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Math Help - changing an integral from rectangular to spherical

  1. #1
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    changing an integral from rectangular to spherical

    I have a triple integral in terms of x,y and z and need to convert it to spherical coordinates.

    it is the integral from 0 to 2 and then from 0 to (4 - x^2)^(1/2) inner bounds from 0 to (4 - x^2 - y^2)^(1/2) of (x^2 + y^2 + z^2)^(1/2) dz dy dx

    Since the integral is of roe spherical will work well I think!!!

    So, we will integrate roe^3 sin phi d roe d phi d theta

    My lower bounds must be all zeros.

    Can someone please go through the thinking process to determine my upper bounds?

    Once I have these I can do the integration no problem. Without them I am stuck! Frostking
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  2. #2
    Super Member Showcase_22's Avatar
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    \int_0^2 \int_0^{\sqrt {4-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \sqrt{x^2+y^2+z^2} dz dy dx

    \phi varies from 0 to \pi..
    \theta varies from 0 to 2 \pi.
    \rho varies from 0 to 2.

    Hence the integral is:

    \int_0^{2\pi} \int_0^{\pi} \int_0^{2} \rho^3 sin \phi \  d \rho d \phi d \theta

    EDIT: I've changed the \rho limits to how they should be.
    Last edited by Showcase_22; March 16th 2009 at 12:07 PM.
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  3. #3
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    Shouldn't you have \rho ranging from 0 to 2?

    --Kevin C.
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  4. #4
    Super Member Showcase_22's Avatar
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    Ah yes, it isn't a unit sphere is it!!

    *doi*
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  5. #5
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    limits of integral

    Yes 0 to 2 pi is what the key has for roe but it has 0 to pi /2 for both the others and I still do not understand how to get any of these??? Any explanation would be very much appreciated. Frostking
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  6. #6
    Super Member Showcase_22's Avatar
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    I've only started doing these myself, hence why I made a mistake!

    The limits of \rho, \theta and \phi are found from the definition of spherical polar co-ordinates (see the video below):

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  7. #7
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    Your only dealing with 1/8 of a sphere (first octant only) and in the xy plane the region is a quarter of a circle so the limits should be

    \int_0^{\pi/2} \int_0^{\pi/2} \int_0^{2} \rho^3 sin \phi \ d \rho d \phi d \theta
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  8. #8
    Super Member Showcase_22's Avatar
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    Sorry, but how is it \frac{1}{4} of a circle?
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  9. #9
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    Quote Originally Posted by Showcase_22 View Post
    Sorry, but how is it \frac{1}{4} of a circle?
    From your earlier post

    \int_0^2 \int_0^{\sqrt {4-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \sqrt{x^2+y^2+z^2} dz dy dx so from your limits

    0 \le x \le 2,\; 0 \le y \le \sqrt{4-x^2}, a circle in the first quadrant, i.e. a 1/4 of a circle.
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  10. #10
    Super Member Showcase_22's Avatar
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    Oh right, sorry. I think my mind shut down there for a while =S
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