# Find the volume of the solid generated by revolving the region described below about $x=4$

Region bounded by $y=\sqrt{x}$, $y=2$ and $x=0$.

I got the answer as $8\pi$, is that correct?

This was a quiz question and I was marked wrong but I am thinking it was because of how I derived the answer. Is the radius supposed to be $r=\pi(\sqrt{x})^2$ or $4\pi - \pi(\sqrt{x})^2$?

• Much appreciated – Scott Feb 3 '16 at 3:42

If you're doing the integral with horizontal discs, it's $$\pi\int_0^24^2-(4-x)^2~dy$$ where $$4-x$$ is the radius of the inner disc you're removing. Since $$y=\sqrt x$$, $$x=y^2$$, so the integral is $$\pi\int_0^24^2-(4-y^2)^2~dy=\int_0^24^2-(4^2-8y^2+y^4)~dy$$ which pops out as $$\dfrac{224\pi}{15}$$. • You're rotating about the line $x=4$. You don't have a radius for any $y$ value: you have two radii: the outer radius of the disc (which is always 4, going from $x=0$ to $x=4$) and the inner radius (which is $4-x$, going from $x$ to $x=4$). Give me a minute and I'll add a diagram. – Frentos Feb 3 '16 at 4:05