I've tried solving the following question two ways, and I got different results.
Find the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 9$ about the line $y = -2$.
1. Disc Method
We have that the spin axis is $-2$, therefore:
$$\int_0^9\pi(\sqrt{x}+2)^2\, dx = \frac{297}{2}\pi.$$
2. Shell Method
Since we are using the shell method that means, our width is $dy$. Solving for $x$ we have $x = y^2$ and now we find the intercepts by setting $y^2 = 9$, which are $3$ and $-3$. Therefore, we have
$$\int_0^3 2\pi (9-y^2)(y+2)\, dy= \frac{225}{2}\pi.$$
I would like to know what I did wrong and how can I fix it. Thank you in advance.
Best Answer
For the disk, you missed that you have a hole of radius $2$ in the middle of the disk. That removes a cylinder of volume $\pi 2^2\cdot 9=36\pi$, which makes it agree with the shell method.