My Math Forum Disk/Washer Integration with ln function

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 September 15th, 2012, 06:08 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)^2-g(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, 07:21 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,642 Thanks: 960 Math Focus: Elementary mathematics and beyond Re: Disk/Washer Integration with ln function Here's a setup: $\pi\int^{5\sqrt{e}}_{5}\ln$$\frac{x^2}{25}$$\,-\,\ln$$\frac{x}{5}$$\,dx\,+\,\pi\int^{5e}_{5\sqrt{ e}}1\,-\,\ln$$\frac{x}{5}$$\,dx \\ u\,=\,\frac{x}{5},\,5\,du\,=\,dx \\ 5\pi\int_{1}^{\sqrt{e}}\ln(u)\,du\,+\,5\pi\int_{\s qrt{e}}^{e}1\,-\,\ln(u)\,du$ Integration by parts for ln(u): $\int\ln(u)\,du \\ w\,=\,\ln(u) \\ dw\,=\,\frac{1}{u}\,du dv\,=\,1\,du \\ v\,=\,u \\ u\ln(u)\,-\,\int 1\,du\,=\,u\ln(u)\,-\,u\,+\,C$ O.k?
 September 16th, 2012, 11:46 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, 02:27 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,642 Thanks: 960 Math Focus: Elementary mathematics and beyond Re: Disk/Washer Integration with ln function My understanding of shells is that we are rotating around the y-axis, 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: $2\pi\int_5^{5\sqrt{e}}x$$\sqrt{\ln\(\frac{x^2}{25}$$}\,-\,\sqrt{\ln$$\frac{x}{5}$$\)\,dx\,+\,2\pi\int_{5\s qrt{e}}^{5e}x$$1\,-\,\sqrt{\ln\(\frac{x}{5}$$\)\,dx$ These are non-elementary integrals; a numerical approximation of the result is 70.82743281.
 September 16th, 2012, 09:20 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Disk/Washer Integration with ln function To set this up using the shell method, we could use: $dV=2\pi y$$5e^{y^2}-5e^{\frac{y^2}{2}}$$\,dy$ $V=10\pi\int_0\,^1y$$e^{y^2}-e^{\frac{y^2}{2}}$$\,dy=5\pi$$\sqrt{e}-1$$^2$ 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 x-coordinates 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 u-substitution at most.

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# integal de ln u

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