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Confusing Surface Integral

  1. Jul 16, 2014 #1
    Hello,

    We know that surface integrals come to the form of a surface integral of a scalar function over a surface and a vector field over a surface. First one is [tex]\oint F(x,y,z)d{S}[/tex] and the second one is [tex]\oint \vec{F}(x,y,z)\cdot d\vec{S}=\oint \vec{F}(x,y,z)\cdot \vec{n}dS[/tex], where [itex]n[/itex] is the unit normal vector of surface [itex]S[/itex].

    Lately, I've seen integrals of the form [tex]\oint \vec{p}dp[/tex], over [itex]p[/itex], where [itex]p[/itex] is a unit vector. I fail to understand the meaning of that, is it surface integral of scalar or vector function? There is no dot product if its a vector function, so what am i missing here?
     
  2. jcsd
  3. Jul 16, 2014 #2
    ##\vec{p}## is a unit vector? I think you need to be more specific about what p is to be able to solve the integral.

    In general though, you can integrate vector quantities. Just remember that the integral means a sum of a bunch of different objects. In the two first integrals you posted, you are taking a sum of a bunch of different scalars. In the third integral, you are taking a sum of a bunch of vectors. So, in the first two, you will end up with a scalar quantity, but in the last one you will end up with a vector quantity. I would be more specific about how to solve the third integral, but I'm not sure I quite understand what ##\vec{p}## is supposed to mean.

    Consider ##\vec{p}(x, y, z)## and assume that you are integrating over some closed surface in ##R^3##, then for every point ##(x, y, z)## on the surface of the object in ##R^3##, you find ##\vec{p}## and sum all such ##\vec{p}##'s together. This is the value of the integral of the vector function ##\vec{p}(x, y, z)## over the surface in ##R^3##.
     
  4. Jul 16, 2014 #3
    Thanks for the fast reply. Ok, let me be more specific, let's say [itex]\vec{p}[/itex] is the unit vector in spherical coordinates of a unit sphere [itex]S[/itex], so [itex]\vec{p}=(sin(phi)cos(theta),sin(theta)sin(phi),cos(phi))[/itex]. Will the[itex]\oint \vec{p}dp[/itex], over [itex]p[/itex], be equal to the surface area of the sphere? Meaning its the same as [itex]\oint dS[/itex], over [itex]S[/itex], which for the unit sphere equals [itex]4π[/itex]?
     
    Last edited: Jul 16, 2014
  5. Jul 16, 2014 #4
    Well, what exactly do you mean "integral over p"? If ##\vec{p}## is the spherical unit vector and you are integrating over the unit sphere, realize that every vector in the sum will be negated by an exactly opposite vector on the other side of the sphere. So I believe that the integral of the spherical unit vector over a unit sphere is actually equal to the zero vector.
     
  6. Jul 16, 2014 #5
    Hmm, let me focus on what you said:

    Our surface integral [itex]I=\oint\vec{p} d\vec{S}[/itex], integrating over [itex]S[/itex], if [itex]\vec{p}[/itex] equals the spherical unit vector and [itex]S[/itex] is the unit sphere, then [itex]I=\oint\vec{p} \vec{n}{dS}[/itex], (since [itex]dS=\vec{n}dS[/itex], where [itex]\vec{n}[/itex] is the unit vector normal to the small surface [itex]dS[/itex] but [itex]\vec{n}[/itex] is identical to the unit vector [itex]\vec{p}[/itex] so it's actually [itex]I=\oint\vec{p}\vec{p}{dS}[/itex]? Is that product [itex]\vec{p}\vec{p}[/itex] dyadic or inner?

    In my previous post, when I said "integrating over [itex]p[/itex]" I meant over [itex]S[/itex], sorry.
     
  7. Jul 19, 2014 #6
    The ##dp## in the final integral isn't a vector, it's a scalar. It represents an infinitesimal area on your surface.

    Look at the difference between the ##dS## in the first integral and the ##d\vec{S}## in the second integral.
     
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