Use the cylindrical shell method to compute the volume obtained by rotating the region under y=9-x^2 and over [0, 3] about the x-axis.

Answer 1

Although it is much easier to calculate this volume by the disk method, it is feasible to use the shell method as well.

First, one must realize that the element of area that is rotated to generate a cylindrical shell must be parallel to the axis of rotation. Since this axis is the x-axis, y must be constant along the element of area. The element will run from x = 0 to x = f^-1(y), where f is the given function 9 - x². Solving y = 9 - x² for x, we have x = √(9 - y). Therefore, the volume of revolution is given by 2π∫y√(9 - y)dy, where the limits of the integral are 0 and 9. I was able to integrate this rather difficult-looking function by using the substitution y = 9 sin²θ, dy = 18 cosθ sinθ, and the integral becomes 972π∫sin³ θ cos² θ dθ. The limits of the new integral are 0 and π/2. Replacing sin² θ with (1 - cos² θ) enables me to write the integrand as sin θ times a polynomial in cos θ, and to integrate the function to -(1/3) cos³ θ + (1/5) cos⁵ θ + C. Since the integral is from 0 to π/2, and is multiplied by 972π, the final answer is 972π ((1/3) - (1/5)) = 648π/5 cubic units.