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Math Help - Surface Integral of a sphere (or part there of :P!)

  1. #1
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    Surface Integral of a sphere (or part there of :P!)

    Hi all,

    would anyone be able to give me a little bit of help with this question.




    for part (a) i did this....

    our sphere is given by  x^2 + y^2 +z^2 = 25

    putting in z = 3 to find out part of the sphere gives:
     x^2 + y^2 = 16

    so we change to the parametric representation of the part of the sphere we are interested in sphere which is

    \vec{r}(\theta,\psi) = (4sin\psi cos\theta) \vec{i} + (4sin\psi sin\theta) \vec{j} +(4cos\psi) \vec{k}

    then

    \vec{r_\theta}(\theta,\psi) = (-4sin\psi sin\theta) \vec{i} + (4sin\psi cos\theta) \vec{j}

    \vec{r_\psi}(\theta,\psi) = (4cos\psi cos\theta) \vec{i} + (4cos\psi sin\theta) \vec{j} +(-4sin\psi) \vec{k}



    does this look like i am heading in the right direction for part (a)?

    cheers!

    -Sarah
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  2. #2
    Junior Member
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    Aug 2006
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    nm, got it!
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