# Thread: Sherical Coordinates

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:

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

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

and the centre of mass is:

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

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

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

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

RonL

3. Hello, John!

The volume of the region given in spherical coordinates

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

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

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

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

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

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

And there is the region . . .

4. Thanks a lot. I got it.