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Surface Integral (two different answers?)

  1. May 1, 2013 #1
    1. The problem statement, all variables and given/known data
    1t5vv4.png


    2. Relevant equations



    3. The attempt at a solution

    Not sure what's wrong with mine or the provided solution..both seems to be right.

    My Solution:
    e01tav.png

    Provided Solution:
    106diye.png
     
  2. jcsd
  3. May 1, 2013 #2

    Dick

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    I don't know why you just decided to use ##dS=r d\phi dz## instead of calculating it like the book did. Your solution is wrong. The dS you are using looks more like a dS for a cylinder than a paraboloid.
     
  4. May 1, 2013 #3
    But it shouldn't matter what coordinates you choose, right? It's not exactly a cylinder, as r is changing with z; r = √(2-z) and the equation x2 + y2 = 2 - z
    is in the form of x2 + y2 = r2..
     
  5. May 1, 2013 #4

    LCKurtz

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    You must use the correct dS for your parameterization. You can either calculate dS in terms of x and y like they did or directly from your parameterization. The parameterization of that surface in cylindrical coordinates is$$
    \vec R(r,\theta)=\langle r\cos\theta,r\sin\theta,2-r^2\rangle$$If you use that approach, the formula for dS is$$
    dS = |\vec R_r \times\vec R_\theta|drd\theta$$
     
  6. May 2, 2013 #5
    Hmm that makes sense. In my geometric derivation of dS, did I assume that R is constant somewhere?

    is Rr = ∂R/∂θ

    and

    Rθ = ∂R/∂r ??
     
  7. May 2, 2013 #6

    Dick

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    No, it's the other way around ##\vec R_r## is the derivative with respect to r. And, yes, if you imagine r constant (like a cylinder) you might write a dS like that down. But it would be wrong.
     
  8. May 2, 2013 #7

    BruceW

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    yes, the equation dS=r d(theta) dz only works for a surface of constant (cylindrical) radius. For a general surface, the equation will be different. Also, it doesn't matter which coordinates you use, in the sense that you could use
    [tex]dS = |\vec R_r \times\vec R_\theta|drd\theta [/tex]
    Or you could use
    [tex]dS = |\vec R_z \times\vec R_\theta|dzd\theta [/tex]
    Either of these equations would give you the correct answer. The second one is the 'correct way' to do the surface you were thinking of (i.e. integrating over theta and z). The main thing is that you need to substitute your 'parameterisation' into [itex]\vec{R}[/itex] before you do the partial integration. i.e. if you do the second equation for dS, then you will need to replace r with a function of z, then differentiate the position vector with respect to z.

    Or, I guess equivalently, you can say that the partial derivative with respect to z is done, while only holding theta constant (i.e. r is allowed to vary).
     
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