y = 9-x^2 , y=0 Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

Given the curves
y=9-x^2,
y=0
we have to find the volume using the shell method.
so , the volume of vertical rotation is given as
V=2*pi int_a^b p(x)h(x) dx
where p(x) is a function of  average radius and h(x) is a function of height
so as the solid is rotating with respect to y=0 ie x axis
sop(x) =x
and height h(x) = 9-x^2
now let us find the range of x on the x axis by the intersection of the curves  y=9-x^2 and y=0
=> 0=9-x^2
=> x= +-3
now the volume is =2*pi int_a^b p(x)h(x) dx
 
= 2pi int_-3^3 (x)(9-x^2) dx
=4pi int_0^3 (9x-x^3) dx
=4*pi *[(9x^2)/2-x^4/4]_0 ^3
= 4*pi *[[(9(3)^2)/2-(3)^4/4]-[0]]
=4*pi*81/4
= 81pi
is the volume