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math_proof
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center of mass of the upper half of the ball x^2+y^2+z^2<16 (for z>0)
 one year ago
 one year ago
math_proof Group Title
center of mass of the upper half of the ball x^2+y^2+z^2<16 (for z>0)
 one year ago
 one year ago

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malevolence19 Group TitleBest ResponseYou've already chosen the best response.1
Assume it has a density rho. Then: \[M= \iiint \rho dV\] It's easiest in spherical so: \[M=\rho \int\limits_0^{2 \pi} \int\limits_0^{\frac{\pi}{2}} \int\limits_0^4 r^2 \sin(\theta) d r d \theta d \phi\] If rho is constant however then you should expect (V is volume of total sphere): \[M= \rho \frac{V}{2}=\frac{2\rho \pi (4)^3 }{3}=\frac{128 \rho \pi}{3}\] So we get: \[M= \rho \int\limits_0^{2 \pi} d \phi \int\limits_0^{4}r^2 dr \int\limits_0^{\frac{\pi}{2}}\sin(\theta) d \theta=\rho (2 \pi)(\frac{4^3}{3})(1)=\frac{128 \rho \pi}{3}\] Which we were expecting
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.1
So now you need the centroid in each direction. You can see from symmetry that the x and y ones will be zero and you only need find where the z one is. So we need: \[\bar{z}=\frac{\iiint z \rho dV}{M}\] So rho is still constant, pull it out and then you have the integral of z (r cos(theta) in spherical). Integrate that and divide by the mass we found and you're done.
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
thanks so much man
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
i had trouble setting up the integration part
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.1
No problem. Thats the hardest part of the process :P
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
how did you get pi/2 for the limit
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
malevolence19 and how would you integrate z bar since it is z in and there are no integration of z so it would not go away?
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.1
You need to replace z by rcos(theta) which is the coordinate transformation from cartesian to spherical. And I got a pi/2 because the theta angle sweeps from the positive z axis down towards the xy, then towards the z axis; you want it to STOP at the xy plane, this corresponds to theta=pi/2.
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.1
dw:1353370276385:dw
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.1
The theta should read "theta=pi/2" not a range.
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
thanks that explains it well
 one year ago
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