642
Chapter 6 | Applications of Integration
Since the solid was formed by revolving the region around the x -axis, the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by f ( x ). Use the formula for the area of the circle: A ( x ) = πr 2 = π ⎡ ⎣ f ( x ) ⎤ ⎦ 2 = π ⎛ ⎝ x 2 −4 x +5 ⎞ ⎠ 2 (step 2).
The volume, then, is (step 3)
V = ∫ a b
A ( x ) dx
4
4 ⎛ ⎝ x 4 −8 x 3 +26 x 2 −40 x +25 ⎞ ⎠ dx
π ⎛ ⎝ x 2 −4 x +5 ⎞ ⎠ 2
= ∫
dx = π ∫
1
1
⎞ ⎠ | 1 4
⎛ ⎝ x 5 5
−2 x 4 + 26 x 3
x 2 +25 x
= π
= 78 5
π .
3 −20
The volume is 78 π /5.
6.7 Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function f ( x ) =1/ x and the x -axis over the interval [1, 2] around the x -axis. See the following figure.
The Disk Method When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function f ( x ) = ( x −1) 2 +1 and the x -axis over the interval [−1, 3] around the x -axis. The graph of the function and a representative disk are shown in Figure 6.18 (a) and (b). The region of revolution and the resulting solid are shown in Figure 6.18 (c) and (d).
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