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Find dS for sphere

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Let S be the piece of ρ=3 that is below Φ=∏/6 and is oriented up. Write one dS (vector) for all of S.


    2. Relevant equations



    3. The attempt at a solution
    It's a sphere of radius 3.

    [itex]S=<x,y,\sqrt{9-x^{2}-y^{2}}>[/itex]

    Therefore

    [itex]dS = < \frac{-x}{\sqrt{9-x^{2}-y^{2}}}, \frac{-y}{\sqrt{9-x^{2}-y^{2}}}, 1 >[/itex]
     
    Last edited: Apr 16, 2014
  2. jcsd
  3. Apr 16, 2014 #2
    Sorry accidentaly pressed the enter key, one moment while I edit the post with the problem.
     
  4. Apr 16, 2014 #3

    LCKurtz

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    First of all, that doesn't look right even in the first octant. There upward normal would be outward and all three components of ##d\vec S## should be positive.

    Aside from that, I don't understand the orientation question. Orientation is to pick one side of the surface as "positive". The way this problem is described, it uses one side on the upper part and the other side on the lower part. I don't get the "one" dS vector for all of S.
     
  5. Apr 17, 2014 #4
    Yeah dude you tell me... This professor loves to throw curve balls at his students. I've never seen anything like this in a book or anything and yes the question confuses me too.
    Without any justification what so ever he puts down the answer as

    [itex]dS=\frac{<x,y,z>}{3} 9Sin\phi d\phi d\theta [/itex]

    where x=ρSin(phi)cosθ

    and the other spherical coordinate definitions....

    I get where the spherical dA transformation comes from, and the 9. But since no work is shown I am completely and utterly confused. I think I'm just gonna memorize what to do for the test and not even care about understanding it. I'm so over this class.
     
  6. Apr 17, 2014 #5

    LCKurtz

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    That answer is simply incorrect. If you parametrize the sphere as$$
    \vec R(\phi,\theta) = \langle 3\sin\phi \cos\theta,3\sin\phi\sin\theta,3\cos \phi\rangle$$and calculate ##\vec R_\phi \times \vec R_\theta## as you normally would to get a normal vector you get$$
    \vec R_\phi \times \vec R_\theta =
    \langle 9\sin^2\phi\cos\theta, 9\sin^2\phi\sin\theta, 9\sin\phi\cos\phi\rangle
    = \langle 3x\sin\phi, 3y\sin\phi, 3z\sin\phi\rangle=\langle x,y,z\rangle 3\sin\phi$$That gives you the expression your prof has, but it represents the outward pointing normal for the sphere, which is downward on the bottom half.

    Perhaps there is confusion in the statement of the problem, wanting oriented outwards instead of oriented upwards.
     
  7. Apr 17, 2014 #6
    I was looking through my notes and I think that's what he's after. He called it "special case two".

    How are you reducing your 9 to a 3? I did it the way you did it and when I factor out the 9 it just remains as a 9.
     
  8. Apr 17, 2014 #7

    LCKurtz

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    The ##x,y,z## each use a factor of ##3## in their formulas.
     
  9. Apr 17, 2014 #8
    Wait, you lost me. When I look at the result of your cross product, they all have a 9 in them.

    Nevermind I got it. Thank you!
     
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