1. ## solids of revolution

find the solid of revolution of the function lnx when it's rotated about the x-axis from 1 to e.

2. Originally Posted by nertil1
find the solid of revolution of the function lnx when it's rotated about the x-axis from 1 to e.
By the disk method, the required volume is given by:

$\displaystyle V = \pi \int_{1}^{e}( \ln x )^2 dx$

3. Originally Posted by Jhevon
By the disk method, the required volume is given by:

$\displaystyle V = \pi \int_{1}^{e}( \ln x )^2 dx$
Now I'm curious. Just how do you go about integrating that?

-Dan

4. I tried this, and I still can't get it.

What I did is $\displaystyle 2 \pi \int_{1}^{e}( \ln x)^2 dx$
and I get $\displaystyle 2\pi$, which I think is the right answer

but I can't seem to do it with the method suggested.

5. Originally Posted by topsquark
Now I'm curious. Just how do you go about integrating that?

-Dan
by parts

6. Originally Posted by nertil1
I tried this, and I still can't get it.

What I did is $\displaystyle 2 \pi \int_{1}^{e}( \ln x)^2 dx$
and I get $\displaystyle 2\pi$, which I think is the right answer

but I can't seem to do it with the method suggested.
here you are trying to use the shell method. in this case you need to change the function from a function of x to a function of y, and change the limits as well. by the way, the lnx should not be squared when using this method, and you have to multiply by the radius instead. personally, i think this is more work than its worth. use the disk method

7. ## Re;

Re:

8. Originally Posted by qbkr21
Re:
thanks for that qbkr21, but you forgot to multiply by $\displaystyle \pi$

9. ## Re:

I feel that this is awkward given that I have neven multiplied a pi times and e before...

-qbkr21

10. Originally Posted by qbkr21
I feel that this is awkward given that I have neven multiplied a pi times and e before...

-qbkr21
not to be a pain, but i think you also evaluated incorrectly. i think it should be $\displaystyle \pi (e - 2)$.

11. Originally Posted by Jhevon
not to be a pain, but i think you also evaluated incorrectly. i think it should be $\displaystyle \pi (e - 2)$.
That agrees with the answer my calculator gave, anyway.

-Dan

12. ## Re:

I agree you both. I made a mistake.

Andrew