Notes: `x = y^2, x = 1 - y^2` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

Sunday, 4 September 2016

$\displaystyle{x}={{y}}^{{2}},{x}={1}-{{y}}^{{2}}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

You need to evaluate the volume using the washer method, such that:

$\displaystyle{V}=\pi\cdot{\int_{{a}}^{{b}}}{\left({{f}}^{{2}}{\left({x}\right)}-{{g}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

You need first to determine the endpoints, hence you need to solve for y the following equation, such that:

$\displaystyle{{y}}^{{2}}={1}-{{y}}^{{2}}$

$\displaystyle{2}{{y}}^{{2}}={1}\Rightarrow{{y}}^{{2}}=\frac{{1}}{{2}}\Rightarrow{y}_{{{1},{2}}}=\pm\frac{{\sqrt{{2}}}}{{2}}$

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

$\displaystyle{V}=\pi\cdot{\int_{{-\frac{{\sqrt{{2}}}}{{2}}}}^{{\frac{{\sqrt{{2}}}}{{2}}}}}{\left({9}-{6}{{y}}^{{2}}+{{y}}^{{4}}-{4}-\ldots\right.}$

You need to evaluate the volume using the washer method, such that:

$\displaystyle{V}=\pi\cdot{\int_{{a}}^{{b}}}{\left({{f}}^{{2}}{\left({x}\right)}-{{g}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

You need first to determine the endpoints, hence you need to solve for y the following equation, such that:

$\displaystyle{{y}}^{{2}}={1}-{{y}}^{{2}}$

$\displaystyle{2}{{y}}^{{2}}={1}\Rightarrow{{y}}^{{2}}=\frac{{1}}{{2}}\Rightarrow{y}_{{{1},{2}}}=\pm\frac{{\sqrt{{2}}}}{{2}}$

$\displaystyle{V}=\pi\cdot{\int_{{-\frac{{\sqrt{{2}}}}{{2}}}}^{{\frac{{\sqrt{{2}}}}{{2}}}}}{\left({9}-{6}{{y}}^{{2}}+{{y}}^{{4}}-{4}-{4}{{y}}^{{2}}-{{y}}^{{4}}\right)}{\left.{d}{y}\right.}$

$\displaystyle{V}=\pi\cdot{\int_{{-\frac{{\sqrt{{2}}}}{{2}}}}^{{\frac{{\sqrt{{2}}}}{{2}}}}}{\left({5}-{10}{{y}}^{{2}}\right)}{\left.{d}{y}\right.}$

$\displaystyle{V}=\pi\cdot{\left({\int_{{-\frac{{\sqrt{{2}}}}{{2}}}}^{{\frac{{\sqrt{{2}}}}{{2}}}}}{5}{\left.{d}{y}\right.}-{\int_{{-\frac{{\sqrt{{2}}}}{{2}}}}^{{\frac{{\sqrt{{2}}}}{{2}}}}}{10}{{y}}^{{2}}{\left.{d}{y}\right.}\right)}$

$\displaystyle{V}=\pi\cdot{\left({5}{y}{{\left|_{{\left(-\frac{{\sqrt{{2}}}}{{2}}\right)}}^{{\frac{{\sqrt{{2}}}}{{2}}}}-{10}\frac{{{y}}^{{3}}}{{3}}\right|}_{{-\frac{{\sqrt{{2}}}}{{2}}}}^{{\frac{{\sqrt{{2}}}}{{2}}}}}\right)}$

$\displaystyle{V}=\pi\cdot{\left({5}{\left(\frac{{\sqrt{{2}}}}{{2}}+\frac{{\sqrt{{2}}}}{{2}}\right)}-\frac{{10}}{{3}}{\left({2}\frac{\sqrt{{2}}}{{8}}+{2}\frac{\sqrt{{2}}}{{8}}\right)}\right)}$

$\displaystyle{V}=\pi\cdot{\left({5}\sqrt{{2}}-{10}\cdot\frac{\sqrt{{2}}}{{6}}\right)}$

$\displaystyle{V}=\frac{{{20}\cdot\pi\cdot\sqrt{{2}}}}{{6}}$

$\displaystyle{V}=\frac{{{10}\cdot\pi\cdot\sqrt{{2}}}}{{3}}$

Hence, evaluating the volume of the solid obtained by rotating the region bounded by the given curves, about x = 3, yields $\displaystyle{V}=\frac{{{10}\cdot\pi\cdot\sqrt{{2}}}}{{3}}.$

Notes

Sunday, 4 September 2016

$\displaystyle{x}={{y}}^{{2}},{x}={1}-{{y}}^{{2}}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

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

Sunday, 4 September 2016

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

No comments:

Post a Comment

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

$\displaystyle{x}={{y}}^{{2}},{x}={1}-{{y}}^{{2}}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

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