Q:

Volume of solids of revoution

Let $R$ be the region bounded by the line $x = 6$ and the curve $9x^2 - y^2 = 9$ . Find $h(y)$ so that integral from $a$ to $b$ is $h(y)dy$ represents the volume of the solid resulting from revolving about the line $x=-9$. 
I already found $a$ and $b$ which are
$a = - \sqrt {315}$
$b = \sqrt {315}$
I need help setting up the problem to find $h(y)$ which is like $a(y)$.
Can anyone help? Thanks

A:

Hint: If you use the "washer" method, the outer radius should be $R = 6-(-9)=15$ and the inner radius should be $r = \sqrt{1 + (y/3)^2} - (-9) = 9 + \sqrt{1+(y/3)^2}$. 

