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Evaluating a surface integral

  1. Sep 18, 2013 #1
    Problem:

    Use the fact that [tex]\int_S \vec{v} \cdot d\vec{S}=\int_S \vec{v} \cdot \frac{\nabla f}{\partial f/\partial x} dy\ dz[/tex]

    to evaluate the integral for ##S=\{(x,y,z):y=x^2 ; 0 \geq x \geq 2; 0 \geq z \geq 3 \}## and ##\vec{v}=(3z^2, 6, 6xz)##.

    Attempt at a Solution:

    I'm having trouble setting up this integral. If I knew what ##f## was, I could easily calculate the gradient, as well as the partial wrt to x. I'd still need to figure out the limits of integration though.
     
  2. jcsd
  3. Sep 18, 2013 #2

    arildno

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    Dearly Missed

    f(x,y,z) ought to be the scalar function describing the surface S through the relationship:
    f(x,y,z)=0
     
  4. Sep 18, 2013 #3
    Hmm. I'm not sure you'd get this from the info given. Any hints?
     
  5. Sep 18, 2013 #4

    arildno

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    Because
    1. It fits. Sort of (I admit I haven't seen closely, though).
    2. Otherwise, it would be utterly meaningless, since you would have no way to evaluate the second expression due to lack of knowledge of f.
     
  6. Sep 18, 2013 #5
    More specifically I meant, how would you find ##f## analytically from this info given?
     
  7. Sep 18, 2013 #6

    pasmith

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    From here:

    If [itex]y = x^2[/itex] then [itex]y - x^2 = 0[/itex].

    (Your inequalities are the wrong way round: you want [itex]0 \leq x \leq 2[/itex] etc.)
     
  8. Sep 18, 2013 #7
    So ##f(x,y,z)=y-x^2##?
     
  9. Sep 18, 2013 #8

    LCKurtz

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    I would parameterize the surface as ##\vec R(x,z)## and use the formula$$
    \iint \vec v\cdot d\vec S =\pm \iint_{x,z}\vec v \cdot \vec R_x\times \vec R_z~dxdz$$where the choice of signs depends on the orientation of the surface which, by the way, you need to specify.
     
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