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Math Help - 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:

    <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
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  3. #3
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    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 . . .

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