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Thread: Sherical Coordinates

  1. #1
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    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
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  2. #2
    Grand Panjandrum
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    Quote 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
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  3. #3
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    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 . . .

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  4. #4
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    Thanks a lot. I got it.
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