Notes: `y = 1/x , y=0 , x=1 , x=3` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the...

Saturday, 24 August 2013

$\displaystyle{y}=\frac{{1}}{{x}},{y}={0},{x}={1},{x}={3}$ 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 $\displaystyle{y}=\frac{{1}}{{x}}$ , $\displaystyle{y}={0}$ , $\displaystyle{x}={1}$ and $\displaystyle{x}={3}$ and revolved about the x-axis, we may apply Disk method. For the Disk method, we consider a perpendicular rectangular strip with the axis of revolution.

As shown on the attached image, the thickness of the rectangular strip is "dx" with a vertical orientation perpendicular to the x-axis (axis of revolution).

We follow the formula for the Disk...

As shown on the attached image, the thickness of the rectangular strip is "dx" with a vertical orientation perpendicular to the x-axis (axis of revolution).

We follow the formula for the Disk method: $\displaystyle{V}={\int_{{a}}^{{b}}}{A}{\left({x}\right)}{\left.{d}{x}\right.}$ where disk base area is $\displaystyle{A}=\pi{{r}}^{{2}}$ with.

Note: r = length of the rectangular strip. We may apply $\displaystyle{r}={y}_{{{a}{b}{o}{v}{e}}}-{y}_{{{b}{e}{l}{o}{w}}}.$

Then $\displaystyle{r}={f{{\left({x}\right)}}}=\frac{{1}}{{x}}-{0}$

$\displaystyle{r}=\frac{{1}}{{x}}$

The boundary values of x is $\displaystyle{a}={1}$ to $\displaystyle{b}={3}$ .

Plug-in the $\displaystyle{f{{\left({x}\right)}}}$ and the boundary values to integral formula, we get:

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

$\displaystyle{V}={\int_{{1}}^{{3}}}\pi\frac{{1}}{{{x}}^{{2}}}{\left.{d}{x}\right.}$

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

$\displaystyle{V}=\pi{\int_{{1}}^{{3}}}\frac{{1}}{{{x}}^{{2}}}{\left.{d}{x}\right.}$

Apply Law of Exponent: $\displaystyle\frac{{1}}{{{x}}^{{n}}}={{x}}^{{-{n}}}$ and Power rule for integration: $\displaystyle\int{{x}}^{{n}}{\left.{d}{y}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}$ .

$\displaystyle{V}=\pi{\int_{{1}}^{{3}}}{{x}}^{{-{2}}}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi\cdot\frac{{{x}}^{{{\left(-{2}+{1}\right)}}}}{{{\left(-{2}+{1}\right)}}}{{\mid}_{{1}}^{{3}}}$

$\displaystyle{V}=\pi\cdot\frac{{{x}}^{{-{1}}}}{{-{1}}}{{\mid}_{{1}}^{{3}}}$

$\displaystyle{V}=\pi\cdot-\frac{{1}}{{x}}{{\left|_{{1}}^{{3}}{\quad\text{or}\quad}-\frac{\pi}{{x}}\right|}_{{1}}^{{3}}}$

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

$\displaystyle{V}={\left(-\frac{\pi}{{3}}\right)}-{\left(-\frac{\pi}{{1}}\right)}$

$\displaystyle{V}=-\frac{\pi}{{3}}+\pi$

$\displaystyle{V}={2}\frac{\pi}{{3}}$

Notes

Saturday, 24 August 2013

$\displaystyle{y}=\frac{{1}}{{x}},{y}={0},{x}={1},{x}={3}$ 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?

Saturday, 24 August 2013

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?

$\displaystyle{y}=\frac{{1}}{{x}},{y}={0},{x}={1},{x}={3}$ 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?