Calculus- Early Transcendentals, 2021a

8.3. Volume 313
Note that we can instead do the calculation with a generic height and radius:
∫ h
π r2
0 h 2 x2 dx = πr2 h 3
h 2 3 = πr2 h
3 ,
giving us the usual formula for the volume of a cone.
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Example 8.10: Volume of an Object with a Hole
Find the volume of the object generated when the area between y = x 2 and y = x is rotated around
the x-axis.
Solution. This solid has a “hole” in the middle; we can compute the volume by subtracting the volume
of the hole from the volume enclosed by the outer surface of the solid. In figure 8.7 we show the region
that is rotated, the resulting solid with the front half cut away, the cone that forms the outer surface, the
horn-shaped hole, and a cross-section perpendicular to the x-axis.
1
.
0
0 1
Figure 8.7: Solid with a hole, showing the outer cone and the shape to be removed to form the hole.
We have already computed the volume of a cone; in this case it is π/3. At a particular value of x, say
x i , the cross-section of the horn is a circle with radius x 2 i , so the volume of the horn is
∫ 1
0
π(x 2 ) 2 dx =
∫ 1
0
πx 4 dx = π 1 5 ,