New Notes Blog: `y = sqrt(x-2) , y=0 , x=4` Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving...

Friday, 18 April 2014

$\displaystyle{y}=\sqrt{{{x}-{2}}},{y}={0},{x}={4}$ Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving...

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 strips, it forms infinite numbers of these hollow pipes or representative cylinders.

In this method, we follow the formula: $\displaystyle{V}={\int_{{a}}^{{b}}}$ (length * height * thickness)

or $\displaystyle{V}={\int_{{a}}^{{b}}}{2}\pi$ * radius*height*thickness

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

$\displaystyle{r}={x}$

$\displaystyle{h}={f{{\left({x}\right)}}}$ or $\displaystyle{h}={y}_{{{a}{b}{o}{v}{e}}}-{y}_{{{b}{e}{l}{o}{w}}}$

h =(sqrt(x-2)) - 0 = sqrt(x-2)

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

For boundary values of x, we have $\displaystyle{a}={2}$ to $\displaystyle{b}={4}$ .

Plug-in the values on $\displaystyle{V}={\int_{{a}}^{{b}}}{2}\pi$ * radius*height*thickness , we get:

$\displaystyle{V}={\int_{{2}}^{{4}}}{2}\pi\cdot{x}\cdot\sqrt{{{x}-{2}}}\cdot{\left.{d}{x}\right.}$

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

$\displaystyle{V}={2}\pi{\int_{{2}}^{{4}}}{x}\cdot\sqrt{{{x}-{2}}}{\left.{d}{x}\right.}$

To determine the indefinite integral, we may apply u-substitution by letting $\displaystyle{u}={x}-{2}$ then $\displaystyle{d}{u}={\left.{d}{x}\right.}$ . The $\displaystyle{u}={x}-{2}$ can be rearranged as $\displaystyle{x}={u}+{2}$ .

The integral becomes:

$\displaystyle{V}={2}\pi\int{\left({u}+{2}\right)}\cdot\sqrt{{{u}}}{\left.{d}{x}\right.}$

Apply Law of Exponent: $\displaystyle\sqrt{{{u}}}={{u}}^{{\frac{{1}}{{2}}}}$ and $\displaystyle{{u}}^{{n}}\cdot{{u}}^{{m}}={{u}}^{{{n}+{m}}}$ .

$\displaystyle{V}={2}\pi\int{\left({u}+{2}\right)}\cdot{{u}}^{{{\left(\frac{{1}}{{2}}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle{V}={2}\pi\int{{u}}^{{{\left(\frac{{3}}{{2}}\right)}}}+{2}{{u}}^{{{\left(\frac{{1}}{{2}}\right)}}}{\left.{d}{x}\right.}$

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

$\displaystyle{V}={2}\pi{\left[\int{{u}}^{{{\left(\frac{{3}}{{2}}\right)}}}{\left.{d}{x}\right.}+\int{2}{{u}}^{{{\left(\frac{{1}}{{2}}\right)}}}{\left.{d}{x}\right.}\right]}$

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

$\displaystyle{V}={2}\pi{\left[\frac{{{u}}^{{\frac{{3}}{{2}}+{1}}}}{{\frac{{3}}{{2}}+{1}}}+{2}\frac{{{u}}^{{\frac{{1}}{{2}}+{1}}}}{{\frac{{1}}{{2}}+{1}}}\right]}$

$\displaystyle{V}={2}\pi{\left[\frac{{{u}}^{{\frac{{5}}{{2}}}}}{{\frac{{5}}{{2}}}}+{2}\cdot\frac{{{u}}^{{\frac{{3}}{{2}}}}}{{\frac{{3}}{{2}}}}\right]}$

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

$\displaystyle{V}={2}\pi{\left[\frac{{{2}{{u}}^{{\frac{{5}}{{2}}}}}}{{5}}+\frac{{{4}{{u}}^{{\frac{{3}}{{2}}}}}}{{3}}\right]}$

Plug -in $\displaystyle{u}={x}-{2}$ , we get:

$\displaystyle{V}={2}\pi{\left[\frac{{{2}{{\left({x}-{2}\right)}}^{{{\left(\frac{{5}}{{2}}\right)}}}}}{{5}}+\frac{{{4}{{\left({x}-{2}\right)}}^{{{\left(\frac{{3}}{{2}}\right)}}}}}{{3}}\right]}$ with boundary values: $\displaystyle{a}={2}$ to $\displaystyle{b}={4}$ .

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

$\displaystyle{V}={2}\pi{\left[\frac{{{2}{{\left({4}{x}-{2}\right)}}^{{{\left(\frac{{5}}{{2}}\right)}}}}}{{5}}+\frac{{{4}{{\left({4}-{2}\right)}}^{{{\left(\frac{{3}}{{2}}\right)}}}}}{{3}}\right]}-{2}\pi{\left[\frac{{{2}{{\left({4}-{2}\right)}}^{{{\left(\frac{{5}}{{2}}\right)}}}}}{{5}}+\frac{{{4}{{\left({4}-{2}\right)}}^{{{\left(\frac{{3}}{{2}}\right)}}}}}{{3}}\right]}$

$\displaystyle{V}={2}\pi{\left[\frac{{{8}\sqrt{{{2}}}}}{{5}}+\frac{{{8}\sqrt{{{2}}}}}{{3}}\right]}-{2}\pi{\left[{0}+{0}\right]}$

$\displaystyle{V}={2}\pi{\left[\frac{{{64}\sqrt{{{2}}}}}{{15}}\right]}-{2}\pi{\left[{0}\right]}$

$\displaystyle{V}=\frac{{{128}\sqrt{{{2}}}\pi}}{{15}}-{0}$

$\displaystyle{V}=\frac{{{128}\sqrt{{{2}}}\pi}}{{15}}$ or $\displaystyle{37.92}$ (approximated value)

New Notes Blog

Friday, 18 April 2014

$\displaystyle{y}=\sqrt{{{x}-{2}}},{y}={0},{x}={4}$ Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving...

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Friday, 18 April 2014

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving...

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{y}=\sqrt{{{x}-{2}}},{y}={0},{x}={4}$ Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving...

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?