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Surface integral for scalar quantities

  1. Feb 27, 2015 #1
    so for surface integral for scalar quantities. Why do we use cross product not dot product in the integral? but can we just add an unit normal vector n to make the direction the same? My question seems really stupid too a lot people, but this is really my confusion to surface integral. please give me a simple but sensible answer. Thank you!
  2. jcsd
  3. Feb 27, 2015 #2


    Staff: Mentor

    I moved this thread because it was not a homework-type problem.
    Last edited: Feb 28, 2015
  4. Feb 27, 2015 #3


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    Suppose we have a surface given parametricaly as ##\vec R(u,v) =\langle x(u,v),y(u,v),z(u,v)\rangle## with ##(u,v)## in some region ##W##. The element of surface are in this setting is ##dS =|\vec R_u\times \vec R_v|~dudv##. If you wanted to calculate the area of such a surface you would use ##A=\iint_W 1~dS##, or if there was a mass density ##\delta(u,v)## its mass would be ##m=\iint_W \delta(u,v)~dS##. There is nothing about orientation in such a problem. Everything is a scalar in the integral and its area or mass doesn't depend on which side of the surface you are looking at.

    When you are considering some kind of flow or flux through a surface, it matters which side of the surface is the "positive" side because the direction matters. Such problems will contain a flux vector ##\vec F(u,v)## and you need the component of the flux perpendicular to the surface, and it matters which direction. To get a unit normal vector ##\hat n## to the surface you can calculate ##\hat n =\pm \frac {\vec R_u \times \vec R_v}{|\vec R_u \times \vec R_v|}## with the choice of the ##\pm## to agree with the orientation of the surface. The component of the flux in the ##\hat n## direction is ##\vec F\cdot \hat n##. The integral over the surface of the component of the flux perpendicular to the surface gives the total flux through the surface.$$
    \text{Flux} = \iint_W \vec F(u,v)\cdot \hat n ~dS$$Notice that the integrand is a scalar and this is a surface integral just like in the first paragraph. Except, you aren't computing mass or area now. With this scalar integrand, you are calculating a total flux. If you substitute the formulas for ##\hat n## and ##dS## in that formula you get$$
    \text{Flux} = \pm \iint_W \vec F(u,v)\cdot \frac {\vec R_u \times \vec R_v}{|\vec R_u \times \vec R_v|}~|\vec R_u \times \vec R_v|~dudv$$ $$
    =\pm\iint_W \vec F(u,v)\cdot \vec R_u \times \vec R_v~dudv$$The quantity ##\vec R_u \times \vec R_v~dudv## is sometimes defined the vector surface element ##d\vec S##, giving the final formula$$
    \text{Flux} = \iint_W \vec F(u,v)\cdot \hat n~dS= \iint_W \vec F(u,v)\cdot d\vec S$$
  5. Feb 28, 2015 #4


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    You question is not clear. To find a surface integral, we need a normal vector to the surface. Since the cross product of two non-collinear vectors is always perpendicular to both vectors, the cross product of two vectors in the tangent plane is often the simplest way to find a normal vector but not the only way so, strictly speaking, not "necessary" to the integral. I can't understand why you ask "why not dot product?" Why dot product? What information would that give us about the surface?

    What information does the cross product give us about the surface? To find the "Riemann sum" for the area of a surface in space, we would divide the area into small sections that would correspond to rectangles on the tangent plane to the surface at the center point of each section. To integrate, we would project down to, say, the xy-plane. But a rectangle projects, in general to a parallelogram since projection does not, in general, preserve right angles. And the area of a parallelogram, having vectors u and v as adjacent sides, is the magnitude of the cross product of those vectors.

    For example, imagine a parallelogram, in the xy-plane, having adjacent sides given by vectors u and v. The area of a parallelogram is given by "height times base". Taking vector v as the "base", its length is |v|. The "height" is not the length of vector u but is measured perpendicular to the base. Dropping a perpendicular from a vertex perpendicular to the base, we have a right triangle with hypotenuse of length |u| and "opposite side", the perpendicular, of length "h". Taking the angle between the two sides to be "[itex]\theta[/itex]", [itex]sin(\theta)= \frac{h}{|u|}[/itex] so [itex]h= |u|sin(\theta)[/itex]. From that, the area of the parallelogram is [itex]bh= |v||u|sin(\theta)[/itex], precisely the length of [itex]u\times v[/itex].
  6. Mar 2, 2015 #5
    well, i am trying to understand surface integral right now(it probably would take me a month!), so i just want to find some intuitions here. And my goal had achieved! Thank you guys for the help:smile:
  7. Mar 2, 2015 #6
    and my confusion is here: why is the element of surface dS=|R⃗ u×R⃗ v|dudv. not dS=R⃗ u ⋅R⃗ v ⋅n(the normal unit vector) perpendicular to any of these vectors
  8. Mar 2, 2015 #7
    your rhetorical questions and your quote of those parallelogram really are intuititive, but i guess i am just not a smart pants, you know, i don't really understand it, but thank you so much for helping me
  9. Mar 2, 2015 #8


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    Look here:


    on the second page under "tangent plane and surface area" for a picture and explanation.
  10. Mar 4, 2015 #9
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