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

$y={e}^{x-4}\phantom{\rule{0ex}{0ex}}y=0\phantom{\rule{0ex}{0ex}}x=4\phantom{\rule{0ex}{0ex}}x=6$

$y={e}^{x-4}\phantom{\rule{0ex}{0ex}}y=0\phantom{\rule{0ex}{0ex}}x=4\phantom{\rule{0ex}{0ex}}x=6$

Joel Reese

Beginner2022-09-08Added 17 answers

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,

$V=\pi {\int}_{4}^{6}({e}^{x-4}{)}^{2}dx\phantom{\rule{0ex}{0ex}}=\pi {\int}_{4}^{6}{e}^{2x-8}dx\phantom{\rule{0ex}{0ex}}=\pi {\left[\frac{{e}^{2x-8}}{2}\right]}_{4}^{6}\phantom{\rule{0ex}{0ex}}=\pi [\frac{{e}^{2(6)-8}}{2}-\frac{{e}^{2(4)-8}}{2}]\phantom{\rule{0ex}{0ex}}=\pi [\frac{{e}^{4}}{2}-\frac{{e}^{0}}{2}]\phantom{\rule{0ex}{0ex}}=\pi [\frac{{e}^{4}}{2}-\frac{1}{2}]$

Now by using graphing utility, we get

$\pi {\int}_{4}^{6}({e}^{(x-4)}{)}^{2}dx\phantom{\rule{0ex}{0ex}}=84.191777951$

$y={e}^{x-4}\phantom{\rule{0ex}{0ex}}y=0\phantom{\rule{0ex}{0ex}}x=4\phantom{\rule{0ex}{0ex}}x=6$

The volume is,

$V=\pi {\int}_{4}^{6}({e}^{x-4}{)}^{2}dx\phantom{\rule{0ex}{0ex}}=\pi {\int}_{4}^{6}{e}^{2x-8}dx\phantom{\rule{0ex}{0ex}}=\pi {\left[\frac{{e}^{2x-8}}{2}\right]}_{4}^{6}\phantom{\rule{0ex}{0ex}}=\pi [\frac{{e}^{2(6)-8}}{2}-\frac{{e}^{2(4)-8}}{2}]\phantom{\rule{0ex}{0ex}}=\pi [\frac{{e}^{4}}{2}-\frac{{e}^{0}}{2}]\phantom{\rule{0ex}{0ex}}=\pi [\frac{{e}^{4}}{2}-\frac{1}{2}]$

Now by using graphing utility, we get

$\pi {\int}_{4}^{6}({e}^{(x-4)}{)}^{2}dx\phantom{\rule{0ex}{0ex}}=84.191777951$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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