1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface Integrals and Gauss's Law

  1. Jan 18, 2015 #1
    When I learned Integrals in Calc III, the formula looked like this
    ∫∫ F(r(s,t))⋅(rs x rt)*dA
    but in physics for Gauss's law it is
    ∫∫E⋅nhat dA
    How are these the same basic formula? I know that nhat is a unit vector, so it is n/|n|, but in the actual equation, it is a dot between the cross product and E.

    The problem is I dont know how they are the same since Gauss' Law wants E dot a magnitude, when a surface integral is F dot the vector cross product? They dont seem to be following the same rules
  2. jcsd
  3. Jan 18, 2015 #2
    Let me guess: rs and rt are perpendicular unit vectors within the surface, correct?

  4. Jan 18, 2015 #3
    Yeah, I get that you need to cross them to get the perpendicular vector of the surface.
    Basically, the surface integral eq in full form is
    ∫∫S F ⋅ (rs x rt)/|(rs x rt)| *(rs x rt) dA
    and for Gauss's Law it is
    ∫∫S E⋅n/|n| dA

    so you can see in the first one it simplifies to the cross product as a vector,
    but the second one simplifies to the unit vector (because of dividing in magnitude) in the direction normal to the surface.

    I dont get how these are equal
  5. Jan 19, 2015 #4
    Are you sure about that first one that you wrote. It looks to me like there is an extra (rs x rt) in there.

  6. Jan 22, 2015 #5


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The general notion is a surface integral over a vector field,
    $$I=\int_{F} \mathrm{d}^2 \vec{f} \cdot \vec{V}(\vec{r}).$$
    Here ##\mathrm{d}^2 \vec{f}## is a vector on a infinitesimal surface element which is perpendicular to the surface and its magnitude is the area of this infinitesimal surface element.

    To evaluate such a surface integral, one possibility is to parametrize the surface with two generalized coordinates
    $$F: \quad \vec{r}=\vec{r}(q_1,q_2).$$
    In the most simple cases the ##q_j## run over a rectangular area in a plane, ##(q_1,q_2) \in [a_1,b_1] \times [a_2,b_2]##. Then
    $$\vec{T}_j=\frac{\partial}{\partial q_j} \vec{r}$$
    defines two tangent vectors along the surface, and the infinitesimal vectors $$\mathrm{d} q_j \vec{T}_j$$ span an infitesimal parallelogram. Then from the geometric meaning of the cross product, it's clear that
    $$\mathrm{d}^2 \vec{f}=\mathrm{d}q_1 \, \mathrm{d} q_2 \, \vec{T}_1 \times \vec{T}_2$$
    are the surface-element vectors according to this parametrization of your surface. You get
    $$I=\int_{a_1}^{b_1} \mathrm{d} q_1 \int_{a_2}^{b_2} \mathrm{d} q_2 [\vec{T}_1(q_1,q_2) \times \vec{T}_2(q_1,q_2)] \cdot \vec{V}[\vec{r}(q_1,q_2)].$$
    It's pretty easy to show that the value of the integral is independent of the parametrization (up to a sign, which you have to carefully specify anyway by choosing the orientation of the surface by your choice of the order of the generalized coordinates).
  7. Jan 22, 2015 #6
    This guy hasn't responded since Sunday. I'll give him an extra day or two, and then I'm closing this thread.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook