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Thread: Triple Coordinates

  1. #1
    Junior Member
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    Aug 2009
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    Unhappy Triple Coordinates

    Use spherical coordinates to find the volume of the region above the cone z = $\displaystyle \sqrt {x^2 + y^2}$ and between the hemispheres z = $\displaystyle \sqrt {16 - x^2 - y^2}$ and z = $\displaystyle \sqrt {4 - x^2 - y^2}$.

    I'm having trouble finding the bounds, would it be

    $\displaystyle
    \int^{2\pi} _{0} \int^{\frac{\pi}{4}} _{0} \int^{4} _{0}
    d\rho d\theta d\phi $
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    In spherical coordinates, $\displaystyle \rho$ is the distance from the origin, $\displaystyle \theta$ is the angle of the projection on the $\displaystyle xy$-plane with the $\displaystyle x$-axis, and $\displaystyle \phi$ is the angle from the $\displaystyle z$-axis. This is relating points to the Cartesian coordinate system, of course.

    Our region is a section of a cone between two spheres. Because this cone, defined by

    $\displaystyle z=\sqrt{x^2+y^2},$

    makes an angle of $\displaystyle 45^{\circ}$ or $\displaystyle \frac{\pi}{4}$ radians with the $\displaystyle z$-axis, our $\displaystyle d\phi$ limits will be $\displaystyle \left(0,\frac{\pi}{4}\right)$. Similarly, our $\displaystyle d\theta$ limits will be $\displaystyle (0,2\pi)$ and our $\displaystyle \rho$ limits will be $\displaystyle (2,4)$.

    Now, an important thing to consider is that incrementally-defined volume regions in spherical coordinates grow larger as $\displaystyle \rho$ grows larger, and also grow smaller as $\displaystyle \phi$ decreases to $\displaystyle 0$ (as the volume regions shrink to wedges near the $\displaystyle z$-axis). Nothing like this happens with uniformly distributed Cartesian coordinates. To make up for it, we introduce the factor $\displaystyle \rho^2\sin \phi$ when calculating triple integrals in spherical coordinates:

    $\displaystyle \int_0^{\pi/4}\int_0^{2\pi}\int_2^4 \rho^2\sin\phi\,d\rho\,d\theta\,d\phi.$
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