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 * [0]
V =(25pi)/2 - 0
V =(25pi)/2 or 39.27 (approximated value)

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