# Volume of the solid of revolution

1. Jul 12, 2007

### benedwards2020

Find the volume of the solid of revolution obtained when the region under the graph of

$$f(x) = \left( \frac{1}{x} \right) e^\frac{1}{x}$$

from x = 1 to x = 6

2. Relevant equations

$$\pi \int (f(x))^2 dx$$

3. The attempt at a solution

Ok, the equation I gave above should be that of a definite integral with a=1 and b=6 (If anyone can tell me how to write that in Latex it would be much appreciated)

So, the volume is

$$\pi \int_1^6 \left( \frac{1}{x} \right) \left( e^\frac{1}{x} \right)^2 dx$$

So, we can simplyfy this to

$$\pi \int_1^6 (x)^{-2} \times e^\frac{2}{x} dx$$

Now I'm a bit stuck as to where to go from here. Do I use the integration by parts method? I think I'm getting bogged down in unnecessary calculations. Can someone give me a hint or point me in the right direction?

Last edited: Jul 12, 2007
2. Jul 12, 2007

### Benny

Try setting u = 1/x or something like that.

3. Jul 12, 2007

### HallsofIvy

Staff Emeritus
?? You didn't finish your sentence! Between what limits? Rotated around what axis?

So you are rotating around the x-axis?

And only the region between x= 1 and x= 6?

No, If $f(x)= \frac{1}{x}e^{1/x}$ then both $f^2(x)= \frac{1}{x^2}e^{2/x}$

Okay, good. Now you have first x squared also. By the way, the code to put the limits of integration in is "\int_1^6". In other words, treat the lower limit as a subscript and the upper limit as a superscript on the integral sign.

Looks to me like the substitution u= 2/x should work nicely.

4. Jul 12, 2007

### benedwards2020

Many thanx for your time and help..... I've managed to solve using substitution as you suggested.

Last edited: Jul 12, 2007