Blog: `y = x, y = 0, x = 2, x = 4` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified...

Sunday, September 6, 2009

$\displaystyle{y}={x},{y}={0},{x}={2},{x}={4}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified...

You need to evaluate the volume of the solid obtained by the rotation of the region bounded by the curves $\displaystyle{y}={x},{y}={0},{x}={2},{x}={4}$ about x = 1, using washer method, such that:

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

You may evaluate the volume

$\displaystyle{V}=\pi\cdot{\int_{{2}}^{{4}}}{{\left({1}-{y}\right)}}^{{2}}{\left.{d}{y}\right.}$

$\displaystyle{V}=\pi\cdot{\int_{{2}}^{{4}}}{\left({1}-{2}{y}+{{y}}^{{2}}\right)}{\left.{d}{y}\right.}$

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

$\displaystyle{V}=\pi\cdot{\left({y}-{2}\cdot\frac{{{y}}^{{2}}}{{2}}+\frac{{{y}}^{{3}}}{{3}}\right)}{{\mid}_{{2}}^{{4}}}$

$\displaystyle{V}=\pi\cdot{\left({y}-{{y}}^{{2}}+\frac{{{y}}^{{3}}}{{3}}\right)}{{\mid}_{{2}}^{{4}}}$

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

$\displaystyle{V}=\pi\cdot{\left(-{10}+\frac{{56}}{{3}}\right)}$

$\displaystyle{V}=\frac{{{26}\pi}}{{3}}$

Hence, evaluating the volume of the solid obtained by the rotation of the region bounded by the curves $\displaystyle{y}={x},{y}={0},{x}={2},{x}={4}$ about x = 1, yields $\displaystyle{V}=\frac{{{26}\pi}}{{3}}.$

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Sunday, September 6, 2009

$\displaystyle{y}={x},{y}={0},{x}={2},{x}={4}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified...

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