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 $4\pi - \pi(\sqrt{x})^2$?

  • $\begingroup$ Much appreciated $\endgroup$ – 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 $\dfrac{224\pi}{15}$.

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    $\begingroup$ 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. $\endgroup$ – Frentos Feb 3 '16 at 4:05
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    $\begingroup$ I've done the diagram but @!#% math.SE is displaying its image-loading popup in such a way that the submit button is off-screen and due to it being a scripting abomination, I can't even tab to it. $\endgroup$ – Frentos Feb 3 '16 at 4:23
  • $\begingroup$ Thanks for the help, I see what my error was $\endgroup$ – Scott Feb 3 '16 at 4:34

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