Notes: `y = x, y = (2x)/(1 + x^3)` (a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

Friday, 15 May 2015

$\displaystyle{y}={x},{y}=\frac{{{2}{x}}}{{{1}+{{x}}^{{3}}}}$ (a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

The shell has the radius $\displaystyle{x}-{\left(-{1}\right)}$ , the cricumference is $\displaystyle{2}\pi\cdot{\left({x}+{1}\right)}$ and the height is $\displaystyle\frac{{{2}{x}}}{{{1}+{{x}}^{{3}}}}-{x},$ hence, the volume can be evaluated, using the method of cylindrical shells, such that:

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

You need to find the endpoints, using the equation $\displaystyle\frac{{{2}{x}}}{{{1}+{{x}}^{{3}}}}-{x}={0}\Rightarrow{2}{x}-{x}-{{x}}^{{4}}={0}\Rightarrow{3}{x}-{{x}}^{{4}}={0}\Rightarrow{x}{\left({3}-{{x}}^{{3}}\right)}={0}\ldots$

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

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

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

Reducing the like terms yields:

$\displaystyle{V}={2}\pi\cdot{\int_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}\frac{{{3}{x}-{{x}}^{{4}}}}{{{{x}}^{{2}}-{x}+{1}}}{\left.{d}{x}\right.}$

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

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

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

$\displaystyle{V}={2}\pi\cdot{\left({2}\cdot{\int_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}\frac{{{2}{x}+{1}-{1}}}{{{{x}}^{{2}}-{x}+{1}}}{\left.{d}{x}\right.}-\frac{{{x}}^{{3}}}{{3}}{{\left|_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}-\frac{{{x}}^{{2}}}{{2}}\right|}_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}\right)}$

$\displaystyle{V}={2}\pi\cdot{\left({2}\cdot{\int_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}\frac{{{2}{x}-{1}}}{{{{x}}^{{2}}-{x}+{1}}}{\left.{d}{x}\right.}+{2}\cdot{\int_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}\frac{{{1}}}{{{{x}}^{{2}}-{x}+{1}}}{\left.{d}{x}\right.}-{1}-\frac{{{\sqrt[{{3}}]{{9}}}}}{{2}}\right)}$

You need to solve $\displaystyle{\int_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}\frac{{{2}{x}-{1}}}{{{{x}}^{{2}}-{x}+{1}}}{\left.{d}{x}\right.}$ using substitution $\displaystyle{{x}}^{{2}}-{x}+{1}={t}\Rightarrow{\left({2}{x}-{1}\right)}{\left.{d}{x}\right.}={\left.{d}{t}\right.}.$

$\displaystyle{V}={2}\pi\cdot{\left({2}\cdot{\ln{{\left({{x}}^{{2}}-{x}+{1}\right)}}}{{\mid}_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}+{2}\cdot{\int_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}\frac{{{1}}}{{{{\left({x}-\frac{{1}}{{2}}\right)}}^{{2}}+{{\left(\frac{{\sqrt{{3}}}}{{2}}\right)}}^{{2}}}}{\left.{d}{x}\right.}-{1}-\frac{{{\sqrt[{{3}}]{{9}}}}}{{2}}\right)}$

$\displaystyle{V}={2}\pi\cdot{\left({2}\cdot{\ln{{\left({\left({\sqrt[{{3}}]{{9}}}\right)}-{\left({\sqrt[{{3}}]{{3}}}\right)}+{1}\right)}}}+{\left(\frac{{4}}{\sqrt{{3}}}\right)}\cdot\frac{{\arctan{{\left({2}{x}-{1}\right)}}}}{\sqrt{{3}}}{{\mid}_{{0}}^{{{\sqrt[{{3}}]{{3}}}}}}-{1}-\frac{{{\sqrt[{{3}}]{{9}}}}}{{2}}\right)}$

$\displaystyle{V}={2}\pi\cdot{\left({2}\cdot{\ln{{\left({\left({\sqrt[{{3}}]{{9}}}\right)}-{\left({\sqrt[{{3}}]{{3}}}\right)}+{1}\right)}}}+{\left(\frac{{4}}{\sqrt{{3}}}\right)}\cdot\frac{{\arctan{{\left({2}{\left({\sqrt[{{3}}]{{3}}}\right)}-{1}\right)}}}}{\sqrt{{3}}}-{\left(\frac{{4}}{\sqrt{{3}}}\right)}\cdot{\left(\frac{\pi}{{6}}\right)}-{1}-\frac{{{\sqrt[{{3}}]{{9}}}}}{{2}}\right)}$

Hence, evaluating the volume, using the method of cylindrical shells, yields $\displaystyle{V}={2}\pi\cdot{\left({2}\cdot{\ln{{\left({\left({\sqrt[{{3}}]{{9}}}\right)}-{\left({\sqrt[{{3}}]{{3}}}\right)}+{1}\right)}}}+{\left(\frac{{4}}{\sqrt{{3}}}\right)}\cdot\frac{{\arctan{{\left({2}{\left({\sqrt[{{3}}]{{3}}}\right)}-{1}\right)}}}}{\sqrt{{3}}}-{\left(\frac{{4}}{\sqrt{{3}}}\right)}\cdot{\left(\frac{\pi}{{6}}\right)}-{1}-\frac{{{\sqrt[{{3}}]{{9}}}}}{{2}}\right)}$

Notes

Friday, 15 May 2015

$\displaystyle{y}={x},{y}=\frac{{{2}{x}}}{{{1}+{{x}}^{{3}}}}$ (a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

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Is there any personification in "The Tell-Tale Heart"?

Friday, 15 May 2015

(a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

No comments:

Post a Comment

Is there any personification in &quot;The Tell-Tale Heart&quot;?

$\displaystyle{y}={x},{y}=\frac{{{2}{x}}}{{{1}+{{x}}^{{3}}}}$ (a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

Is there any personification in "The Tell-Tale Heart"?