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Math Help - Surface Area

  1. #1
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    Surface Area

    Hi, just having some issues with this problem...

    Integral over the unit sphere of (x^2 + y^2 - 2z^2) dA

    I changed all the x, y, z into spherical coordinates (using a and b for the angles) with r = 1
    Calculated the determinant to be (sin a)

    I set up my work like this:
    double integral of (sin a) ((sin a)^2 - 2(cos a)^2) db * da
    taking b from 2pi to 0, and a from pi to 0

    Skipping the work I did (I presume I did something wrong in the setup...)
    My answer came out to be
    2pi * ((cos pi)^3 - cos pi - (cos 0)^3 + cos 0)

    But this gives me zero. I don't understand...

    The answer given is
    (2pi/3) * ((1 + a^2)^(3/2) - 1)

    I suppose I don't understand something because I'm not sure where this a is supposed to come from//be since its obviously not the same as what I did.

    Help appreciated thanks!
    Last edited by mr fantastic; March 28th 2011 at 06:45 PM. Reason: Moved from another thread and re-titled.
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  2. #2
    MHF Contributor

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    jeanie's thread on Surface Area

    jeanie wrote
    Hi, just having some issues with this problem...

    Integral over the unit sphere of (x^2 + y^2 - 2z^2) dA

    I changed all the x, y, z into spherical coordinates (using a and b for the angles) with r = 1
    Calculated the determinant to be (sin a)
    Okay, the "differential of surface area" for the unit sphere is sin(\phi)d\theta\d\phi or, in your symbols, sin(a)dbda

    I set up my work like this:
    double integral of (sin a) ((sin a)^2 - 2(cos a)^2) db * da
    Which is the same as (sin a)sin^2(a)+ cos^2(a)- 3 cos^2(a)= 1- 3cos^2(a)[/tex]. The way I looked at it was x^2+ y^2- 2z^2= x^2+ y^2+ z^2- 3z^2= 1- 3 cos^2(a).

    taking b from 2pi to 0, and a from pi to 0

    Skipping the work I did (I presume I did something wrong in the setup...)
    My answer came out to be
    2pi * ((cos pi)^3 - cos pi - (cos 0)^3 + cos 0)

    But this gives me zero. I don't understand...
    No, your setup is correct so you must have integrated wrong- your error is in the part you did did not post.

    The answer given is
    (2pi/3) * ((1 + a^2)^(3/2) - 1)
    What? That's non-sense. You are asked to find an integral over the unit sphere. The answer must be a number, not a function of a (or \phi.

    I suppose I don't understand something because I'm not sure where this a is supposed to come from//be since its obviously not the same as what I did.
    Please go back, read the problem again and post exactly what it said.

    Help appreciated thanks!
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  3. #3
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    my apologies, i am an idiot. The homework question was copied from a different edition than the edition of solution manual I have. Question numbers got juggled - it did equal zero, and i am not insane... Sorry for your trouble!
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