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Let $\alpha>0$. Given the cylinder $$B_1=\{(x,y,z)\in\mathbb R^3\;|\;x^2+y^2\leq\alpha x\}$$ and the sphere $$B_2=\{(x,y,z)\in\mathbb R^3\;|\;x^2+y^2+z^2\leq\alpha^2\},$$ how can one find the volume of the intersection $B_1\cap B_2$?

The following image shows the situation (for $\alpha=1$).

I tried the substition $x'=x-\frac\alpha2$ to center the coordinate system $(x',y,z)$ around the center of the cylinder and using cylindrical coordinates to get $$V=2\int_0^{2\pi}\mathrm d\varphi\int_0^{\alpha/2} \rho\,\mathrm d\rho\int_0^{z(\varphi,\rho)}\mathrm d z$$ where $z(\varphi,\rho)=\sqrt{\alpha^2-x^2-y^2}=\sqrt{\alpha^2-\left(x'+\frac\alpha2\right)^2-y^2}$ from the equation of the sphere. In cylindrical coordinates, this would be equal to $z(\varphi,\rho)=\sqrt{\alpha^2-\left(\rho\cos\varphi+\frac\alpha2\right)^2-\rho^2\sin^2\varphi}=\sqrt{\frac34\alpha^2-\alpha\rho\cos\varphi-\rho^2}$

This results in the integral $$V=2\int_0^{2\pi}\mathrm d\varphi\int_0^{\alpha/2} \rho\sqrt{\frac34\alpha^2-\alpha\rho\cos\varphi-\rho^2}\,\mathrm d\rho.$$ This integral is pretty difficult to solve, however.

Is there an easier way to solve this, by using symmetries I did not see etc.?

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  • $\begingroup$ I think the radius of the cylinder is $\alpha/2$ not $\alpha$. $\endgroup$ – Andrei May 14 at 18:44
  • $\begingroup$ Thanks, you're right. I have edited my post. $\endgroup$ – st.math May 14 at 18:45
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    $\begingroup$ @st.math This is what you need, simply replace $2$ by $\alpha \over 2$ $\endgroup$ – user376343 May 14 at 18:53
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The volume can be calculated as follow

$$Vol=\int\int_{D}\int_{-\sqrt{\alpha^2-x^2-y^2}}^{\sqrt{\alpha^2-x^2-y^2}}1 dz\,dx\,dy=\int\int_{D}2\sqrt{\alpha^2-x^2-y^2}\,dx\,dy$$

with the domain $D\equiv x^2+y^2-\alpha x\leq 0$. The double integral can be solved with polar coordenates

$$Vol=\int_{-\pi/2}^{\pi/2}\int_0^{\alpha\,\cos\theta}2r\sqrt{\alpha^2-r^2}dr\,d\theta=\left.\int_{-\pi/2}^{\pi/2}\frac{-2}{3}(\alpha^2-r^2)^{3/2}\right|_0^{\alpha\cos\theta}d\theta=$$

$$=\frac{2}{3}\alpha^3\int_{-\pi/2}^{\pi/2}1-|\sin\theta|^3 d\theta=\frac{4}{3}\alpha^3\int_{0}^{\pi/2}1-\sin^3\theta\,\, d\theta=\frac{2\alpha^3}{9}(3\pi-4)$$

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    $\begingroup$ Thank you. Your approach seems to be correct, but I think you made a mistake in the last integral. I get the result $$V=\frac29\alpha^3(3\pi-4).$$ $\endgroup$ – st.math May 15 at 17:54
  • $\begingroup$ Yes, it's true. I've corrected the mistake. $\endgroup$ – popi May 15 at 18:09

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