Blog: `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....

Tuesday, April 30, 2013

$\displaystyle{x}={{y}}^{{2}},{x}={1}-{{y}}^{{2}}$ 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:

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

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

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

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

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

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

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

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

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

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

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

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

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Tuesday, April 30, 2013

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

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