Blog: `y = sqrt(x - 1), y = 0, x = 5` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified...

Wednesday, June 27, 2012

$\displaystyle{y}=\sqrt{{{x}-{1}}},{y}={0},{x}={5}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified...

The volume of the solid obtained by rotating the region bounded by the curves $\displaystyle{y}=\sqrt{{{x}-{1}}},{y}={0},{x}={5}$ , about x axis, can be evaluated using the washer method, such that:

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

Since the problem provides you the endpoint x = 5, you need to find the other endpoint of interval, hence, you need to solve for x the following equation, such that:

$\displaystyle\sqrt{{{x}-{1}}}={0}\Rightarrow{x}-{1}={0}\Rightarrow{x}={1}$

$\displaystyle{V}={\int_{{1}}^{{5}}}\pi\cdot{{\left(\sqrt{{{x}-{1}}}-{0}\right)}}^{{2}}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi\cdot{\int_{{1}}^{{5}}}{\left({x}-{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi\cdot{\int_{{1}}^{{5}}}{\left({x}\right)}{\left.{d}{x}\right.}-\pi\cdot{\int_{{1}}^{{5}}}{\left.{d}{x}\right.}$

$\displaystyle{V}={\left(\pi\cdot\frac{{{x}}^{{2}}}{{2}}-\pi\cdot{x}\right)}{{\mid}_{{1}}^{{5}}}$

$\displaystyle{V}={\left(\pi\cdot\frac{{{5}}^{{2}}}{{2}}-\pi\cdot{5}-\pi\cdot\frac{{{1}}^{{2}}}{{2}}+\pi\cdot{1}\right)}$

$\displaystyle{V}=\frac{{{25}\pi}}{{2}}-\frac{\pi}{{2}}-{4}\pi$

$\displaystyle{V}={12}\pi-{4}\pi$

$\displaystyle{V}={8}\pi$

Hence, evaluating the volume of the solid obtained by rotating the region bounded by the curves $\displaystyle{y}=\sqrt{{{x}-{1}}},{y}={0},{x}={5}$ , about x axis , using the washer method, yields $\displaystyle{V}={8}\pi.$

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Wednesday, June 27, 2012

$\displaystyle{y}=\sqrt{{{x}-{1}}},{y}={0},{x}={5}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified...

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