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 September 4th, 2017, 07:29 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Another Surface Integral problem... Find the center of mass of an object that occupies the surface z = sqrt(x^2+y^2), 1 <= z <= 4 and has density z * x^2. Official Answer: ( 0, 0, 2275/682 ) Here is my solution: $\displaystyle x = r cos \theta$ $\displaystyle y = r sin \theta$ $\displaystyle z = \sqrt { r^2 cos^2 \theta + r^2 sin^2 \theta } = r$ $\displaystyle G(\theta, r) = < r cos \theta , r sin \theta , r >$ $\displaystyle dG / d \theta = < -r sin \theta , r cos \theta , 0 >$ $\displaystyle dG / dr = < cos \theta , sin \theta , 1 >$ $\displaystyle | dG / d \theta ~cross~product~ dG / dr | = r * \sqrt {2}$ $\displaystyle General~Mass = \int_{0}^{2\pi} \int_{1}^{4} \sqrt {2} r^4 cos^2 \theta ~dS$ $\displaystyle Mass~along~Z~Axis = \int_{0}^{2\pi} \int_{1}^{4} \sqrt {2} r^5 cos^2 \theta ~dS$ My answer: ( 0, 0, 3.4292131 ). My answer is wrong.... Any idea? September 4th, 2017, 09:39 AM #2 Global Moderator   Joined: May 2007 Posts: 6,852 Thanks: 743 I got z=3.335777126099707. This matches the right answer. Your error is in the arithmetic at the end, since I used your integrals to get the result. Thanks from zollen Last edited by mathman; September 4th, 2017 at 09:42 AM. September 4th, 2017, 10:05 AM   #3
Senior Member

Joined: Jan 2017
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Posts: 209
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Thanks. I think it is my arithmetic too...

Quote:
 Originally Posted by mathman I got z=3.335777126099707. This matches the right answer. Your error is in the arithmetic at the end, since I used your integrals to get the result. Tags integral, problem, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MMath Algebra 1 June 9th, 2016 06:08 AM henrymerrild Calculus 2 May 1st, 2014 11:33 AM kriko Calculus 1 August 21st, 2010 11:55 AM sansar Calculus 1 May 10th, 2009 05:21 AM TTB3 Real Analysis 1 April 22nd, 2009 06:28 PM

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