Russell Marsh

2022-09-07

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify your result using the integration capabilities of a graphing utility
$y={e}^{x-4}\phantom{\rule{0ex}{0ex}}y=0\phantom{\rule{0ex}{0ex}}x=4\phantom{\rule{0ex}{0ex}}x=6$

Joel Reese

Given
$y={e}^{x-4}\phantom{\rule{0ex}{0ex}}y=0\phantom{\rule{0ex}{0ex}}x=4\phantom{\rule{0ex}{0ex}}x=6$
The volume is,
$V=\pi {\int }_{4}^{6}\left({e}^{x-4}{\right)}^{2}dx\phantom{\rule{0ex}{0ex}}=\pi {\int }_{4}^{6}{e}^{2x-8}dx\phantom{\rule{0ex}{0ex}}=\pi {\left[\frac{{e}^{2x-8}}{2}\right]}_{4}^{6}\phantom{\rule{0ex}{0ex}}=\pi \left[\frac{{e}^{2\left(6\right)-8}}{2}-\frac{{e}^{2\left(4\right)-8}}{2}\right]\phantom{\rule{0ex}{0ex}}=\pi \left[\frac{{e}^{4}}{2}-\frac{{e}^{0}}{2}\right]\phantom{\rule{0ex}{0ex}}=\pi \left[\frac{{e}^{4}}{2}-\frac{1}{2}\right]$
Now by using graphing utility, we get
$\pi {\int }_{4}^{6}\left({e}^{\left(x-4\right)}{\right)}^{2}dx\phantom{\rule{0ex}{0ex}}=84.191777951$

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