# Solid hemisphere center of mass in spherical coordinates

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?

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SteamKing
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How did you derive your equation for the Cm?

haruspex
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$\int$ $\rho$ r^3 sin $\theta$ dr d$\theta$ d$\varphi$
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.

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?

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haruspex
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1/M $\sum$ mi ri
1/M$\int$ $\rho$(r) r dV - this is a volume integral.
What exactly does ri stand for in the first line? What is the appropriate formula to replace it in the second?

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.

haruspex
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ri is the vector position of a mass in some direction.
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.
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.
Multiplying by sin or cos of theta (depending on which way it's measured) makes sense to me.

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.

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