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A solid lies between planes perpendicular to the y-axis at $ y=0$ and $y=1$. The cross-sections perpendicular to the y-axis are circular disks with diameters running from the y-axis to the parabola $x=\sqrt{11}y^2$. Find the volume of the solid.

The wording of this question really throws me off, should my integral end up being $\int_0^1 π(\sqrt{11}y^2)^2$. Or am I doing this completely wrong?

Thank you for your input!

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    $\begingroup$ I'd be curious to see someone make a nice Mathematica plot of this. $\endgroup$ – Semiclassical Jul 19 '14 at 3:42
  • $\begingroup$ Wait what does that mean? $\endgroup$ – Jerry Jul 19 '14 at 3:47
  • $\begingroup$ As in, make a picture in the program Mathematica that displays this solid. $\endgroup$ – Semiclassical Jul 19 '14 at 3:47
  • $\begingroup$ Oh okay, do you know if my setup is right? Or am I completely wrong? $\endgroup$ – Jerry Jul 19 '14 at 3:48
  • $\begingroup$ @Semiclassical, is this the plot you are looking for? I've used the equation $$\sqrt{(x-\sqrt{11}y^2/2)^2+z^2}=\sqrt{11}y^2/2$$ $\endgroup$ – cutculus Jul 19 '14 at 4:12
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The diameter runs from the $y$-axis to the parabola $x=\sqrt{11}y^2$. So the radius of cross-section at height $y$ is $\frac{\sqrt{11}y^2}{2}$.

From this radius, you can calculate the area $A(y)$ of cross-section at height $y$, and then find the volume in the usual way.

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