Notes: `y = x^2 , y = x^5` Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

Saturday, 8 August 2015

$\displaystyle{y}={{x}}^{{2}},{y}={{x}}^{{5}}$ Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

To be able to use the Shell method, a rectangular strip from the bounded plane region should be parallel to the axis of revolution.

By revolving multiple rectangular strip, it forms infinite numbers of hollow pipes or representative cylinders.

In this method, we follow the formula: $\displaystyle{V}={\int_{{a}}^{{b}}}{2}\pi{r}{h}{\left.{d}{y}\right.}$

where:

radius (r)= distance of the rectangular strip to the axis of revolution

height (h) = length of the rectangular strip

thickness = width of...

To be able to use the Shell method, a rectangular strip from the bounded plane region should be parallel to the axis of revolution.

By revolving multiple rectangular strip, it forms infinite numbers of hollow pipes or representative cylinders.

In this method, we follow the formula: $\displaystyle{V}={\int_{{a}}^{{b}}}{2}\pi{r}{h}{\left.{d}{y}\right.}$

where:

radius (r)= distance of the rectangular strip to the axis of revolution

height (h) = length of the rectangular strip

thickness = width of the rectangular strip as $\displaystyle{\left.{d}{x}\right.}$ or $\displaystyle{\left.{d}{y}\right.}$ .

For the bounded region, as shown on the attached image, the rectangular strip is parallel to x-axis (axis of rotation). We can let:

$\displaystyle{r}={y}$

$\displaystyle{h}={f{{\left({y}\right)}}}$ or $\displaystyle{h}={x}_{{2}}-{x}_{{1}}$

$\displaystyle{h}={{y}}^{{\frac{{1}}{{5}}}}-{{y}}^{{\frac{{1}}{{2}}}}$

Note: $\displaystyle{y}={{x}}^{{5}}$ is expressed as $\displaystyle{x}={{y}}^{{\frac{{1}}{{5}}}}$ and $\displaystyle{y}={{x}}^{{2}}$ is expressed as $\displaystyle{x}={{y}}^{{\frac{{1}}{{2}}}}$ .

thickness $\displaystyle={\left.{d}{y}\right.}$

Boundary values of y: $\displaystyle{a}={0}$ to $\displaystyle{b}={1}$ .

Plug-in the values on $\displaystyle{V}={\int_{{a}}^{{b}}}{2}\pi{r}{h}{\left.{d}{y}\right.}$

$\displaystyle{V}={\int_{{0}}^{{1}}}{2}\pi\cdot{y}\cdot{\left({{y}}^{{\frac{{1}}{{5}}}}-{{y}}^{{\frac{{1}}{{2}}}}\right)}{\left.{d}{y}\right.}$

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

V = 2pi int_0^1 y( y^(1/5)-y^(1/2)) dy

Apply Law of Exponent: y^n*x^m = y^((n+m)).

V = 2pi int_0^1 (y^((1/5+1))-y^((1/2+1))) dy

V = 2pi int_0^1 (y^(6/5)-y^(3/2)) dy

Apply basic integration property: $\displaystyle\int{\left({u}-{v}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}-\int{\left({v}\right)}{\left.{d}{x}\right.}$ .

$\displaystyle{V}={2}\pi{\left[{\int_{{0}}^{{1}}}{\left({{y}}^{{\frac{{6}}{{5}}}}\right)}{\left.{d}{y}\right.}-{\int_{{0}}^{{1}}}{\left({{y}}^{{\frac{{3}}{{2}}}}\right)}{\left.{d}{y}\right.}\right]}$

Apply Power rule for integration: $\displaystyle\int{{y}}^{{n}}{\left.{d}{y}\right.}=\frac{{{y}}^{{{n}+{1}}}}{{{n}+{1}}}.$

$\displaystyle{V}={2}\pi{\left[\frac{{{y}}^{{{\left(\frac{{6}}{{5}}+{1}\right)}}}}{{{\left(\frac{{6}}{{5}}+{1}\right)}}}-\frac{{{y}}^{{{\left(\frac{{3}}{{2}}+{1}\right)}}}}{{{\left(\frac{{3}}{{2}}+{1}\right)}}}\right]}{{\mid}_{{0}}^{{1}}}$

$\displaystyle{V}={2}\pi{\left[\frac{{{y}}^{{{\left(\frac{{11}}{{5}}\right)}}}}{{{\left(\frac{{11}}{{5}}\right)}}}-\frac{{{y}}^{{{\left(\frac{{5}}{{2}}\right)}}}}{{{\left(\frac{{5}}{{2}}\right)}}}\right]}{{\mid}_{{0}}^{{1}}}$

$\displaystyle{V}={2}\pi{\left[{{y}}^{{\frac{{11}}{{5}}}}\cdot{\left(\frac{{5}}{{11}}\right)}-{{y}}^{{\frac{{5}}{{2}}}}\cdot{\left(\frac{{2}}{{5}}\right)}\right]}{{\mid}_{{0}}^{{1}}}$

$\displaystyle{V}={2}\pi{\left[\frac{{{5}{{y}}^{{\frac{{11}}{{5}}}}}}{{11}}-\frac{{{2}{{y}}^{{\frac{{5}}{{2}}}}}}{{5}}\right]}{{\mid}_{{0}}^{{1}}}$

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

$\displaystyle{V}={2}\pi{\left[\frac{{{5}{{\left({1}\right)}}^{{\frac{{11}}{{5}}}}}}{{11}}-\frac{{{2}{{\left({1}\right)}}^{{\frac{{5}}{{2}}}}}}{{5}}\right]}-{2}\pi{\left[\frac{{{5}{{\left({0}\right)}}^{{\frac{{11}}{{5}}}}}}{{11}}-\frac{{{2}{{\left({0}\right)}}^{{\frac{{5}}{{2}}}}}}{{5}}\right]}$

$\displaystyle{V}={2}\pi{\left[\frac{{5}}{{11}}-\frac{{2}}{{5}}\right]}-{2}\pi{\left[{0}-{0}\right]}$

$\displaystyle{V}={2}\pi{\left[\frac{{3}}{{55}}\right]}-{2}\pi{\left[{0}\right]}$

$\displaystyle{V}=\frac{{{6}\pi}}{{55}}-{0}$

$\displaystyle{V}=\frac{{{6}\pi}}{{55}}$ or $\displaystyle{0.343}$ (approximated value)

Notes

Saturday, 8 August 2015

$\displaystyle{y}={{x}}^{{2}},{y}={{x}}^{{5}}$ 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, 8 August 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?

$\displaystyle{y}={{x}}^{{2}},{y}={{x}}^{{5}}$ 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?