Notes: `(x-3)^2+y^2 =1 ` Find the volume of the torus generated by revolving the region bounded by the graph of the circle about the y-axis.

Sunday, 19 January 2014

$\displaystyle{{\left({x}-{3}\right)}}^{{2}}+{{y}}^{{2}}={1}$ Find the volume of the torus generated by revolving the region bounded by the graph of the circle about the y-axis.

To find the volume of the torus generated by revolving the region bounded by the graph of circle about the y-axis, we may apply Washer method. In this method, rectangular strip representation that is perpendicular to the axis of rotation. We follow the formula for Washer method: $\displaystyle{V}=\pi{\int_{{a}}^{{b}}}{\left[{{\left({f{{\left({y}\right)}}}\right)}}^{{2}}-{{\left({g{{\left({y}\right)}}}\right)}}^{{2}}\right]}{\left.{d}{y}\right.}$ using a horizontal rectangular strip representation with thickness of $\displaystyle{\left.{d}{y}\right.}$ .

The given equation: $\displaystyle{{\left({x}-{3}\right)}}^{{2}}+{{y}}^{{2}}={1}$ is in a form of $\displaystyle{{\left({x}-{R}\right)}}^{{2}}+{{y}}^{{2}}={{r}}^{{2}}$ .

We set up the function of each radius based on the following formula:

inner radius: $\displaystyle{f{{\left({y}\right)}}}={R}-\sqrt{{{{r}}^{{2}}-{{y}}^{{2}}}}$

Then the function for the graph from x=3 to x=4 will be: $\displaystyle{f{{\left({y}\right)}}}={3}-\sqrt{{{1}-{{y}}^{{2}}}}$

outer radius: $\displaystyle{g{{\left({y}\right)}}}={R}+\sqrt{{{{r}}^{{2}}-{y}}}$

Then the function for the graph from x=2 to x=3 will be: $\displaystyle{g{{\left({y}\right)}}}={3}+\sqrt{{{1}-{{y}}^{{2}}}}$

From the attached image, the boundary values of y are: $\displaystyle{a}=-{1}$ and $\displaystyle{b}={1}$ .

Plug-in the values on the formula, we set up:

$\displaystyle{V}=\pi{\int_{{-{1}}}^{{{1}}}}{\left[{{\left({3}+\sqrt{{{1}-{{y}}^{{2}}}}\right)}}^{{2}}-{{\left({3}-\sqrt{{{1}-{{y}}^{{2}}}}\right)}}^{{2}}\right]}{\left.{d}{y}\right.}$

$\displaystyle=\pi{\int_{{-{1}}}^{{{1}}}}{\left[{\left({3}+{6}\sqrt{{{1}-{{y}}^{{2}}}}+{1}-{{y}}^{{2}}\right)}-{\left({3}-{6}\sqrt{{{1}-{{y}}^{{2}}}}+{1}-{{y}}^{{2}}\right)}\right]}{\left.{d}{y}\right.}$

$\displaystyle=\pi{\int_{{-{1}}}^{{{1}}}}{\left[{3}+{6}\sqrt{{{1}-{{y}}^{{2}}}}+{1}-{{y}}^{{2}}-{3}+{6}\sqrt{{{1}-{{y}}^{{2}}}}-{1}+{{y}}^{{2}}\right]}{\left.{d}{y}\right.}$

$\displaystyle=\pi{\int_{{-{1}}}^{{{1}}}}{\left[{12}\sqrt{{{1}-{{y}}^{{2}}}}\right]}{\left.{d}{y}\right.}$

Apply the basic integration property: $\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}-{c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle{V}={12}\pi{\int_{{-{1}}}^{{{1}}}}{\left[\sqrt{{{1}-{{y}}^{{2}}}}\right]}{\left.{d}{y}\right.}$

From integration table, we may apply the integral formula for function with roots:

$\displaystyle\int\sqrt{{{{a}}^{{2}}-{{u}}^{{2}}}}{d}{u}=\frac{{{u}\cdot\sqrt{{{{a}}^{{2}}-{{u}}^{{2}}}}}}{{2}}+\frac{{{a}}^{{2}}}{{2}}{\arcsin{{\left(\frac{{u}}{{a}}\right)}}}$

Then,

$\displaystyle{V}={12}\pi{\int_{{-{1}}}^{{{1}}}}{\left[\sqrt{{{1}-{{y}}^{{2}}}}\right]}{\left.{d}{y}\right.}$

$\displaystyle={12}\pi\cdot{\left[\frac{{{y}\cdot\sqrt{{{1}-{{y}}^{{2}}}}}}{{2}}+\frac{{1}}{{2}}{\arcsin{{\left(\frac{{y}}{{1}}\right)}}}\right]}{{\mid}_{{-{1}}}^{{{1}}}}$

$\displaystyle={12}\pi\cdot{\left[\frac{{{y}\sqrt{{{1}-{{y}}^{{2}}}}}}{{2}}+\frac{{\arcsin{{\left({y}\right)}}}}{{2}}\right]}{{\mid}_{{-{1}}}^{{{1}}}}$

$\displaystyle={\left[{6}\pi{y}\sqrt{{{1}-{{y}}^{{2}}}}+{6}\pi{\arcsin{{\left({y}\right)}}}\right]}{{\mid}_{{-{1}}}^{{{1}}}}$

Apply definite integral formula: .

$\displaystyle{V}={\left[{6}\pi{y}\sqrt{{{1}-{{y}}^{{2}}}}+{6}\pi{\arcsin{{\left({y}\right)}}}\right]}{{\mid}_{{-{1}}}^{{{1}}}}$

$\displaystyle={\left[{6}\pi{\left({1}\right)}\sqrt{{{1}-{{1}}^{{2}}}}+{6}\pi{\arcsin{{\left({1}\right)}}}\right]}-{\left[{6}\pi{\left(-{1}\right)}\sqrt{{{1}-{{\left(-{1}\right)}}^{{2}}}}+{6}\pi{\arcsin{{\left(-{1}\right)}}}\right]}$

$\displaystyle={\left[{6}\pi\sqrt{{{1}-{1}}}+{6}\pi{\arcsin{{\left({1}\right)}}}\right]}-{\left[-{6}\pi\sqrt{{{1}-{1}}}+{6}\pi{\arcsin{{\left(-{1}\right)}}}\right]}$

$\displaystyle={\left[{6}\pi\sqrt{{{0}}}+{6}\pi{\arcsin{{\left({1}\right)}}}\right]}-{\left[-{6}\pi\sqrt{{{0}}}+{6}\pi{\arcsin{{\left(-{1}\right)}}}\right]}$

$\displaystyle={\left[{6}\pi\cdot{0}+{6}\pi\cdot{\left(\frac{\pi}{{2}}\right)}\right]}-{\left[-{6}\pi\cdot{0}+{6}\pi\cdot{\left(-\frac{\pi}{{2}}\right)}\right)}{\]}$