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Ted123
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Homework Statement
Calculate the z-component of the centre of mass for a northern hemisphere of radius R with constant density \rho_0 > 0 using spherical coordinates (r,\theta, \varphi ) defined by:
x(r,\theta, \varphi) = r\sin\theta\cos\varphi \;\;\;\;\;\;0 \leq r < \infty
y(r,\theta, \varphi) = r\sin\theta\sin\varphi \;\;\;\;\;\;\, 0 \leq \theta \leq \pi
z(r,\theta, \varphi) = r\cos\theta \;\;\;\;\;\;\;\;\;\;\;\;\;\; 0 \leq \varphi < 2\pi
Homework Equations
For a solid with density \rho (\bf{r}) occupying a region \cal{R},
z_{cm} = \frac{1}{M} \iiint_{\cal{R}} z \rho (\bf{r})\;dV
where M= \iiint_{\cal{R}} \rho (\bf{r})\;dV
The Attempt at a Solution
I have the solution but I'm wondering why the limits of \theta is [0,\pi /2] for calculating M then it changes to [0,\pi] when calculating z_{cm} ?
M = \iiint _{\cal{R}} \rho_0\;dV = \int_0^R dr \int_0^{\frac{\pi}{2}} d\theta \int_0^{2\pi} d\varphi \; r^2\sin\theta = \frac{2\pi}{3}R^3 \rho_0.
Mz_{cm} = \iiint _{\cal{R}} z \rho_0\;dV = \int_0^R dr \int_0^{\pi} d\theta \int_0^{2\pi} d\varphi \; r\cos\theta r^2\sin\theta = M \frac{3}{8} R.
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