There are a couple things I'm unclear on regarding the disc method of approximating shape volume. Given $x=y^2$ and $x=4$, determine the approximate volume by revolving the shape around the line $x=5$. This particular question contained a couple curveballs and I ended up checking the solution work to see where I went wrong, and there are a couple things I'm not clear about.
First, the solution gives two $R$ formulas, while I thought there was to be only one, since the line x=4 is a boundary. Second, I was clearly off when it came to the bounds of the definite integral.
My original integral was $\pi\int_0^4(5-y^2)^2 dy$
However, the solution integral was
$\pi\int_{-2}^2(5-y^2)^2-1 dy$
and as you can see I was a little off. The solution integral follows the formula
$\pi\int_a^b(R(y))^2-(r(y))^2 dy$
I got the first R formula, but why is the r(y) represented as 1 instead of zero? Since this shape is revolving around a coordinate axis, why is it necessary to subtract a secondary radius?
Also, why are the bounds of the integral different in the book solution than my answer? Is this an issue that is essential to solving $\int_a^b(R(y))^2dy$ problems?