# Thread:

1. ## Homework Problem

Wondering if someone could help me get this answer. I don't get spherical coordinates at all.

The volume of the region given in spherical coordinates by the inequalities
3 less than or equal to rho less than or equal to 5
0 less than or equal to phi less than or equal to pi/6
-pi/6 less than or equal to theta less than or equal to pi/6
is filled with uniform material. Find the x-coordinate of the centre of mass.

Thanks for any help.

John

2. Originally Posted by OntarioStud
Wondering if someone could help me get this answer. I don't get spherical coordinates at all.

The volume of the region given in spherical coordinates by the inequalities
3 less than or equal to rho less than or equal to 5
0 less than or equal to phi less than or equal to pi/6
-pi/6 less than or equal to theta less than or equal to pi/6
is filled with uniform material. Find the x-coordinate of the centre of mass.

Thanks for any help.

John
The mass in the region is:

$<br /> M=\int_{\rho=3}^5 \int_{\phi=0}^{\pi/6} \int_{\theta=-\pi/6}^{\pi/6} \kappa(\bold{r})\ \rho^2 \sin(\phi)\ d\theta \ d\phi \ d\rho<br />$

(Note $\rho^2 \sin(\phi)\ d\theta \ d\phi \ d\rho$ is the volume element in spherical polars)

and the centre of mass is:

$<br /> \bold{R}=\frac{1}{M} \int_{\rho=3}^5 \int_{\phi=0}^{\pi/6} \int_{\theta=-\pi/6}^{\pi/6} \kappa(\bold{r})\ \bold{r}\ \rho^2 \sin(\phi)\ d\theta \ d\phi \ d\rho<br />$

where $\kappa(\bold{r})$ is the density at $\bold{r}$, which we are told is a constant,
and so independent of $\bold{r}$.

Now the $x$ component of the centre of mass is obtained from the above by replacing
$\bold{r}$ by its $x$ component $\bold{r}_x=\rho \sin(\phi) \cos(\theta)$:

$<br /> \bold{R_x}=\frac{1}{M} \int_{\rho=3}^5 \int_{\phi=0}^{\pi/6} \int_{\theta=-\pi/6}^{\pi/6} \kappa\ \rho \sin(\phi) \cos(\theta) \ \rho^2 \sin(\phi)\ d\theta \ d\phi \ d\rho<br />$

RonL

3. Hello, John!

The volume of the region given in spherical coordinates

by the inequalities: . $\begin{array}{ccc}3 \leq \rho \leq 5 \\ 0 \leq \phi \leq \frac{\pi}{6} \\ \text{-}\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}\end{array}$ is filled with uniform material.

Find the $x$-coordinate of the centre of mass.

A good sketch might help ... or try to visualize the region.

$3 \leq \rho \leq 5$ . . .We have a hollow sphere at the origin: inner radius 3, outer radius 5.

$0 \leq \phi \leq \frac{\pi}{6}$ . . The upper polar region only, cut off by a cone with vertex angle $\frac{\pi}{3}\;(60^o)$
. . . . . . . . . .The region is like an angel-food cake.

$\text{-}\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}$ . Cut a $60^o$ slice of the "cake".

And there is the region . . .

4. Thanks a lot. I got it.
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