Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 32

Let's use the shell method for finding the volume of the solid.
The volume of the solid (V) generated by revolving about the y-axis the region between the x-axis and the graph of the continuous functiony=f(x), a <= x<= b is,
V=int_a^b2pi(shell radius) (shell height)dx
V=int_a^b2pixf(x)dx
Given ,y=9-x^2 , y=0 , x=2 , x=3
V=int_2^3(2pi)x(9-x^2)dx
V=2piint_2^3(9x-x^3)dx
V=2pi[9x^2/2-x^4/4]_2^3
V=2pi{(9(3)^2/2-3^4/4)-(9/2(2)^2-2^4/4)}
V=2pi{(81/2-81/4)-(18-4)}
V=2pi(81/4-14)
V=2pi((81-56)/4)
V=2pi(25/4)
V=(25pi)/2