Question

The region bounded below by the $xy$-plane, bounded above by the sphere with radius $2$and centered at the origin, and outside the cylinder with equation$x^2+y^2=1$

Short answer

The volume is$2\sqrt{3}\mathrm{\pi }$units.

Step-by-step solution

Given Information

Radius if sphere is $2$and is centered at origin and outside cylinder with equation $x^2+y^2=1$

Simplification and evaluation of limits

Relationship between cylindrical and rectangular coordinates is given by

$r=\sqrt{x^2+y^2},\tan \theta =\frac{y}{x},z=z$

and

$x=r\cos \theta ,y=r\sin \theta ,z=z$

Equation of sphere in terms of rectangular coordinates is $x^2+y^2+z^2=4$

In terms of cylindrical coordinates, equation is $r^2+z^2=4$

$\Rightarrow z=\sqrt{4-r^2}$

Region in $xy$plane above which lies the surface is

$x^2+y^2=1\Rightarrow x^2+y^2=2y$

And

$r^2=2r\sin \theta ,\Rightarrow r=2\sin \theta$

Limits in cylindrical coordinates are

$2\sin \theta \le z\le \sqrt{4-r^2},1\le r\le 2,0\le \theta \le \mathrm{\pi }$

Evaluating the volume

Required volume is given by

$V={\int }_0^{2\mathrm{\pi }}{\int }_0^2{\int }_{2\sin \theta }^{\sqrt{4-r^2}}rdzdrd\theta$

$\Rightarrow V={\int }_0^{2\mathrm{\pi }}{\int }_0^{2}r\left(\sqrt{4-r^2}-2\sin \theta \right)drd\theta$

$\Rightarrow V=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}{\left[\frac{{\left(4-r^2\right)}^{{\displaystyle \frac{3}{2}}}}{-{\displaystyle \frac{3}{2}}}\right]}_1^{2}d\theta -{\int }_0^{2\mathrm{\pi }}2\sin \theta {\left(r\right)}_1^{2}d\theta$

$=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}\left(-\frac{2}{3}\right)\left(0-3\sqrt{3}\right)d\theta +2{\left(\cos \theta \right)}_0^{2\mathrm{\pi }}$

Simplifying we get

$=2\sqrt{3}\mathrm{\pi }$