1. ## Solid of Revolution about x = 4

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

I went about it as follows:

$x = y^2$ therefore...
$\int_0^2 \pi R(y)^2dy - \int_0^2 \pi r(y)^2dy$

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

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

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

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 $y = \sqrt{x}$ and the lines $y = 2$ and $x = 0$ is rotated about the line $x = 4$

I went about it as follows:

$x = y^2$ therefore...

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

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

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

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

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.