Notes: `y = 3-x , y=0, y=2, x=0` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the...

Wednesday, 10 May 2017

$\displaystyle{y}={3}-{x},{y}={0},{y}={2},{x}={0}$ Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the...

For the region bounded by $\displaystyle{y}={3}-{x}$ , $\displaystyle{y}={0}$ , $\displaystyle{y}={2}$ , and $\displaystyle{x}={0}$ 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...

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 has 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 $\displaystyle{x}_{{2}}={5}$ since it a distance between the axis of rotation and each boundary graph.

For the inner radius, we have: $\displaystyle{g{{\left({y}\right)}}}={5}-{\left({3}-{y}\right)}$ simplified to $\displaystyle{g{{\left({y}\right)}}}={2}+{y}$

Note: $\displaystyle{y}_{{{b}{e}{l}{o}{w}}}$ for the inner radius is based from $\displaystyle{y}={3}-{x}$ rearrange into $\displaystyle{x}={3}-{y}$

For the outer radius, we have: $\displaystyle{f{{\left({y}\right)}}}={5}-{0}$ simplified to $\displaystyle{f{{\left({y}\right)}}}={5}$ .

Then the boundary values of y is $\displaystyle{a}={0}$ and $\displaystyle{b}={2}$ .

Then the integral will be:

$\displaystyle{V}=\pi{\int_{{0}}^{{2}}}{\left[{{\left({5}\right)}}^{{2}}-{{\left({2}+{y}\right)}}^{{2}}\right]}{\left.{d}{y}\right.}$

Expand using the FOIL method on:

$\displaystyle{{\left({2}+{y}\right)}}^{{2}}={\left({2}+{y}\right)}{\left({2}+{y}\right)}={4}+{4}{y}+{{y}}^{{2}}{\quad\text{and}\quad}{{5}}^{{2}}={25}$ .

The integral becomes:

$\displaystyle{V}=\pi{\int_{{0}}^{{2}}}{\left[{25}-{\left({4}+{4}{y}+{{y}}^{{2}}\right)}\right]}{\left.{d}{y}\right.}$

Simplify:

$\displaystyle{V}=\pi{\int_{{0}}^{{2}}}{\left[{25}-{4}-{4}{y}-{{y}}^{{2}}\right]}{\left.{d}{y}\right.}$

$\displaystyle{V}=\pi{\int_{{0}}^{{2}}}{\left[{21}-{4}{y}-{{y}}^{{2}}\right]}{\left.{d}{y}\right.}$

Apply basic integration property:

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

$\displaystyle{V}=\pi{\left[{\int_{{0}}^{{2}}}{21}{\left.{d}{y}\right.}-{\int_{{0}}^{{2}}}{4}{y}{\left.{d}{y}\right.}-{\int_{{0}}^{{2}}}{{y}}^{{2}}{\left.{d}{y}\right.}\right]}$

For the integration of $\displaystyle\int{21}{\left.{d}{y}\right.}$ , we apply basic integration property: $\displaystyle\int{c}{\left.{d}{x}\right.}={c}{x}$ .

For the integration of $\displaystyle{\int_{{0}}^{{2}}}{4}{y}{\left.{d}{y}\right.}$ and $\displaystyle{\int_{{0}}^{{2}}}{{y}}^{{2}}{\left.{d}{y}\right.}$ , we apply the Power rule for integration: $\displaystyle\int{{y}}^{{n}}{\left.{d}{x}\right.}=\frac{{{y}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

$\displaystyle{V}=\pi{\left[{21}{y}-{4}\cdot\frac{{{y}}^{{2}}}{{2}}-\frac{{{y}}^{{3}}}{{3}}\right]}{{\mid}_{{0}}^{{2}}}$

$\displaystyle{V}=\pi{\left[{21}{y}-{2}{{y}}^{{2}}-\frac{{{y}}^{{3}}}{{3}}\right]}{{\mid}_{{0}}^{{2}}}$

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

$\displaystyle{V}=\pi{\left[{21}{\left({2}\right)}-{2}{{\left({2}\right)}}^{{2}}-\frac{{{\left({2}\right)}}^{{3}}}{{3}}\right]}-\pi{\left[{21}{\left({0}\right)}-{2}{{\left({0}\right)}}^{{2}}-\frac{{{\left({0}\right)}}^{{3}}}{{3}}\right]}$

$\displaystyle{V}=\pi{\left[{42}-{8}-\frac{{8}}{{3}}\right]}-\pi{\left[{0}-{0}-{0}\right]}$

$\displaystyle{V}=\pi{\left[\frac{{94}}{{3}}\right]}-\pi{\left[{0}\right]}$

$\displaystyle{V}=\frac{{{94}\pi}}{{3}}$ or $\displaystyle{98.44}$ (approximated value)

Notes

Wednesday, 10 May 2017

$\displaystyle{y}={3}-{x},{y}={0},{y}={2},{x}={0}$ Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the...

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

Wednesday, 10 May 2017

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the...

No comments:

Post a Comment

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

$\displaystyle{y}={3}-{x},{y}={0},{y}={2},{x}={0}$ Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the...

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