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

[MarkFL]

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.

 

[MarkFL]

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:

$$dV=\pi r^2\,dx$$

where:

$$r=y=xe^x$$

and so we have:

$$dV=\pi x^2e^{2x}\,dx$$

Summing the disks by integration, we then have:

$$V=\pi\int_0^1 x^2e^{2x}\,dx$$

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

$$u=x^2\,\therefore\,du=2x\,dx$$

$$dv=e^{2x}\,dx\,\therefore\,v=\frac{1}{2}e^{2x}$$

and so we have:

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

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

Now, using integration by parts again:

$$u=x\,\therefore\,du=dx$$

$$dv=e^{2x}\,dx\,\therefore\,v=\frac{1}{2}e^{2x}$$

and we have:

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

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

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

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 http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.