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September 4th, 2017, 06:29 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, 08:39 AM  #2 
Global Moderator Joined: May 2007 Posts: 6,786 Thanks: 708 
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.
Last edited by mathman; September 4th, 2017 at 08:42 AM. 
September 4th, 2017, 09:05 AM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  

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integral, problem, surface 
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