# Find the center of mass with tripl-integral

1. Jan 10, 2010

### ppitsme

A being is on a sphere like planet. Sudently hey grabs a GPS and starts diging straight on a diemeter, heading to the center. We can theoretically say tha he is carrying a level with him. Then tha sphere is theoretically cut in two section , A and B. I have(well...with help) found the volume of the two sections with a tripl-integral using Spherical coordinate system.We know that the sphere has the same density all over (an inside) the sphere. I neeeeeeeeeeeeed to find the center of mass. Odviously the center of mass is going to be on the z Axle. Can someone help me find the center of mass?

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2. Jan 10, 2010

### ideasrule

If you can find the mass, why can't you find the center of mass? It's just the integral of rho*dV*z whereas the mass is the integral of rho*dV*z.

3. Jan 10, 2010

### Fightfish

I don't quite understand what the question is about, with the digging and stuff o.0
For a sphere with uniform density, clearly the centre of mass must be at the centre of the sphere.

4. Jan 10, 2010

### ppitsme

Well what i ment is that i need to find the center of mass of each section , A and then B.

5. Jan 10, 2010

### ppitsme

ideasrule, i fail to see the difrens betwen the two integral, also rho=?

6. Jan 10, 2010

### Fightfish

Oh ok I get it now. For each region just apply the usual formulas with the appropriate parameters:
$$\overline{x} = \frac{1}{M} \iiint x \rho (x,y,z) dV$$
$$\overline{y} = \frac{1}{M} \iiint y \rho (x,y,z) dV$$
$$\overline{z} = \frac{1}{M} \iiint z \rho (x,y,z) dV$$

7. Jan 10, 2010

### ppitsme

I can understand that i am only going to use the z formula. I alsow understand it is going to be a tripl-integral. What i cant find is from where to where am i going to integrat, and what coordinate system would be most suted

8. Jan 10, 2010

### Fightfish

Since it is a sphere, definitely we will be using spherical coordinates:
$$\psi (\rho, \theta, \phi) = (\rho\,sin \phi\, cos \theta, \rho\, sin \phi\, sin \theta, \rho\, cos \phi)$$
for both regions A and B.
However, the parameters have different ranges.
For A:
$$0 \leq \rho \leq R\,\, , 0\leq \theta \leq 2\pi\,\, , 0\leq \phi \leq a$$
For B:
$$0 \leq \rho \leq R\,\, , 0\leq \theta \leq 2\pi\,\, , a\leq \phi \leq \pi$$
I presume you are able to convert the triple integral in rectangular coordinates into spherical coordinates?

--Scrap this, i made a terrible mistake--

Last edited: Jan 10, 2010
9. Jan 10, 2010

### ppitsme

you meen dV=dz*dy*dx=r^2*sinθ*dr*dθ*dφ ?? Also by a you meen the corner acording to the depth?

10. Jan 10, 2010

### Fightfish

Yup. And remember to convert z as well.
Wait I think I made a blunder in my parameters...oops I did a cone instead...gimme a while to fix it

11. Jan 10, 2010

### ppitsme

"Yup. And remember to convert z as well" what do you meen?

12. Jan 10, 2010

### Fightfish

Scrap those, we'll return to good old rectangular coordinates for now. If I'm not mistaken, the region A can be represented as
$$\int_{-\sqrt{R^2-a^2}}^{\sqrt{R^2-a^2}}\,\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\,\int_{a}^{\sqrt{R-x^2-y^2}}\,\,dz\,dy\,dx$$
where a = height of base of region above the origin.
Looks pretty ugly though, so maybe someone else wants to help out here?

13. Jan 10, 2010

### ppitsme

if this helps, when calculating the volume i had to solve this:
triple-integral{(r from b/Cosθ to R,θ from 0 to ArcCos[b/R],φ from 0 to 2π)r^2 sinθ dr dθ dφ} R = radius, b = R - depth of being

14. Jan 10, 2010

### Fightfish

Sadly, my calculus remains rather weak.
Ah..So those will be the limits of the centre of mass integral as well. Just substitute z with the parameterised equivalent.