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September 15th, 2012, 05:08 PM  #1 
Newbie Joined: Sep 2012 Posts: 9 Thanks: 0  Disk/Washer Integration with ln function
Hi, me again. I have a region R bounded by y=(ln(x/5))^1/2, y=(ln((x^2)/25)^1/2, y=1 and revolved around the x axis. I feel completely stuck on this one. I know one of the bounds will be 5, at x=5, they are both y=0. The other will be x=5e, so it's from 5 > 5e. With the formula, it's the integral of pi*(f(x)^2g(x)^2)dx, which gets rid of the square root sign. What I'm confused on is, how am I supposed to take the integral of the ln? I tried deriving it by parts, but I get a funky answer. The books answer is 5pi(e^(1/2)  1)^2 Help would be much appreciated 
September 15th, 2012, 06:21 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond  Re: Disk/Washer Integration with ln function
Here's a setup: Integration by parts for ln(u): O.k? 
September 16th, 2012, 10:46 AM  #3 
Newbie Joined: Sep 2012 Posts: 9 Thanks: 0  Re: Disk/Washer Integration with ln function
Thanks a bunch, if you wouldn't mind, how would I set this up for the shell method? The fact that they're not being squared throws me off. Also would I have to set up 2 different integrals for the shell method also?

September 16th, 2012, 01:27 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond  Re: Disk/Washer Integration with ln function
My understanding of shells is that we are rotating around the yaxis, so the answer will be different than that above. Does the book give an answer for this problem? If so, what is it? Here's a setup: These are nonelementary integrals; a numerical approximation of the result is 70.82743281. 
September 16th, 2012, 08:20 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,163 Thanks: 472 Math Focus: Calculus/ODEs  Re: Disk/Washer Integration with ln function
To set this up using the shell method, we could use: This result agrees with the setup given by [color=#008000]greg1313[/color] using the washer method. We wouldn't change the axis of rotation, but our "slices" would be horizontal and of thickness dy. The radius of each cylindrical shell would be y, and the height of each shell would be the difference in the xcoordinates of the two curves. This method is more straightforward for this volume, as it leads to a single integral, and doesn't require integration by parts to evaluate, just a simple usubstitution at most. 

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