664
Chapter 6 | Applications of Integration
Solution First, graph the region R and the associated solid of revolution, as shown in the following figure.
Figure 6.33 (a) The region R between the graph of f ( x ) and the graph of g ( x ) over the interval [1, 4]. (b) The solid of revolution generated by revolving R around the y -axis.
Note that the axis of revolution is the y -axis, so the radius of a shell is given simply by x . We don’t need to make any adjustments to the x -term of our integrand. The height of a shell, though, is given by f ( x )− g ( x ), so in this case we need to adjust the f ( x ) term of the integrand. Then the volume of the solid is given by V = ∫ 1 4 ⎛ ⎝ 2 πx ⎛ ⎝ f ( x )− g ( x ) ⎞ ⎠ ⎞ ⎠ dx
4 ⎛
4 ⎛
⎛ ⎝ x − 1 x
⎞ ⎠ ⎞ ⎠ dx =2 π ∫
⎝ x 3/2 −1 ⎞
= ∫
⎝ 2 πx
⎠ dx
1
1
⎤ ⎦ | 1 4
⎡ ⎣ 2 x 5/2 5
= 94 π 5
units 3 .
=2 π
− x
6.16 Define R as the region bounded above by the graph of f ( x ) = x and below by the graph of g ( x ) = x 2 over the interval [0, 1]. Find the volume of the solid of revolution formed by revolving R around the y -axis.
Which Method Should We Use? We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. Figure 6.34 describes the different approaches for solids of revolution around the x -axis. It’s up to you to develop the analogous table for solids of revolution around the y -axis.
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