# Solid hemisphere center of mass in spherical coordinates

1. Nov 8, 2012

### Ledamien

Hello,

I am struggling with what was supposed to be the simplest calc problem in spherical coordinates. I am trying to fid the center of mass of a solid hemisphere with a constant density, and I get a weird result.
First, I compute the mass, then apply the center of mass formula. I divide both and voila, obviously wrong result. What is wrong here?

Cm = 1 / M $\int$ $\rho$ r^3 sin $\theta$ dr d$\theta$ d$\varphi$

M = ρ V → 1/M = 1/ρV

V = 2 $\pi$ R^3 / 3

Cm = 3/ (2 $\pi$ R^3) $\int$ r^3 sin $\theta$ dr d$\theta$ d$\varphi$

Integrated over:
r → 0 to R
$\theta$ → 0 to $\pi$/2
$\varphi$ → 0 to 2 $\pi$

Cm = 3/ (2 $\pi$ R^3) * $\pi$ R^4 / 2
Cm = 3R / 4

So, what do I do wrong?

Last edited: Nov 8, 2012
2. Nov 8, 2012

### SteamKing

Staff Emeritus
How did you derive your equation for the Cm?

3. Nov 9, 2012

### haruspex

Isn't there a trig term missing in that integral? There should be one from the Jacobian and another for the fact that you're interested in the displacement in only one Cartesian coordinate.

4. Nov 9, 2012

### Ledamien

1/M $\sum$ mi ri
1/M$\int$ $\rho$(r) r dV - this is a volume integral.
1/M$\int$ [$\rho$(r) r] r^2 dr sin $\theta$) d$\theta$ d$\varphi$ - this is a volume integral in spherical coordinates.
My idea was that a simple volume integral for a hemisphere, going from 0 to r, 0 to $\pi$/2 and 0 to 2$\pi$ would be sufficient. As the density is constant I can pull it out of the integral, and $\rho$/M is simply 1/V, one over the total volume - the same integral but with r^2 instead of r^3. I seem to be missing a 1/2 somewhere, but where?

Last edited: Nov 9, 2012
5. Nov 9, 2012

### haruspex

What exactly does ri stand for in the first line? What is the appropriate formula to replace it in the second?

6. Nov 9, 2012

### Ledamien

ri is the vector position of a mass in some direction. I know that by symmetry it should be on the z axis. Should I multiply the whole thing be z hat in spherical (cos$\theta$ - sin$\theta$). That does not make much sense, and it does not work.

7. Nov 9, 2012

### haruspex

Yes, it's a vector. But you have replaced it by r, the magnitude of the vector. You cannot do that because in the integral the components of the vector in the x and y directions would cancel.
Multiplying by sin or cos of theta (depending on which way it's measured) makes sense to me.

8. Nov 9, 2012

### Ledamien

OK I got it, I skipped a step:
Cm = 1/V$\int$ r (r hat) dV
As I know it's on the z axis:
Cm = 1/V$\int$ z (z hat) dV

z = r cos$\theta$
dV = r^2 sin$\theta$dr d$\theta$ d$\varphi$

Cm = 1/V$\int$r cos$\theta$ (z hat) r^2 sin$\theta$dr d$\theta$ d$\varphi$

I can pull out z hat from the integral.

With V = (2$\pi$R^3)/3, this gives me 3R/8 (z hat).

Well, that was silly...

Thank you guys for your help.

Last edited: Nov 9, 2012