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albear · Jun 17, 2007, 06:20 AM

$\y=x{e}^x$

finite region R which is bounded by the curve y=xe^x the line x=1 and the line x=3 and the x axis
the region R is rotated through 360 degrees (2pi) about the x axis
use integration by parts to find an exact value for the volume of the solid generated

i think i can do the parts but it's the 360 thing I'm stuck with

i end up with the equation $\{pie}[x{e}^{x}+e^x] with boundaries 3 and 1$

albear · Jun 17, 2007, 06:45 AM

Anybody

galactus · Jun 17, 2007, 07:12 AM

Use washers.

${\pi}\int_{1}^{3}(xe^{x})^{2}dx$

Also, never spell 'pi' as 'pie' again. Bad Albear. ;) :rolleyes:

albear · Jun 17, 2007, 07:21 AM

I understand why there is an integral sign with 3 and 1 as boundaries but I don't understand why there is a pi and ^2 involved, and what do you mean by washers? (I didn't know weather to use pi or pie in the math linguo but you got my meaning )

galactus · Jun 17, 2007, 07:23 AM

I am sorry Albear, if you're that lost you should see your instructor and/or read your calc book. That is basic solids of revolution technique. I can not teach all that here.
Good luck.

albear · Jun 17, 2007, 08:20 AM

could you just tell me why you have put a pi in front and why y has been squared

galactus · Jun 17, 2007, 08:46 AM

Let f be continuous and nonnegative on [a,b], and let R be the region bounded above by the graph of f, below by the x-axis, and on the sides by the lines x=a and x=b. When this region is revolved about the x-axis, it generates a solid having circular cross sections. Since the cross sections at x has radius f(x), the cross sectional area is given by $A(x)={\pi}\int_{a}^{b}[f(x)]^{2}dx$ . Because the cross sections are circular or disk-shaped, it's known as method of washers or disks

I would suggest looking it up in a good calc book. To explain it well is more than I am willing to get into here.

Besides, there's shells as well as washers. $2{\pi}\int_{a}^{b}xf(x)dx$