# Thread: Solid of Revolution about x = 4

1. ## Solid of Revolution about x = 4

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?

2. 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?
You are correctly finding the surface area of washers and integrating along the y-axis.
Your first line doesn't really require the y in brackets part.
You just need to know the two radii, which you show you do.

3. Seems good to me.

4. Awesome thanks very much for the quick responses. Wasn't sure if I had gone about this the right way.