# `y = 9-x^2 , y = 0 , x = 2 , x = 3` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis. Aside from Shell method,we may apply the Washer method using the formula:

`V =pi int_a^b ( (f(y))^2 -(g(y))^2) dy`

where `f(y)` as function of the outer radius

`g(y)` as function of the inner radius

In washer method, the rectangular strip is perpendicular to the axis of rotation.

For this problem,  we may  let:

Boundary values of y: ` a=0 ` to` b=5` .

Note: the highest value of y for the bounded region is along `x=2` .

Plug-in `x=2` on `y =9-x^2` , we get `y=9-(2)^2=9-4 =5` .

Inner radius: g(y) = 2-0 = 2

Outer radius: `f(y) = sqrt(9-y) -0 = sqrt(9-y)`

Note: We can rearrange` y = 9-x^2` as `x^2 =9-y.` Take the square root on both side to express it as `x= sqrt(9-y)`

Then, wee may set-up the integral as:

`V =pi int_0^5 ( (sqrt(9-y))^2 -(2)^2) dy`

Simplify:

`V =pi int_0^5 ( 9-y -4) dy`

`V =pi int_0^5 ( 5-y ) dy`

Apply basic integration property: `int (u-v)dy = int (u)dy-int (v)dy` .

`V =pi*[ int_0^5 ( 5)dy-int_0^5(y ) dy]`

`V = pi * [5y -y^2/2]|_0^5`

Apply definite integration formula: `int_a^b f(y) dy= F(b)-F(a)` .

`V = pi * [5(5) -(5)^2/2]-pi * [5(0)-(0)^2/2]`

`V = pi * [25 -25/2]-pi * [0-0]`

`V = pi * [25/2]-pi * `

`V =(25pi)/2 - 0`

`V =(25pi)/2` or `39.27` (approximated value)

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Approved by eNotes Editorial Team 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 function`y=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`

Approved by eNotes Editorial Team