Q 53. - Calculus

Question

Find the volume using integrals:

The region in the next figure which is bounded below by the $xy$-plane, bounded above by the hyperboloid with equation $x^2+y^2-z^2=1$and inside the cylinder with equation $x^2+y^2=5$.

Step-by-Step Solution

Answer

The required Volume is $V=\frac{16\mathrm{\pi }}{3}$ units.

Step 1: Given Information

The given equations are $x^2+y^2-z^2=1$ and $x^2+y^2=5$.

Step 2: Evaluation of limits

The relation between rectangular and spherical coordinates are as below:

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

And

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

Rectangular coordinates are $x^2+y^2-z^2=1$ and $x^2+y^2=5$

Cylindrical coordinates are $r^2-z^2=1$ and $r^2=5$

Limits for Cartesian coordinates are:

$x^2+y^2-z^2=1 \Rightarrow z=\sqrt{r^2-1}$ (Equation for $xy$ plane is $z=0$)

$x^2+y^2=5 \Rightarrow r=\sqrt{5}$ and $z=0 \Rightarrow r=1$

Limits for Cylindrical coordinates are:

$0\le z\le \sqrt{r^2-1}, 1\le r\le \sqrt{5},0\le \theta \le 2\pi$

Step 3: Calculation of Volume

Required Volume is given by

$V={\int }_{\theta =0}^{2\pi }{\int }_{r=1}^{\sqrt{5}}{\int }_{z=0}^{\sqrt{r^2-1}}rdzdrd\theta$

$V={\int }_{\theta =0}^{2\pi }\left(\frac{1}{2}\right){\int }_{r=1}^{\sqrt{5}}2r\sqrt{r^2-1}drd\theta$

$V={\left.{\int }_{\theta =0}^{2\pi }\left(\frac{1}{2}\right)\left(\frac{{\left(r^2-1\right)}^{3/2}}{3/2}\right)\right|}_{r=1}^{r=\sqrt{5}}d\theta$

$V=\left(\frac{1}{3}\right){\int }_{\theta =0}^{2\pi }(8-0)d\theta$

$V={\left.\left(\frac{8}{3}\right)\left(\theta \right)\right|}_0^{2\pi }$

$V=\left(\frac{8}{3}\right)2\pi$

Hence, $V=\frac{16\mathrm{\pi }}{3}$units