`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.

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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...

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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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