

dx

2



2 


1
2 x  2
2x 
2 x 2 
 2
 6
5
2 x
 x6
x
x
1 
x2
1 
47
 2  

  3 dx  2 

 2  12 

1
6
16
8x 
 12 12 4 4 x 
 72
3. (20 pts.) Find the volume of the solid formed by revolving the region bounded by x 
and y = 8, x = 0 about x = 2. Sketch the area.
ANS:




V  2  p( x)h( x)dx  2  2  x  8  x 3 dx  2  16  8x  2 x 3  x 4 dx
2
2
0
0

x4
x 5  2 144
0 
 2 16 x  4 x 2 


4
5 
5

2
0
3
y
4. (20 pts.) Solve the initial value problem: y   2 xy  2 x 3 y , y(0) = 3.
ANS:
dy
dy
 2 x 3 y  2 xy  2 y ( x 3  x) 
 2( x 3  x)dx 
dx
y
1
 ln y 
1
dy
 2 ( x 3  x)dx
y
1
x4  x2  C
x4  x2
x4  x2
1 4
x  x2  C  y  e 2
 e2
e C  Ce 2
.
2
When x = 0, y = 3  3  Ce  C  3 . Thus, y  3e
0
5. (20 pts.) Solve the differential equation: y  xe y 
ANS:
dy
xe y e
dx

x
dy y
 1 
e 
dx
1
xe
x

e
x
x
x
1 4
x  x2
2
 1 .
 e dy 
y
.
 e
x
x
dx 
e
To find the integral of the left hand side use substitution: let u   x , du 
y
dy 
1
2 x



x
dx .
x
dx . So, we
have
y
u
y
u
y
u
u

 e dy  2 e du  e  2e  C  e  2e  C  y  ln 2e  C  ln 2e

 e
x
C
