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Homework Help: CalcIII questions

  1. Dec 8, 2007 #1

    F(x,y,z) = x*i + y*j - 2z*k

    S: Z = [tex]\sqrt{}a^2 - x^2 -y^2[/tex]

    I solved for delG and doted F to del G. then I converted it polar since I had x^2 + y^2

    I got:

    [tex]\int\int[/tex]r^3/(a^2-r^2)^(1/2)drdo - 2[tex]\int\int[/tex](a^2-r^2)rdrdo

    I evaluated the double integral and got 4/3pi(a^2+a^3)

    sorry for the shortcut guys, its just so much stuff to type if I try to some the work. Plz, tell me if my answer is wrong.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 9, 2007 #2


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    Uh, what is delG? What is G here?
  4. Dec 9, 2007 #3

    del G is fxi + fyj + fzk which is

    del G = x/sqrt(a^2-x^2-y^2)*i + y/sqrt(a^2-x^2-y^2)*j -2k

    then the formula says I have to dot del G to F(x,y,z) which gives me

    (x/sqrt(a^2-x^2-y^2)*i + y/sqrt(a^2-x^2-y^2)*j -2k) * (x*i + y*j - 2z*k)

    delG*F = (x^2+y^2)/(sqrt(a^2-x^2-y^2) -2sqrt(a^2-x^2-y^2)

    then I converted it to polar to make the integral easier

    [tex]\int\int[/tex]delG*Fds = [tex]\int\int[/tex](r^2/(a^2-r^2) - 2sqrt(a^2-r^2))rdrd[tex]\Theta[/tex]
  5. Dec 9, 2007 #4


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    What you have failed to say is that your surface is given by G(x,y,z)= [itex]\sqrt{a^2- x^2- y^2}- z[/itex]= 0.
  6. Dec 9, 2007 #5
    how is it help me?
  7. Dec 9, 2007 #6
    oh, ya. I remember my question now. Thanks for pointing out the surface. Actually I got up to that part...z-sqrt(a^2-x^2-y^2)..my question was the limits of integrating in this problem. Thanks for pointing out the surface euclide, however, I knew what the surface was its just the graphing, and finding the limits of integrating I am worrying about.
  8. Dec 10, 2007 #7


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    It helps us by telling us what you are talking about! You initially said "I solved for delG and doted F to del G" without saying what you meant by "G"!

    Defennnder asked "Uh, what is delG? What is G here?" and you still did not say what G was.

    [itex]z= \sqrt{a^2- x^2- y^2}[/itex] is the upper half of the sphere [itex]x^2+ y^2+ z^2= a^2[/itex] Since you don't say anything about integrating over a portion of that, I must assume (hesitantly, given your reluctance to clearly state the entire problem) that you are to integrate over the entire hemisphere. You might want to [itex]x= 2cos(\theta)sin(\phi)[/itex], [itex]y= 2sin(\theta)sin(\phi)[/itex], [itex]z= 2cos(\phi)[/itex], spherical coordinates with [itex]\rho[/itex] set equal to 2, as parameters. To cover the entire sphere we take [itex]\theta= 0[/itex] to [itex]2\pi[/itex], [itex]\phi= 0 to \pi[/itex]. What should the limits be to get only the upper hemisphere?
  9. Dec 10, 2007 #8
    thanks euclid. Sorry for the several grammatical errors. I have 4 finals coming up tomorrow, tuesday, so I am sorta stressng a bit. And not to mention that it is my first semester in college. ;D

    anyways, I have to convert it to spherical? hmm, I usually convert things into polar coordinates, and my instructor rushed the discussion on the last 3 sections of our text, which are the surface integrals, divergence and stoke's. I'm really having troule in those sections (mostly algebraic problems). and thanks for stating the regions of integration.
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