## anonymous 3 years ago center of mass of the upper half of the ball x^2+y^2+z^2<16 (for z>0)

1. anonymous

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

2. anonymous

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.

3. anonymous

thanks so much man

4. anonymous

i had trouble setting up the integration part

5. anonymous

No problem. Thats the hardest part of the process :P

6. anonymous

how did you get pi/2 for the limit

7. anonymous

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?

8. anonymous

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 x-y, then towards the -z axis; you want it to STOP at the x-y plane, this corresponds to theta=pi/2.

9. anonymous

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10. anonymous

The theta should read "theta=pi/2" not a range.

11. anonymous

thanks that explains it well