Notes: `y = (1/4)x^2, y = 5 - x^2` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

Monday, 12 October 2015

$\displaystyle{y}={\left(\frac{{1}}{{4}}\right)}{{x}}^{{2}},{y}={5}-{{x}}^{{2}}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

$\displaystyle{y}=\frac{{{x}}^{{2}}}{{4}},{y}={5}-{{x}}^{{2}}$

Refer the attached image. Graph of $\displaystyle{y}=\frac{{{x}}^{{2}}}{{4}}$ is plotted in red color and graph of $\displaystyle{y}={5}-{{x}}^{{2}}$ is plotted in blue color.

From the graph, the curves intersects at x=-2 , x=2.

Using washer method,

Inner radius of washer= $\displaystyle\frac{{{x}}^{{2}}}{{4}}$

Outer radius of washer= $\displaystyle{5}-{{x}}^{{2}}$

Cross sectional area of washer A(x)= $\displaystyle\pi{\left({{\left({5}-{{x}}^{{2}}\right)}}^{{2}}-{{\left(\frac{{{x}}^{{2}}}{{4}}\right)}}^{{2}}\right)}$

$\displaystyle{A}{\left({x}\right)}=\pi{\left({25}+{{x}}^{{4}}-{10}{{x}}^{{2}}-\frac{{{x}}^{{4}}}{{16}}\right)}$

$\displaystyle{A}{\left({x}\right)}=\pi{\left(\frac{{15}}{{16}}{{x}}^{{4}}-{10}{{x}}^{{2}}+{25}\right)}$

Volume of the solid obtained by rotating the region bounded by the curves about the x=axis ,V= $\displaystyle{\int_{{-{2}}}^{{2}}}{A}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}={\int_{{-{2}}}^{{2}}}\pi{\left(\frac{{15}}{{16}}{{x}}^{{4}}-{10}{{x}}^{{2}}+{25}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi{{\left[\frac{{15}}{{16}}\cdot\frac{{{x}}^{{5}}}{{5}}-{10}\cdot\frac{{{x}}^{{3}}}{{3}}+{25}{x}\right]}_{{-{2}}}^{{2}}}$

$\displaystyle{V}=\pi{{\left[\frac{{3}}{{16}}\cdot{{x}}^{{5}}-\frac{{10}}{{3}}\cdot{{x}}^{{3}}+{25}{x}\right]}_{{-{2}}}^{{2}}}$

$\displaystyle{V}=\pi{\left({\left(\frac{{3}}{{16}}\cdot{{2}}^{{5}}-\frac{{10}}{{3}}\cdot{{2}}^{{3}}+{25}\cdot{2}\right)}-{\left(\frac{{3}}{{16}}\cdot{{\left(-{2}\right)}}^{{5}}-\frac{{10}}{{3}}\cdot{{\left(-{2}\right)}}^{{3}}+{25}\cdot{\left(-{2}\right)}\right)}\right)}$

$\displaystyle{V}=\pi{\left({\left({6}-\frac{{80}}{{3}}+{50}\right)}-{\left(-{6}+\frac{{80}}{{3}}-{50}\right)}\right)}$

$\displaystyle{V}=\pi{\left({6}-\frac{{80}}{{3}}+{50}+{6}-\frac{{80}}{{3}}+{50}\right)}$

$\displaystyle{V}=\pi{\left({112}-\frac{{160}}{{3}}\right)}$

$\displaystyle{V}=\pi\frac{{{336}-{160}}}{{3}}$

$\displaystyle{V}=\pi{\left(\frac{{176}}{{3}}\right)}$

$\displaystyle{V}=\frac{{{176}\pi}}{{3}}$

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