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Frank Rodgriguez's question at Yahoo! Answers regarding a solid of revolution

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MarkFL

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Feb 24, 2012
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Here is the question:

Find the volume of the solid obtained by revolving the region enclosed by y = xe^x , y = 0 and x = 1?

About the x-axis. I am having a hard time with this problem. My professor says the answer should be pi/2 (e^2 - 1). He also mentions that he could be wrong. Can someone please show me HOW to solve this problem?
Here is a link to the question:

Find the volume of the solid obtained by revolving the region enclosed by y = xe^x , y = 0 and x = 1? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
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Hello Frank Rodriguez,

The first thing I like to do is plot the region to be revolved:

Let's use the disk method. The volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dx\)

where:

\(\displaystyle r=y=xe^x\)

and so we have:

\(\displaystyle dV=\pi x^2e^{2x}\,dx\)

Summing the disks by integration, we then have:

\(\displaystyle V=\pi\int_0^1 x^2e^{2x}\,dx\)

To evaluate this definite integral, let's try integration by parts:

\(\displaystyle u=x^2\,\therefore\,du=2x\,dx\)

\(\displaystyle dv=e^{2x}\,dx\,\therefore\,v=\frac{1}{2}e^{2x}\)

and so we have:

\(\displaystyle \frac{V}{\pi}=\left[\frac{1}{2}x^2e^{2x} \right]_0^1-\int_0^1 xe^{2x}\,dx\)

\(\displaystyle \frac{V}{\pi}=\frac{1}{2}e^{2}-\int_0^1 xe^{2x}\,dx\)

Now, using integration by parts again:

\(\displaystyle u=x\,\therefore\,du=dx\)

\(\displaystyle dv=e^{2x}\,dx\,\therefore\,v=\frac{1}{2}e^{2x}\)

and we have:

\(\displaystyle \frac{V}{\pi}=\frac{1}{2}e^{2}-\left(\left[\frac{1}{2}xe^{2x} \right]_0^1-\frac{1}{2}\int_0^1 e^{2x}\,dx \right)\)

\(\displaystyle \frac{V}{\pi}=\frac{1}{2}e^{2}-\left(\frac{1}{2}e^{2}-\frac{1}{4}\left[e^{2x} \right]_0^1 \right)\)

\(\displaystyle V=\frac{\pi}{4}\left(e^{2}-1 \right)\)

Normally, if practical, I like to also use the shell method to check my work, however solving $y=xe^x$ for $x$ requires the use of the Lambert-W function, and so we shall leave it at that.

To Frank Rodgriguez and any other guests viewing this topic, I invite and encourage you to post other calculus problems in our Calculus forum.

Best Regards,

Mark.