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

Thursday, 22 December 2016

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

For a region bounded by $\displaystyle{y}={x}$ (slant line), $\displaystyle{x}={0}$ (along the vertical axis), and $\displaystyle{y}={3}$ (horizontal line) revolved about the line $\displaystyle{y}={4}$ , we may apply the Washer Method. It considers multiple disc with a hole. Basically, it can be just two disc method in which we take the difference of the volume of the bigger and smaller disc.

It follows the formula: $\displaystyle{A}=\pi{{\left({R}_{{{o}{u}{t}{e}{r}}}\right)}}^{{2}}-{{\left({r}_{{{i}.{n}.{n}.{e}.{r}}}\right)}}^{{2}}{\]}$

Let $\displaystyle{R}_{{{o}{u}{t}{e}{r}}}$ as a function of $\displaystyle{f{}}$ ...

It follows the formula: $\displaystyle{A}=\pi{{\left({R}_{{{o}{u}{t}{e}{r}}}\right)}}^{{2}}-{{\left({r}_{{{i}.{n}.{n}.{e}.{r}}}\right)}}^{{2}}{\]}$

Let $\displaystyle{R}_{{{o}{u}{t}{e}{r}}}$ as a function of $\displaystyle{f{}}$ and $\displaystyle{r}_{{{i}.{n}.{n}.{e}.{r}}}$ as function of $\displaystyle{g{}}$ .

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

We use a rectangular strip representation that is perpendicular to the axis of rotation as shown on the attached image.

For the inner radius, we have:

$\displaystyle{g{{\left({x}\right)}}}={4}-{3}={1}$

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

The boundary values of $\displaystyle{x}$ will be $\displaystyle{a}={0}$ to $\displaystyle{b}={3}$ .

The integral to approximate the volume of the solid is:

$\displaystyle{V}=\pi{\int_{{0}}^{{3}}}{\left[{{\left({4}-{x}\right)}}^{{2}}-{{1}}^{{2}}\right]}{\left.{d}{x}\right.}$

Expand using FOIL method on $\displaystyle{{\left({4}-{x}\right)}}^{{2}}={\left({4}-{x}\right)}\cdot{\left({4}-{x}\right)}={16}-{8}{x}+{{x}}^{{2}}$ and $\displaystyle{{1}}^{{2}}={1}$ .

The integral becomes:

$\displaystyle{V}=\pi{\int_{{0}}^{{3}}}{\left[{16}-{8}{x}+{{x}}^{{2}}-{1}\right]}{\left.{d}{x}\right.}$

Simplify: $\displaystyle{V}=\pi{\int_{{0}}^{{3}}}{\left[{15}-{8}{x}+{{x}}^{{2}}\right]}{\left.{d}{x}\right.}$

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

$\displaystyle{V}=\pi{\left[{\int_{{0}}^{{3}}}{15}{\left.{d}{x}\right.}-{\int_{{0}}^{{3}}}{8}{x}{d}+{\int_{{0}}^{{3}}}{{x}}^{{2}}{\left.{d}{x}\right.}\right]}$

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

$\displaystyle{V}=\pi{\left[{15}{x}-{8}\cdot\frac{{{x}}^{{2}}}{{2}}+\frac{{{x}}^{{3}}}{{3}}\right]}{{\mid}_{{0}}^{{3}}}$

$\displaystyle{V}=\pi{\left[{15}{x}-{4}{{x}}^{{2}}+\frac{{{x}}^{{3}}}{{3}}\right]}{{\mid}_{{0}}^{{3}}}$

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

$\displaystyle{V}=\pi{\left[{15}\cdot{\left({3}\right)}-{4}{{\left({3}\right)}}^{{2}}+\frac{{{\left({3}\right)}}^{{3}}}{{3}}\right]}-\pi{\left[{15}{\left({0}\right)}-{4}{{\left({0}\right)}}^{{2}}+\frac{{{\left({0}\right)}}^{{3}}}{{3}}\right]}$

$\displaystyle{V}=\pi{\left[{45}-{36}+{9}\right]}-\pi{\left[{0}-{0}+{0}\right]}$

$\displaystyle{V}={18}\pi$ or $\displaystyle{56.55}$ (approximated value)

Notes

Thursday, 22 December 2016

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

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

Thursday, 22 December 2016

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

No comments:

Post a Comment

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

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

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