Notes: `y = 2 , y = 4-x^2/4` Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

Saturday, 20 June 2015

$\displaystyle{y}={2},{y}={4}-\frac{{{x}}^{{2}}}{{4}}$ Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

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

As shown on the attached image, we are using vertical rectangular strip that is perpendicular to the x-axis (axis of revolution) with a thickness of $\displaystyle\text{dx}$ . In line with this, we will consider the formula for the Washer Method as:

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

where $\displaystyle{f{{\left({x}\right)}}}$ as function of the outer radius, $\displaystyle{R}$

$\displaystyle{g{{\left({x}\right)}}}$ as a function of the inner radius, $\displaystyle{r}$

For each radius, we follow the $\displaystyle{y}_{{{a}{b}{o}{v}{e}}}-{y}_{{{b}{e}{l}{o}{w}}}$ , we have $\displaystyle{y}_{{{b}{e}{l}{o}{w}}}={0}$ since it a distance between the axis of rotation and each boundary graph.

For the inner radius, we have: $\displaystyle{g{{\left({x}\right)}}}={2}-{0}={2}$

For the outer radius, we have: $\displaystyle{f{{\left({x}\right)}}}={\left({4}-\frac{{{x}}^{{2}}}{{4}}\right)}-{0}={4}-\frac{{{x}}^{{2}}}{{4}}$

To determine the boundary values of x, we equate the two values of y's:

$\displaystyle{4}-\frac{{{x}}^{{2}}}{{4}}={2}$

$\displaystyle-\frac{{{x}}^{{2}}}{{4}}={2}-{4}$

$\displaystyle-\frac{{{x}}^{{2}}}{{4}}=-{2}$

$\displaystyle{\left(-{4}\right)}{\left(-\frac{{{x}}^{{2}}}{{4}}\right)}={\left(-{4}\right)}{\left(-{2}\right)}$

$\displaystyle{{x}}^{{2}}={8}$ then $\displaystyle{x}=\pm\sqrt{{{8}}}$ or $\displaystyle+{2}\sqrt{{{2}}}$ and $\displaystyle-{2}\sqrt{{{2}}}$

Then, boundary values of x: $\displaystyle{a}=-{2}\sqrt{{{2}}}$ and $\displaystyle{b}={2}\sqrt{{{2}}}$ .

Plug-in the values in the formula $\displaystyle{V}=\pi{\int_{{a}}^{{b}}}{\left({{\left({f{{\left({x}\right)}}}\right)}}^{{2}}-{{\left({g{{\left({x}\right)}}}\right)}}^{{2}}\right)}{\left.{d}{x}\right.}$ , we get:

$\displaystyle{V}=\pi{\int_{{-{2}\sqrt{{{2}}}}}^{{{2}\sqrt{{{2}}}}}}{\left[{{\left({4}-\frac{{{x}}^{{2}}}{{4}}\right)}}^{{2}}-{{2}}^{{2}}\right]}{\left.{d}{x}\right.}$ .

Expand using the FOIL method on: $\displaystyle{{\left({4}-\frac{{{x}}^{{2}}}{{4}}\right)}}^{{2}}={\left({4}-\frac{{{x}}^{{2}}}{{4}}\right)}{\left({4}-\frac{{{x}}^{{2}}}{{4}}\right)}={16}-{2}{{x}}^{{2}}+\frac{{{x}}^{{4}}}{{16}}$ and $\displaystyle{{2}}^{{2}}={4}$ .

The integral becomes:

$\displaystyle{V}=\pi{\int_{{-{2}\sqrt{{{2}}}}}^{{{2}\sqrt{{{2}}}}}}{\left[{16}-{2}{{x}}^{{2}}+\frac{{{x}}^{{4}}}{{16}}-{4}\right]}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi{\int_{{-{2}\sqrt{{{2}}}}}^{{{2}\sqrt{{{2}}}}}}{\left[{12}-{2}{{x}}^{{2}}+\frac{{{x}}^{{4}}}{{16}}\right]}{\left.{d}{x}\right.}$

Apply basic integration property: $\displaystyle\int{\left({u}\pm{v}\pm{w}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}\pm\int{\left({v}\right)}{\left.{d}{x}\right.}\pm\int{\left({w}\right)}{\left.{d}{x}\right.}$ to be able to integrate them separately using Power rule for integration: $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

$\displaystyle{V}=\pi\cdot{\left[{\int_{{-{2}\sqrt{{{2}}}}}^{{{2}\sqrt{{{2}}}}}}{\left({12}\right)}{\left.{d}{x}\right.}-{\int_{{-{2}\sqrt{{{2}}}}}^{{{2}\sqrt{{{2}}}}}}{\left({2}{{x}}^{{2}}\right)}{\left.{d}{x}\right.}+{\int_{{-{2}\sqrt{{{2}}}}}^{{{2}\sqrt{{{2}}}}}}{\left(\frac{{{x}}^{{4}}}{{16}}\right)}{\left.{d}{x}\right.}\right]}$

$\displaystyle{V}=\pi\cdot{\left[{12}{x}-{2}\cdot\frac{{{x}}^{{3}}}{{3}}+\frac{{1}}{{16}}\cdot\frac{{{x}}^{{5}}}{{5}}\right]}{{\mid}_{{-{2}\sqrt{{{2}}}}}^{{{2}\sqrt{{{2}}}}}}$

$\displaystyle{V}=\pi\cdot{\left[{12}{x}-\frac{{{2}{{x}}^{{3}}}}{{3}}+\frac{{{x}}^{{5}}}{{80}}\right]}{{\mid}_{{-{2}\sqrt{{{2}}}}}^{{{2}\sqrt{{{2}}}}}}$

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

$\displaystyle{V}=\pi\cdot{\left[{12}{\left({2}\sqrt{{{2}}}\right)}-\frac{{{2}{{\left({2}\sqrt{{{2}}}\right)}}^{{3}}}}{{3}}+\frac{{{\left({2}\sqrt{{{2}}}\right)}}^{{5}}}{{80}}\right]}-\pi\cdot{\left[{12}{\left(-{2}\sqrt{{{2}}}\right)}-\frac{{{2}{{\left(-{2}\sqrt{{{2}}}\right)}}^{{3}}}}{{3}}+\frac{{{\left(-{2}\sqrt{{{2}}}\right)}}^{{5}}}{{80}}\right]}$

$\displaystyle{V}=\pi\cdot{\left[{24}\sqrt{{{2}}}-\frac{{{32}\sqrt{{{2}}}}}{{3}}+\frac{{{8}\sqrt{{{2}}}}}{{5}}\right]}-\pi\cdot{\left[-{24}\sqrt{{{2}}}+\frac{{{32}\sqrt{{{2}}}}}{{3}}-\frac{{{8}\sqrt{{{2}}}}}{{5}}\right]}$

$\displaystyle{V}=\frac{{{224}\sqrt{{{2}}}\pi}}{{15}}-\frac{{-{224}\sqrt{{{2}}}\pi}}{{15}}$

$\displaystyle{V}=\frac{{{224}\sqrt{{{2}}}\pi}}{{15}}+\frac{{{224}\sqrt{{{2}}}\pi}}{{15}}$

$\displaystyle{V}=\frac{{{448}\sqrt{{{2}}}\pi}}{{15}}$ or $\displaystyle{132.69}$ (approximated value)

Notes

Saturday, 20 June 2015

$\displaystyle{y}={2},{y}={4}-\frac{{{x}}^{{2}}}{{4}}$ Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

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

Saturday, 20 June 2015

Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

No comments:

Post a Comment

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

$\displaystyle{y}={2},{y}={4}-\frac{{{x}}^{{2}}}{{4}}$ Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

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