Volume of the solid of revolution

[benedwards2020]

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

[tex]f(x) = \left( \frac{1}{x} \right) e^\frac{1}{x} [/tex]

from x = 1 to x = 6

Homework Equations

[tex] \pi \int (f(x))^2 dx [/tex]

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

[tex]\pi \int_1^6 \left( \frac{1}{x} \right) \left( e^\frac{1}{x} \right)^2 dx[/tex]


So, we can simplyfy this to

[tex]\pi \int_1^6 (x)^{-2} \times e^\frac{2}{x} dx[/tex]

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?

 

[Benny]

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

 

[HallsofIvy]

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

[tex]f(x) = \left( \frac{1}{x} \right) e^\frac{1}{x} [/tex]

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

Homework Equations

[tex] \pi \int (f(x))^2 dx [/tex]

So you are rotating around the x-axis?

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)

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

So, the volume is

[tex]\pi \int \left( \frac{1}{x} \right) \left( e^\frac{1}{x} \right)^2 dx[/tex]

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

With the values a=1 and b=6

So, we can simplyfy this to

[tex]\pi \int (x)^{-2} \times e^\frac{2}{x} dx[/tex]

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.

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?

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

 

[benedwards2020]

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