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I want to calculate the volume of the solid determined by this tho surfaces: $$S_1=\{(x,y,z)\in\mathbb{R}:x^2+y^2+z^2=R^2\}$$ $$S_2=\{(x,y,z)\in\mathbb{R}:x^2+y^2=Rx\}$$ The solid is the intersection of a sphere of radius $R$ ($S_1$) and a cylinder of diameter $R$ (centered in $(R/2,0,0)$)($S_2$)

I guess i must change to spherical or cylindrical coordinates, and that's what i have problems with. I'm stucked in finding the new values of the variables. Also, which coordinate system will work better for this problem? Spherical or cylindrical? I will thank any help.

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I use cylindrical coordinate to obtain the answer.

For $S_2$, if $x^2+y^2=Rx$ a, then we have $r^2=Rr\cos \theta$, hence $r=R\cos \theta$.

The intersection region involves the first and fourth quadrant.

Hence, we want to evaluate

\begin{align} \int_{-\frac{\pi}2}^\frac{\pi}2 \int_0^{R\cos \theta} \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} r\, dz \,dr \, d\theta \end{align}

By using symmetry, we can simplify the expression to

\begin{align} &4\int_{0}^\frac{\pi}2 \int_0^{R\cos \theta} \int_{0}^{\sqrt{R^2-r^2}} r\, dz \,dr \, d\theta \\ &= 4\int_{0}^\frac{\pi}2 \int_0^{R\cos \theta} r\sqrt{R^2-r^2} \,dr \, d\theta \\ &=-2\int_{0}^\frac{\pi}2 \int_0^{R\cos \theta} (-2r)\sqrt{R^2-r^2} \,dr \, d\theta \\ &=-\frac43\int_{0}^\frac{\pi}2 \left[(R^2-r^2)^\frac32 \right]_0^{R\cos \theta} \, d\theta \\ &= - \frac43 \int_0^\frac{\pi}2 (R^3\sin^3 \theta - R^3) \, d\theta \\ &= \frac{4}{3}R^3 \int_0^\frac{\pi}2 (1-\sin^3 \theta) \, d\theta \\ &=\frac{4}{3}R^3 \int_0^\frac{\pi}2 (1-\sin \theta(1-\cos^2\theta)) \, d\theta \\ &= \frac{4}{3}R^3 \int_0^\frac{\pi}2 (1-\sin \theta- (-\sin \theta)\cos^2\theta) \, d\theta \\ &= \frac43 R^3\left[ \theta +\cos \theta- \frac{\cos^3 \theta}{3}\right]_0^\frac{\pi}2 \\ &= \frac43 R^3\left[\frac{\pi}2-1+\frac13 \right] \\ &= \frac{2(3\pi-4)}9 R^3 \end{align}

Remark: In the event that $R$ is not specified to be a nonnegative quantity such as radius or diameter, that is if $R$ can take negative value, by symmetry, the answer is $\frac{2(3\pi-4)}9 |R|^3$

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  • $\begingroup$ Are you sure your answer is correct? Other user solved it too and he gets your same result but multiplied by $1/2$. Is it 100% correct? $\endgroup$ – Alejandro Bergasa Alonso Jun 1 at 18:30
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    $\begingroup$ If we project it to the $x-y$ plane. part of the region is in the fourth region and part of the region is in the first quadrant. Hence the angle should be from $-\frac{\pi}2$ to $\frac{\pi}2$ right? $\endgroup$ – Siong Thye Goh Jun 1 at 18:38
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Alejandro Bergasa Alonso Jun 1 at 18:48
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Note that $S_2$ is cylinder $$ (x-\frac{R}{2})^2+ y^2 = (\frac{R}{2})^2 $$

We will use cylindrical coordinate : Here the curve $ z=0,\ (x-\frac{R}{2})^2+ y^2 = (\frac{R}{2})^2$ is parametrized by $$x=r\cos\ \theta,\ y=r\sin\ \theta,\ r=R\cos\ \theta,\ -\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}$$

That cylinder has top and bottom roofs : $$z=\sqrt{R^2-x^2-y^2},\ z=-\sqrt{R^2-x^2-y^2}$$

Hence \begin{align*} dxdy &= rdrd\theta \\ V&= 2\int_0^{\frac{\pi}{2}}\int^{R\cos\ \theta}_0\ z \cdot r drd\theta \\&=2\int_0^{\frac{\pi}{2}}\int^{R\cos\ \theta}_0\ \sqrt{R^2-r^2} \cdot r drd\theta \\&=2\int_0^{\frac{\pi}{2}}\int^{R^2\sin^2 \theta}_{R^2}\ \sqrt{T} \frac{dT}{(-2)} d\theta \\&= \int_0^{\frac{\pi}{2}} \ (\frac{-2}{3}) R^3 \{ \sin^3 \theta -1\}\ d\theta\\&= \frac{3\pi -4}{9} R^3 \end{align*} since $\int\ \sin^3\theta = -\frac{1}{3}\cos\ \theta(\sin^2\theta+2)$

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