Notes: `(x-h)^2+y^2=r^2 , h>r` Find the volume of the torus generated by revolving the region bounded by the graph of the circle about the y-axis.

Tuesday, 29 July 2014

$\displaystyle{{\left({x}-{h}\right)}}^{{2}}+{{y}}^{{2}}={{r}}^{{2}},{h}>{r}$ Find the volume of the torus generated by revolving the region bounded by the graph of the circle about the y-axis.

Volume of a shape bounded by curve $\displaystyle{y}={f{{\left({x}\right)}}}$ and $\displaystyle{x}$ -axis between $\displaystyle{a}\leq{x}\leq{b}$ revolving about $\displaystyle{y}$ -axis is given by

$\displaystyle{V}_{{y}}={2}\pi{\int_{{a}}^{{b}}}{x}{y}{\left.{d}{x}\right.}$

In order to use the above formula we first need to write $\displaystyle{y}$ as a function of $\displaystyle{x}.$

$\displaystyle{{\left({x}-{h}\right)}}^{{2}}+{{y}}^{{2}}={{r}}^{{2}}$

$\displaystyle{y}=\pm\sqrt{{{{r}}^{{2}}-{{\left({x}-{h}\right)}}^{{2}}}}$

The positive part describes upper half of the circle (blue) while the negative part (red) describes the lower semicircle.

In the graph above $\displaystyle{r}={2}$ and $\displaystyle{h}={5}.$

Since the both halves have equal ares the resulting volumes will also be equal for each half. Therefore, we can calculate volume of whole torus as two times the semi-torus (solid obtained by revolving a semicircle).

Bounds of integration will be points where the semicircle touches the $\displaystyle{x}$ -axis.

$\displaystyle{4}\pi{\int_{{{h}-{r}}}^{{{h}+{r}}}}{x}\sqrt{{{{r}}^{{2}}-{{\left({x}-{h}\right)}}^{{2}}}}{\left.{d}{x}\right.}=$

Substitute $\displaystyle{x}-{h}={r}{\sin{{t}}}$ $\displaystyle\Rightarrow$ $\displaystyle{x}={r}{\sin{{t}}}+{h}$ $\displaystyle\Rightarrow$ $\displaystyle{\left.{d}{x}\right.}={r}{\cos{{t}}}{\left.{d}{t}\right.},$ $\displaystyle{t}_{{l}}=-\frac{\pi}{{2}},$ $\displaystyle{t}_{{u}}=\frac{\pi}{{2}}$

$\displaystyle{t}_{{l}}$ and $\displaystyle{t}_{{u}}$ denote new lower and upper bounds of integration.

$\displaystyle{4}\pi{r}{\int_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}{\left({r}{\sin{{t}}}+{h}\right)}\sqrt{{{{r}}^{{2}}-{{r}}^{{2}}{{\sin}}^{{2}}{t}}}{\cos{{t}}}{\left.{d}{t}\right.}=$

$\displaystyle{4}\pi{r}{\int_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}{\left({r}{\sin{{t}}}+{h}\right)}{r}\sqrt{{{1}-{{\sin}}^{{2}}{t}}}{\cos{{t}}}{\left.{d}{t}\right.}=$

Use the fact that $\displaystyle\sqrt{{{1}-{{\sin}}^{{2}}{t}}}={\cos{{t}}}.$

$\displaystyle{4}\pi{{r}}^{{2}}{\int_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}{\left({r}{\sin{{t}}}+{h}\right)}{{\cos}}^{{2}}{t}{\left.{d}{t}\right.}=$

$\displaystyle{4}\pi{{r}}^{{3}}{\int_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}{{\cos}}^{{2}}{t}{\sin{{t}}}{\left.{d}{t}\right.}+{4}\pi{{r}}^{{2}}{h}{\int_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}{{\cos}}^{{2}}{t}{\left.{d}{t}\right.}=$

Let us calculate each integral separately

$\displaystyle{I}_{{1}}={4}\pi{{r}}^{{3}}{\int_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}{{\cos}}^{{2}}{t}{\sin{{t}}}{\left.{d}{t}\right.}=$

Substitute $\displaystyle{u}={\cos{{t}}}$ $\displaystyle\Rightarrow$ $\displaystyle{d}{u}={\sin{{t}}}{\left.{d}{t}\right.},$ $\displaystyle{u}_{{l}}={0},$ $\displaystyle{u}_{{u}}={0}.$

$\displaystyle{4}\pi{{r}}^{{3}}{\int_{{0}}^{{0}}}\frac{{{u}}^{{3}}}{{3}}{d}{u}={0}$

$\displaystyle{I}_{{2}}={4}\pi{{r}}^{{2}}{h}{\int_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}{{\cos}}^{{2}}{t}{\left.{d}{t}\right.}=$

Rewrite the integral using the following formula $\displaystyle{{\cos}}^{{2}}\theta=\frac{{{1}+{\cos{{2}}}\theta}}{{2}}.$

$\displaystyle{2}\pi{{r}}^{{2}}{h}{\int_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}{\left({1}+{\cos{{2}}}{t}\right)}{\left.{d}{t}\right.}={2}\pi{{r}}^{{2}}{h}{\left({t}+\frac{{1}}{{2}}{\sin{{2}}}{t}\right)}{{\mid}_{{-\frac{\pi}{{2}}}}^{{\frac{\pi}{{2}}}}}=$

$\displaystyle{2}\pi{{r}}^{{2}}{h}{\left(\frac{\pi}{{2}}+{0}+\frac{\pi}{{2}}-{0}\right)}={2}{\pi}^{{2}}{{r}}^{{2}}{h}$

The volume of the torus generated by revolving the given region about $\displaystyle{y}$ -axis is $\displaystyle{2}{\pi}^{{2}}{{r}}^{{2}}{h}.$

The image below shows the torus generated by revolving region bounded by circle $\displaystyle{{\left({x}-{5}\right)}}^{{2}}+{{y}}^{{2}}={{2}}^{{2}}$ i.e. $\displaystyle{r}={2},$ $\displaystyle{h}={5}$ about $\displaystyle{y}$ -axis. The part generated by revolving $\displaystyle{y}=\sqrt{{{{r}}^{{2}}-{{\left({x}-{h}\right)}}^{{2}}}}$ is colored blue while the negative part is colored red.