# Disc method of approximating volume

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?

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For the limits of integration: We are integrating with respet to $y$. So the integral is over the range of travel of $y$. For why subtract: The shape has a cylindrical hole in it. You could instead calculate without the hole, if you remember to subtract the volume of the cylinder afterwards. – André Nicolas Jul 29 '11 at 4:23

The limits come from the intersection of $x=4$ and $x=y^2$. The area to rotate is the area between those two and it goes from $x=-2$ to $x=2$. The $-1$ comes from the fact that you are revolving around the $x=5$ axis and removes the effect of revolving the rectangles with corners $(-2,4)$ and $(2,5)$, which is not included in the area you should be rotating.