Determining the volume from a 2D graph Problem:

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line $y=8$.
$$\begin{align}y=\sqrt{4-x}\\x=0\\y=0\end{align}$$

My graph:

Edit: the equation is y = $\sqrt{4-x}$ labelled wrong on the graph.
I know that $R(x)$ is $8$, and $r(x) = 8 - \sqrt{4-x}$, so shouldn't the integral be:

I've been stuck on this one for a while, so help would be greatly appreciated.
 A: It's not clear what area you are trying to find. Is it the shaded purple area? If so, then you solve it by simple integration. Also you're integral limits are not correct, going to x = 2 would cut short before the line reaches y = 0. I believe you intend for it to go until x = 4, which would represent the shaded area in your graph:
$$\int_{0}^{4}{\sqrt(4-x)dx} = -\frac{2}{3}(4-x)^{3/2}\Big|_0^4$$
$$-\frac{2}{3}((0)^{3/2} - (4)^{3/2}) = \frac{16}{3} = 5.333$$
Note: $dx$ is $-1$ here, so you add a negative to the front to equalize. You could also do a $u$ substitution here, which would give the same result.
A: The volume of revolution of the shaded region is the volume of the appropriate cylinder minus the volume of revolution of the radius $r(x)$
$$
\pi 8^2 h - \int_0^h \pi r(x)^2 dx 
$$
$h$ is the 'height' of the cylinder, defined by where your curve intersects the x-axis ($x=4$).
A: The curve is $y = \sqrt{4-x}$ and at the intersection of the curve and x-axis, $y = 0 = \sqrt{4-x} \implies x = 4$. That is your mistake. So the integral should be,
$V =  \displaystyle \pi \int_0^4 \left (64 - (8 - \sqrt{4-x})^2 \right) ~dx$
Also note that $y = \sqrt{4-x}$ can be written as $x = 4-y^2$. Now you can use the shell method where we take the horizontal shells and width of shell at any $y$ is $4 - y^2$. Now the volume element of the shell as we rotate about $y = 8~$ is $~dV =  2 \pi \cdot (8-y) \cdot (4-y^2) ~ dy$. As $0 \leq y \leq 2$, the integral becomes,
$\displaystyle V = \int_0^2 2 \pi \cdot (8-y) \cdot (4-y^2)  ~dy$
