MAT 254 – Winter Quarter 2002
Test 4 – Answers
NAME_______________________________________________________
Show work and write clearly.
Make sure graphs show approximating rectangle, R(x), r(x), p(x) and h(x).
1. (30 pts.) Use the disk method to find the volume of the solid obtained by rotating the region
bounded by x  y  y 2 and x  y 2  3 about x = – 4.
ANS:
The purple function is x  y  y 2 and the red function is x  y 2  3 . To find the intersection points
set the functions equal to one another, y  y 2  y 2  3 and use the quadratic formula: y = –1, 3/2.
The outer radius, Ro or R(y): the function furthest to right minus the axis of rotation, i.e.,
y  y 2  (4)  y  y 2  4 .
The inner radius, Ri or r(y): the function to left minus the axis of rotation, i.e., y 2  3  (4)  y 2  1 .
So, V = 
 R( y)
3/2
2

 r ( y ) dy  
2
1
 y  y
3/2
2
4
  y
2
2

2
 1 dy . Be careful to distribute fully
1
the functions when squaring them:
V 
  7 y

3/2
2
 2 y 3  8 y  y 4  16  ( y 4  2 y 2  1) dy
1
3/2
 y4

875
 3 y 3  4 y 2  15 y 


    2 y  9 y  8 y  15dy    
2
32


1
1
3/2
3
2
2. (30 pts.) Find the following integrals:
 tan
a.
1

x dx
ANS:
So,
U
tan 1 x
dV
1
1
1  x2
x
 tan
 tan
1
1

x dx = x tan  1 x 

x dx = x tan 1 x 
b.
 x
4
x
1 x
2
dx . Now use substitution, let t = 1 + x2, dt = 2xdx. So,


1 1
1
1
dt  x tan 1 x  ln t  C  x tan  1 x  ln 1  x 2  C .

2 t
2
2

ln x dx
ANS:
So,
 x
U
ln x
dV
x4
1
x
(1/5) x5
4
1
ln x dx = x
5
 x e
1
c.
3 x2
5
1 x5
1
1
1
1 1
ln x   dx  x 5 ln x   x 4 dx  x 5 ln x    x 5  C .
5 x
5
5
5
5 5
dx
0
ANS: Let t = x2, dt = 2xdx.
1
1
1
1
1
3 x2
2 t
So,  x e dx   x e dt   tet dt .
20
20
0
U
dV
et
t



 x e dx
1
1
3 x2
0
 
et
1
So,


 

1
1
tet dt  tet  e t

20
2

1
0

1
.
2
3. (30 pts.) Use the shell method to find the volume of the solid obtained by rotating the region
bounded by y  x 2 , y = 1 and x = 2 about y = –3.
ANS:
1

 2  6  2 y  3 y
2

 2  6  1  2  
5

1
0
0


y  y  3dy
1

2


 y
dy  2  6 y  y 2  2 y 3 / 2  y 5 / 2 
5

0
46
0
.
5
1/ 2
0
1
y . So, V = 2   p( y)h( y) dy  2  2 
h( y)  2  x  2 
4. (10 pts.) Solve: y  
3/ 2
xy
.
3 ln y
dy
xy
ln y
1
ln y
1


dy  xdx  
dy   xdx. . To find the integral on the left hand
dx 3 ln y
y
3
y
3
side, use substitution, let u = ln y, du = dy/y. So,
ln y
1
1
u2
1 x2
1
2
dy

xdx

udu

xdx


 C  u 2  ln y   x 2  C
 y



3
3
2
3 2
3
ANS:
 ln y 
1 2
x  C  e ln y  e
3
1 2
x C
3
 y e
1 2
x C
3
.