Question

The region bounded below by the plane with equation $z=c$and bounded above by the sphere with equation $x^2+y^2+z^2=R^2$where $c,R$ are constants such that$0<c<R$

Short answer

Volume$V=\left(\frac{1}{3}\pi (R-c)\right)\left(2R^2-cR-c^2\right)$

Step-by-step solution

Given Information

The given equations are $z=C$and $x^2+y^2+z^2=R^2$.

$0<c<R$

Simplification and limit evaluation

We know the relation as:

$x=\rho \sin \varphi \cos \theta ,y=\rho \sin \varphi \sin \theta ,y=\rho \cos \varphi$

and

$\rho =\sqrt{x^2+y^2+z^2},\tan \theta =\frac{y}{2},\cos \varphi =\frac{2}{\rho },dxdydz={\rho }^2\sin \varphi d\rho d\varphi d\theta$

For the given equation $x^2+y^2+z^2=R^2$, in terms of spherical coordinates, $\rho =R$

For equation $z=c$, in terms of spherical coordinates, $z=\rho \cos \varphi =c\Rightarrow \rho =c\sec \varphi$

Therefore the limits are:

role="math" localid="1652362819417" $0\le \theta \le 2\pi ,0\le \varphi \le {\cos }^{-1}\left(\frac{c}{\rho }\right),\frac{c}{\cos \varphi }\le \rho \le R$

Calculation of Volume

The volume is evaluated as:

$V={\iiint }_{E}dxdydz$

Mentioning limits

$={\int }_{\theta =0}^{2\pi }{\int }_{\varphi =0}^{{\cos }^{-1}(c/R)}{\int }_{\rho =c/\cos \varphi }^R{\rho }^2\sin \varphi d\rho d\varphi d\theta$

$=\left\{{\int }_{\theta =0}^{2\pi }d\theta \right\}\times \left\{{\int }_{\varphi =0}^{{\cos }^{-1}(c/R)}{\int }_{\rho =c/\cos \varphi }^R{\rho }^{2}d\rho \sin \varphi d\varphi \right\}$

$={\left.\left(\theta \right)\right|}_{\theta =0}^{2\pi }\times \left\{{\left.{\int }_{\varphi =0}^{{\cos }^{-1}(c/R)}\left(\frac{{\rho }^3}{3}\right)\right|}_{\rho =c/\cos \varphi }^R\sin \varphi d\varphi \right\}$

$=\left(2\pi \right)\times \left\{{\left(\frac{1}{3}\right)}^{{\cos }^{-1}(c/R)}{\int }_{\varphi =0}\left(R^3-\frac{c^3}{{\cos }^3\varphi }\right)\sin \varphi d\varphi \right\}$

$=\left(\frac{2\pi }{3}\right)R^3\left\{{\int }_{\varphi =0}^{{\cos }^{-1}(c/R)}\sin \varphi d\varphi \right\}-\left(\frac{2\pi }{3}\right)c^3\left\{{\int }_{\varphi =0}^{{\cos }^{-1}(c/R)}\frac{c^3\sin \varphi }{{\cos }^3\varphi }d\varphi \right\}$

Simplifying further yields

$={\left.\left(\frac{2\pi R^3}{3}\right)\{-\cos \varphi \}\right|}_{\varphi =0}^{{\cos }^{-1}(c/R)}-\left(\frac{2\pi c^3}{3}\right)\left\{(-1){\int }_{\varphi =0}^{{\cos }^{-1}(c/R)}\frac{(-\sin \varphi )}{{\cos }^3\varphi }d\varphi \right\}$

$=\left(\frac{2\pi R^3}{3}\right)\left\{-\frac{c}{R}+1\right\}-\left(\frac{2\pi c^3}{3}\right)\left\{{\left.(-1)\left(\frac{1}{-2{\cos }^2\varphi }\right)\right|}_{\varphi =0}^{{\cos }^{-1}(c/R)}\right\}$

Now, we will apply the limits

Volume becomes
$V=\left(\frac{2\pi R^3}{3}\right)\left\{\frac{R-c}{R}\right\}-\left(\frac{2\pi c^3}{3}\right)\left(\frac{1}{2(c/R{)}^2}-\frac{1}{2}\right)$

$V=\left(\frac{2\pi R^3(R-c)}{3R}\right)-\left(\frac{2\pi c^3}{3}\right)\left(\frac{R^2}{2c^2}-\frac{1}{2}\right)$

$V=\left(\frac{2\pi R^3(R-c)}{3R}\right)-\left(\frac{\pi c^3}{3}\right)\left(\frac{R^2-c^2}{c^2}\right)$

$V=\left(\frac{2}{3}\pi R^2(R-c)\right)-\left(\frac{1}{3}\pi c(R-c)(R+c)\right)$

Hence,

$V=\left(\frac{1}{3}\pi (R-c)\right)\left(2R^2-cR-c^2\right)$