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Math Help - spherical coordinates!!!

  1. #1
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    spherical coordinates!!!

    Hi,
    For the following problem, how would I setup the triple integral? Do I have to convert it to spherical coordinates first:

    many thanks....

    Problem:

    Integrate the function over the solid above z=0 and contained between spheres centered on the origin with radii 1 and 4, excluding everything in the first octant.
    integral =
    (You should find the actual value when you evaluate the integral!)
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  2. #2
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    Quote Originally Posted by althaemenes View Post
    Hi,
    For the following problem, how would I setup the triple integral? Do I have to convert it to spherical coordinates first:

    many thanks....

    Problem:

    Integrate the function over the solid above z=0 and contained between spheres centered on the origin with radii 1 and 4, excluding everything in the first octant.
    integral =
    (You should find the actual value when you evaluate the integral!)
    Because of the spherical symmetry, the limits of integration in spherical coordinates are very simple. We want to go completely around so \theta goes from 0 to 2\pi. We want to go from the z axis down to the xy-plane so \phi goes from 0 to \pi/2. Finally, of course, \rho goes from 1 to 4.

    I presume you know that, in polar coordinates, x= \rho sin(\phi)cos(\theta) and y= \rho sin(\phi)sin(\theta) so -5x+ 5y= -5\rho sin(\phi)cos(\theta)+ 5 sin(\phi)sin(\theta) = 5sin(\phi)(sin(\theta)- cos(\theta))

    Of course, the differential of volume in spherical coordinates is \rho^2 sin(\phi)d\rho d\theta d\phi.
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  3. #3
    Super Member Showcase_22's Avatar
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    \int_1^4 \int_0^\frac{\pi}{2} \int_0^{2\pi} 5sin (\phi) (sin (\theta)-cos (\theta))  \rho^2 sin(\phi)d \rho d\phi d\theta

    I was having trouble deciding what the \rho limits should be until I read HallsofIvy's post. Thanks!
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  4. #4
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    HallsofIvy and Showcase_22 !!! Thanks guys!!



    => (105/2)*Pi-(105/2)*Pi*sin(Pi)*cos(Pi)-(105/2)*Pi*cos(Pi)^2

    => 0
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