The solid lies between planes perpendicular to the x -axis at x = -1 and x = 1. The cross sections perpendicular to the x - axis between these planes are squares whose bases run from the semicircle 1−x2 to the semicircle y=1−x2.

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Khribechy · Accepted Answer

This problem is a continuation of Exersice 1(b) so we can use the area function we wrote for that exercise. In Exercise 1(b), we got an are function of A(x)=4(1−x2)  The bounds of the region are x=-1 and x=1 so integrate the area function from -1 to 1 find the volume  V=∫−11A(x)dx  =∫−114(1−x2)dx  =4(x−13x3)∣−11  =4(1−13(1)3)−4(−1−13(−1)3)  =4(1−13)−4(−1+13)  =4(23)−4(−23)  =83+83  =163  Result: 163

The solid lies between planes perpendicular to the x -axis at x = -1 and x = 1.

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