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Homework Help: Confusion on sign convention for surface integrals

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Compute the surface integral for F = [3x^2, y^22, 0] and S being a portion of the plane r(u,v)=[u,v,2u+3v], 0≤u≤2, −1≤v≤1.

    3. The attempt at a solution

    I managed to get the correct answer, because with some luck I defined the normal in the correct direction. I am just confused on when this answer is negative or positive? Doesn't it really depend only on which of the two possible ways you defined the normal?

    When I search around the textbook, an "outward pointing" normal will always give you a positive answer. But I think that's referring to a closed surface. For this question this is an open surface.

    Also, I stumbled across the right hand screw rule idea, but you use this when you are performing the line integral.(Stoke's Theorem) I want to do it with surface integrals.
    Last edited: May 12, 2012
  2. jcsd
  3. May 12, 2012 #2
    It's exactly as you wrote in the title, it's a "convention", so there's really no way to determine which sign is better. If the problem doesn't give you an hint, you assign a pointing direction, the worse is that you'll eventually flip the sign.
  4. May 12, 2012 #3
    I still don't understand though. If I used Stoke's theorem and applied the line integral there will be a fixed answer, using the right hand rule, defining your normal either way will get you the same answer, can't be either positive or negative.

    However when you apply the surface integral it can be positive or negative?? That wouldn't work.
  5. May 12, 2012 #4


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    In order to specify a surface integral, you also need to specify the orientation of the surface, normally by specifying one of the two possible directions of the unit normal vector to the surface.

    If you apply Stokes theorem to integrate around the boundary of the region, you also have to specify which direction you are integrating on the boundary.

    Stokes theorem specifically requires that if your surface is oriented "positively" then you integrate around the boundary of the region "counter clockwise". Which direction is counter clockwise depends upon the direction of the normal vector. One way of expressing that is that you integrate around the boundary in the direction such that if you were walking around the boundary with your left hand toward the interior of the region, your head is in the direction of the normal to the surface.

    That is, each side of Stokes theorem requires a choice of "orientation"- the orientation of the surface being given by the direction of the normal, the orientationof the path being given by the direction of integration. The two sides will be equal as long the choices are consistent.
  6. May 12, 2012 #5
    Ok I think I'm almost understanding this. So the answer to my original question can be either positive or negative?
  7. May 12, 2012 #6


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    Yes. The information given in
    "Compute the surface integral for F = [3x^2, y^22, 0] and S being a portion of the plane r(u,v)=[u,v,2u+3v], 0≤u≤2, −1≤v≤1."
    does not specify the orientation of the surface.

    A standard method for integrating on surfaces, given by parametric equations as here, is to form the "fundamental vector product" for the surface:
    We have r(u,v)= [u, v, 2u+ 3v] so the derivatives of that position vector with respect to the parameters are [itex]r_u= [1, 0, 2][/itex] and [itex]r_v= [0, 1, 3][/itex] which are in the plane itself. Their cross product, the "fundamental vector product",. is normal to the surface and reflects the "differential of area".

    But the order of that cross product is important:
    [tex]r_u\times r_v= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 2 \\ 0 & 1 & 3\end{array}\right|= [-2, -3, 1][/tex]
    so that [itex]d\vec{S}= [-2, -3, 1]dudv[/itex]
    [tex]r_v\times r_u= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k}\\ 0 & 1 & 3 \\ 1 & 0 & 2 \end{array}\right|= [2, 3, -1][/tex]
    so that [itex]d\vec{S}= [2, 3, -1]dudv[/itex].

    Those have different signs: [itex]\vec{u}\times\vec{v}= -\vec{v}\times\vec{u}[/itex].

    Typically problems like this will include "oriented upward" (so the z-component is positive as in [-2, -3, 1]) or "oriented downward" (so the z-component is negative as in [2, 3, -1]). Closed surfaces may be given as "oriented outward" or "oriented inward".
  8. May 12, 2012 #7
    Ok thanks so much! Makes sense now.
  9. May 12, 2012 #8


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    I have seen calculus texts that take the position that if a surface is given parametrically as ##\vec R = \vec R(u,v)## and no other information is given about the orientation, then the parameterization itself defines the orientation by assuming the orientation given by specifying ##\vec R_u\times\vec R_v## as the orienting normal. Personally, I don't care for that convention, but it would settle the issue if the text adopted that convention.
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