Q 49. - Calculus

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 $x^2+y^2=1$

Step-by-Step Solution

Answer

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

Step 1: Given Information

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

Step 2: 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

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

Step 3: 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

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