Originally Posted by

**blackhug** I have the following question:

Determine the volume of the solid when the region bounded by $\displaystyle y = \sqrt{x}$ and the lines $\displaystyle y = 2$ and $\displaystyle x = 0$ is rotated about the line $\displaystyle x = 4$

I went about it as follows:

$\displaystyle x = y^2$ therefore...

$\displaystyle \int_0^2 \pi R(y)^2dy - \int_0^2 \pi r(y)^2dy$

$\displaystyle \int_0^2 \pi4^2 dy - \int_0^2 \pi(4 - y^2)^2 dy$

$\displaystyle \pi[16y]_0^2 - \pi[16y - \frac{8}{3}y^3 + \frac{1}{5}y^5]_0^2$

giving me a final answer of $\displaystyle \frac{224\pi}{15}$

Have I done anything wrong?