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