`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...
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
so`p(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
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