Notes: `x=y^2 , x=4` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5

Monday, 6 March 2017

$\displaystyle{x}={{y}}^{{2}},{x}={4}$ Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5

For the region bounded by $\displaystyle{x}={{y}}^{{2}}$ and $\displaystyle{x}={4}$ revolved about the line $\displaystyle{x}={5}$ , we may apply Washer method for the integral application for the volume of a solid.

The formula for the Washer Method is:

$\displaystyle{V}=\pi{\int_{{a}}^{{b}}}{\left[{{\left({f{{\left({x}\right)}}}\right)}}^{{2}}-{{\left({g{{\left({x}\right)}}}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi{\int_{{a}}^{{b}}}{\left[{{\left({f{{\left({y}\right)}}}\right)}}^{{2}}-{{\left({g{{\left({y}\right)}}}\right)}}^{{2}}\right]}{\left.{d}{y}\right.}$

where f as function of the outer radius

g as a function of the inner radius

To determine which form we use, we...

The formula for the Washer Method is:

$\displaystyle{V}=\pi{\int_{{a}}^{{b}}}{\left[{{\left({f{{\left({x}\right)}}}\right)}}^{{2}}-{{\left({g{{\left({x}\right)}}}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi{\int_{{a}}^{{b}}}{\left[{{\left({f{{\left({y}\right)}}}\right)}}^{{2}}-{{\left({g{{\left({y}\right)}}}\right)}}^{{2}}\right]}{\left.{d}{y}\right.}$

where f as function of the outer radius

g as a function of the inner radius

To determine which form we use, we consider the horizontal rectangular strip representation that is perpendicular to the axis of rotation as shown on the attached image. The given strip have a thickness of " $\displaystyle{\left.{d}{y}\right.}$ " which is our clue to use the formula:

$\displaystyle{V}=\pi{\int_{{a}}^{{b}}}{\left[{{\left({f{{\left({y}\right)}}}\right)}}^{{2}}-{{\left({g{{\left({y}\right)}}}\right)}}^{{2}}\right]}{\left.{d}{y}\right.}$

For each radius, we follow the $\displaystyle{x}_{{2}}-{x}_{{1}}$ . We have $\displaystyle{x}_{{2}}={5}$ on both radius since it is a distance between the axis of rotation and each boundary graph.

For the inner radius, we have: $\displaystyle{g{{\left({y}\right)}}}={5}-{4}={1}$ .

For the outer radius, we have: $\displaystyle{f{{\left({y}\right)}}}={5}-{{y}}^{{2}}$ .

Then the boundary values of y are $\displaystyle{a}=-{2}$ and $\displaystyle{b}={2}$ .

Then the integral will be:

$\displaystyle{V}=\pi{\int_{{-{2}}}^{{2}}}{\left[{{\left({5}-{{y}}^{{2}}\right)}}^{{2}}-{{\left({1}\right)}}^{{2}}\right]}{\left.{d}{y}\right.}$

Exapnd using the FOIL method on:

$\displaystyle{{\left({5}-{{y}}^{{2}}\right)}}^{{2}}={\left({5}-{{y}}^{{2}}\right)}{\left({5}-{{y}}^{{2}}\right)}={25}-{10}{{y}}^{{2}}+{{y}}^{{4}}$ and $\displaystyle{{1}}^{{2}}={1}$

The integral becomes:

$\displaystyle{V}=\pi{\int_{{-{2}}}^{{2}}}{\left[{25}-{10}{{y}}^{{2}}+{{y}}^{{4}}-{1}\right]}{\left.{d}{y}\right.}$

Simplify: $\displaystyle{V}=\pi{\int_{{-{2}}}^{{2}}}{\left[{24}-{10}{{y}}^{{2}}+{{y}}^{{4}}\right]}{\left.{d}{y}\right.}$

Apply basic integration property:

$\displaystyle\int{\left({u}\pm{v}\pm{w}\right)}{\left.{d}{y}\right.}=\int{\left({u}\right)}{\left.{d}{y}\right.}\pm\int{\left({u}\right)}{\left.{d}{y}\right.}\pm\int{\left({v}\right)}{\left.{d}{y}\right.}\pm\int{\left({w}\right)}{\left.{d}{y}\right.}$

$\displaystyle{V}=\pi{\left[{\int_{{-{2}}}^{{2}}}{\left({24}\right)}{\left.{d}{y}\right.}-{\int_{{-{2}}}^{{2}}}{\left({10}{{y}}^{{2}}\right)}{\left.{d}{y}\right.}+{\int_{{-{2}}}^{{2}}}{\left({{y}}^{{4}}\right)}{\left.{d}{y}\right.}\right]}$

Apply basic integration property: $\displaystyle\int{c}{\left.{d}{x}\right.}={c}{x}$ , $\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ , and Power rule for integration: $\displaystyle\int{{y}}^{{n}}{\left.{d}{x}\right.}=\frac{{{y}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

$\displaystyle{V}=\pi{\left[{24}{y}-{10}\frac{{{y}}^{{3}}}{{3}}+\frac{{{y}}^{{5}}}{{5}}\right]}{{\mid}_{{-{2}}}^{{2}}}$

Apply the definite integral formula: $\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle{V}=\pi{\left[{24}{\left({2}\right)}-{10}\frac{{{\left({2}\right)}}^{{3}}}{{3}}+\frac{{{\left({2}\right)}}^{{5}}}{{5}}\right]}-\pi{\left[{24}{\left(-{2}\right)}-{10}\frac{{{\left(-{2}\right)}}^{{3}}}{{3}}+\frac{{{\left(-{2}\right)}}^{{5}}}{{5}}\right]}$

$\displaystyle{V}=\pi{\left[{48}-\frac{{80}}{{3}}+\frac{{32}}{{5}}\right]}-\pi{\left[-{48}-\frac{{-{80}}}{{3}}+\frac{{-{32}}}{{5}}\right]}$

$\displaystyle{V}=\pi{\left[{48}-\frac{{80}}{{3}}+\frac{{32}}{{5}}+{48}-\frac{{80}}{{3}}+\frac{{32}}{{5}}\right]}$

$\displaystyle{V}=\pi{\left[{96}-\frac{{160}}{{3}}+\frac{{64}}{{5}}\right]}$

$\displaystyle{V}=\pi{\left[\frac{{1440}}{{15}}-\frac{{800}}{{13}}+\frac{{192}}{{15}}\right]}$

$\displaystyle{V}=\frac{{832}}{{15}}\pi$

or $\displaystyle{V}={174.25}$ (approximated value)

Notes

Monday, 6 March 2017

$\displaystyle{x}={{y}}^{{2}},{x}={4}$ Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Monday, 6 March 2017

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{x}={{y}}^{{2}},{x}={4}$ Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?