Volume of Solid of Revolution: Find Solution Here!

[MacLaddy1]

Hey guys and gals. Hoping someone can help out with a problem I am finding myself stuck on.

The question goes as follows.

Solids of revolution. Find the volume of the solid of revolution. The region bounded by \(y= \frac{ln(x)}{\sqrt(x)}\), y=0 and x=2, revolved about the x-axis.

The problem I am having is trying to figure out how to separate the top and bottom of this fraction. These are some of the things I've looked at, but I don't think I can integrate any of them.

\([x^{\frac{1}{2}} * ln(x)]^2\)

\(\frac{[ln(x)]}{x}\)

And a few others.

I know that I have to integrate with \(\int \pi[\frac{[ln(x)]}{x}]dx\) from 1 to 2 ( I don't know how to get the limits on the integral)

Any help would be very much appreciated. This doesn't seem to be too hard of a problem, but I can't figure out how to get these separated so I can integrate.

Thanks,
Mac

 

[Jameson]

Hi MacLaddy!

I agree that the limits of integration are from x=1 to x=2, however you need to integrate \(\displaystyle \int \pi [f(x)]^2 dx\). It looks like you forgot to square the numerator as well or weren't able to write it in Latex. Either way I get that in this case \(\displaystyle [f(x)]^2=\left( \frac{\ln(x)}{\sqrt{x}} \right)^2= \left( \frac{[\ln(x)]^2}{x} \right)\). From here let \(\displaystyle u=\ln(x)\) and \(\displaystyle du=\frac{1}{x}dx\). From there it's doing a standard u-substitution integral and plugging everything in.

 

[MacLaddy1]

Hi MacLaddy!

I agree that the limits of integration are from x=1 to x=2, however you need to integrate \(\displaystyle \int \pi [f(x)]^2 dx\). It looks like you forgot to square the numerator as well or weren't able to write it in Latex. Either way I get that in this case \(\displaystyle [f(x)]^2=\left( \frac{\ln(x)}{\sqrt{x}} \right)^2= \left( \frac{[\ln(x)]^2}{x} \right)\). From here let \(\displaystyle u=\ln(x)\) and \(\displaystyle du=\frac{1}{x}dx\). From there it's doing a standard u-substitution integral and plugging everything in.

Ahh, I knew I was missing something simple. I didn't even notice the \(\frac{1}{x}\) u-substitution connection. I did remember to square the ln(x), I just forgot to write it in the Latex.

So we end up with \(\pi[\frac{[ln(x)]^3}{3}]^2_1\) = \(\frac{\pi(ln2)^3}{3} \approx 0.349\)

Does that look about right to you? (seems strange that I can't get a better answer. I think I still might be making a mistake)

Thank you very much. That helped a ton.
Mac

(by the way, how do you make your Latex readable? I need a magnifying glass for mine)

 

[Jameson]

If you use the tags [math][/math] then the Latex output will appear larger. If you want to keep using the current tags you are using then add \displaystyle after the opening tag and before any of the Latex code.

You're close but made a small error. If \(\displaystyle u=\ln(x)\) then our integral becomes \(\displaystyle \pi \int u^2du=\pi \frac{u^3}{3}\). From there you plug back in ln(x) for u and evaluate everything using the bounds.

EDIT: I see you've already caught your own error. Well done :)

I never make any promises that I'm 100% correct but your solution looks good to me. Hopefully I won't wake up to find others correcting a mistake I made but I can't find any in your work.

 

[MacLaddy1]

If you use the tags [math][/math] then the Latex output will appear larger. If you want to keep using the current tags you are using then add \displaystyle after the opening tag and before any of the Latex code.

You're close but made a small error. If \(\displaystyle u=\ln(x)\) then our integral becomes \(\displaystyle \pi \int u^2du=\pi \frac{u^3}{3}\). From there you plug back in ln(x) for u and evaluate everything using the bounds.

EDIT: I see you've already caught your own error. Well done :)

I never make any promises that I'm 100% correct but your solution looks good to me. Hopefully I won't wake up to find others correcting a mistake I made but I can't find any in your work.

The Math tags look simpler to use for me. More like the itex tags I am used to.

Thank you again, I really appreciate your help. That was my last question for the night, now I can get some sleep.

Mac