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I want to find the volume of the region inside the ellipsoid $$\frac{x^2}{4}+\frac{y^2}{4}+z^2=1$$ and the cylinder $$x^2+(y-1)^2=1$$ enter image description here I tried shifting the axes so that the cylinder was centered at the origin, then evaluating an integral in cylindrical/polar coordinates. $$\int^{2\pi}_{0}\int^{1}_{0}\int^{\frac{1}{2}\sqrt{4-r^2-2rsin(\theta)-1}}_{-\frac{1}{2}\sqrt{4-r^2-2rsin(\theta)-1}}r dzdrd\theta$$ $$\int^{2\pi}_{0}\int^{1}_{0}r\sqrt{3-r^2-2rsin(\theta)} drd\theta$$

However, this integral gets messy. Is there an easier method of finding the volume?

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You can do it using ordinary cylindrical coordinates. Since the projection of that cylinder onto the plane $z=0$ is the circle centered at $(0,1)$ with radius $1$, which is the union if the first and the second quadrants, $0\leqslant\theta\leqslant\pi$. The condition$$\frac{x^2}4+\frac{y^2}4+z^2\leqslant1\quad\text{becomes}\quad\frac{r^2}4+z^2\leqslant1$$and the condition$$x^2+(y-1)^2\leqslant1\quad\text{becomes}\quad r^2-2r\sin\theta\leqslant0.$$So, compute the integral$$\int_0^\pi\int_0^{2\sin\theta}\int_{-\sqrt{1-r^2/4}}^{\sqrt{1-r^2/4}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$You should get $\dfrac83\pi-\dfrac{32}9$.

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  • $\begingroup$ Thank you, that's a lot easier than I though it would be. $\endgroup$ Oct 24 at 16:28

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