Calculating Centre of Mass for Northern Hemisphere Using Spherical Coordinates

[Ted123]

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 $z_{cm}$ ?

$M = \iiint _{\cal{R}} \rho_0\;dV = \int_0^R dr \int_0^{\frac{\pi}{2}} d\theta \int_0^{2\pi} d\varphi \; r^2\sin\theta = \frac{2\pi}{3}R^3 \rho_0$.

$Mz_{cm} = \iiint _{\cal{R}} z \rho_0\;dV = \int_0^R dr \int_0^{\pi} d\theta \int_0^{2\pi} d\varphi \; r\cos\theta r^2\sin\theta = M \frac{3}{8} R$.

 

[lanedance]

have tried carrying through the integration? i think it should be pi/2 as well...

 

[tiny-tim]

Hi Ted123!

It's wrong …

it must be [0,π/2] for both. ​