Page 204 - How To Solve Word Problems In Calculus

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4
2 2 5/2
= 2π y − y
5
0

2
= 2π 16 − 4 5/2 − 0
5

64
= 2π 16 −
5

16
= 2π
5
32
= π
5
The nature of the shell method makes it well suited for
finding volumes of “hollow” solids.

EXAMPLE 9
Find the volume obtained if the region bounded by
2
y = x and y = 2x is rotated about the x axis. (This problem
was solved previously using the washer method. See Example 7
for comparison.)

Solution

h
(x , y) (x , y)
(2, 4) 1 2
4 4
y
h = x − x
2 1
r = y
(x,y)
1 (x,y)
2

2 2

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