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
Thursday, February 28, 2019
Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 32
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