My Math Forum Another Surface Integral problem...

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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
From: Toronto

Posts: 209
Thanks: 3

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.

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