Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The normal in Stoke's theorem?

  1. Jun 2, 2006 #1
    When integrating over the suface, you have (curl F dot n) in the centre of the integral. Is that n or normal the normal to the boundary of the surface? Or is it the normal of the whole surface, in which case there will be many different ones so seems wrong because n should be definite.

    I also like to ask about the normal (I assume it should be constant as well) in Gauss' Divergence theorem (again replacing (F dot ds) by (F dot n dS). How do you determine that? If I have the bottom hemisphere of a ball, z<=0. What would be its normal vector?
    Last edited: Jun 2, 2006
  2. jcsd
  3. Jun 2, 2006 #2
    Stokes' theorem says:

    \oint_{\partial A} \vec{F} \cdot \vec{dl} =
    \iint_A \left(\vec{\nabla}\times\vec{F}\right) \cdot\vec{ds}.[/tex]

    On the LHS, we have the value of the vector field F at a point on the boundary of the surface A dotted with the infinitesimal line vector at that point and summed over the line that forms the boundary of A.

    On the RHS, we have the value of the curl of F at a point on the surface A dotted with an infinitesimal area vector at that point (i.e. a vector who has a value of the size of the area, and a direction of the outward normal) summed over the area A.

    Which term in particular are you having trouble with?

    P.S. There is a more general Stokes' theorem, but it looks to me like you are referring to this one.

    **EDIT**: Post has been corrected.
    Last edited: Jun 3, 2006
  4. Jun 2, 2006 #3
    When you are integrating over the surface (the left eqn), you can also do (F dot n dS). Is the n the normal vector of the boundary or the normal to the whole surface? If the latter than there can be many normal vectors.
  5. Jun 2, 2006 #4
    I think pivoxa15 is referring to Stokes' Theorem in the following form:

    [tex] \int_C \vec F \cdot d\vec r = \iint_S \nabla \times \vec F \cdot d\vec S [/tex]

    [tex] d\vec S = \hat n dS [/tex]

    and is concerned with [itex] \hat n [/itex] on the surface [itex] S [/itex]

    I also believe pivoxa15 is asking if [itex] \hat n [/itex] is constant, or if it is a function that is related to position, such as [itex] \hat n(x,y,z) [/itex].

    If it is a function, then [itex] \hat n [/itex] will point in a direction depending on x,y,z where x,y,z are points on the surface S.

    My understanding is that the normal vector is a vector of length 1. It is also perpindicular to the surface S. So I believe that in general [itex] \hat n [/itex] should be thought of as a function [itex] \hat n (x,y,z) [/itex] with properties:
    [itex] |\hat n| =1 [/itex]
    and n is perpindicular to S

    So if S is flat, then [itex] \hat n [/itex] is "constant" and does not change based on position, so you would have something like:
    [itex] \hat n = (a,b,c) [/itex]

    Now I could be wrong here, so take it with a grain of salt.
    Last edited: Jun 2, 2006
  6. Jun 3, 2006 #5


    User Avatar
    Science Advisor

    It should be obvious that when you are integrating [itex]\vec{f}\cdot\vec{n}dS[/itex] over a surface then the normal is the normal to the surface. Of course there are "many normal vectors"- the normal at each point is a vector function and the dot product is the function you are integrating.
  7. Jun 3, 2006 #6
    Sorry, FrogPad is right, of course; thanks for the correction. The post has been edited.
  8. Jun 4, 2006 #7
    I have realised that for the Stoke's theorem, the normal is the normal to the boundary of the surface which is normally a disk. So the normal should be a constant in this case.

    But with Green's theorem, you are integrating the surface of a closed solid so in that case, the normal would be a variable.
  9. Jun 4, 2006 #8


    User Avatar
    Science Advisor
    Homework Helper

    think about it: stokes says the flux across the boundary of the region equals the integral of the divergence over the interior region bouned by the surface, hence the normal must be to the boundary surface of the region.
  10. Jun 5, 2006 #9
    How does divergence come into Stokes? I thought it was only the curl?
  11. Jun 5, 2006 #10


    User Avatar
    Science Advisor

    Now I'm getting confused! The "generalized Stoke's theorem"
    says that [itex]\int_{\partial M}\omega= \int_M d\omega[itex] where M is a manifold, [itex]\partial M[/itex] is its boundary, [itex]\omega[/itex] is a differential form on [\partial M[/itex], and [itex]d\omega[/itex] is its differential. If we take [itex]\omega[/itex] to be f(x,y,z)dS on the boundary of some 3 dimensional manifold, we get the divergence theorem and if we take it to be f(x,y,z)ds on the boundary of a 2 dimensional manifold, then we get the original "Stoke's theorem".

    A lot of detail is given here:

    But, once again, whether the normal is to the surface or the curve bounding the surface, it is typically not constant.

    One specific question that was asked was "If I have the bottom hemisphere of a ball, z<=0. What would be its normal vector?"

    We can parametrize the sphere, of radius R, centered at the origin, using x, y themselves: x= x, y= y, z= [itex]-\sqrt{R^2- x^2- y^2}[/itex]
    Then the "fundamental vector product" (see http://www.math.duke.edu/education/ccp/materials/mvcalc/parasurfs/para3.html)
    [tex](i+ \frac{x}{\sqrt{R^2-x^2-y^2}}k)X(j\frac{x}{\sqrt{R^2-x^2-y^2}}k)[/itex]
    is a normal vector to the surface. That time dxdy is the vector differential.

    It would, perhaps be easier to use "spherical coordinates" with [itex]\rho= R[/itex]: [itex]x= Rcos(\theta)sin(\phi), y= Rsin(\theta)sin(\phi), z= Rcos(\phi)[/itex].

    Now the fundamental vector product is
    [tex](-Rsin(\theta)sin(\phi)i+ Rcos(\theta)sin(\phi)j)X(Rcos(\theta)cos(\phi)i+ Rsin(\theta)cos(\phi)j- Rsin(\phi)k)[/tex]
    Again, that is a vector normal to the ball. [itex]d\vec{S}[/itex] would be that times [itex]d\thetad\phi[/itex]. To integrate over the bottom, [itex]\phi[/itex] would range from [itex]\frac{\pi}{2}[/itex] to [itex]\pi[/itex]. [itex]\theta[/itex], of course, ranges from 0 to [itex]2\pi[/itex].
  12. Jun 5, 2006 #11
    I see that you have done the standard surface integral of curl F and I understand how that is done.

    When I was talking about a constant normal, I was trying to link the line integral in stokes (around the boundary) to Green's theorem. But that can only happen if the normal (of the boundary disc) has only z component and is flat. So in the case of the half hemisphere, it works. Offcourse, I did not understand this at the time of my post but things are slowly coming to me.
    Last edited: Jun 5, 2006
  13. Jun 6, 2006 #12


    User Avatar
    Science Advisor
    Homework Helper

    in the general statement, there is no difference between greens, stokes, and divergence theorems. abstractly they are all the same. in the plane there is a unique way to pass from a tangent vector field to a curve to a normal vector field (via 90degree rotation), so curl, which measures the integral of the tangential component as you go around a curve, is easily translated into divergence, the integral of the outward normal component.

    i will gladly give more detail, but you need to tell me which of the several equivalent theorems you think of as "stokes" theorem. my advice is: just learn green's theorem, that's basically all there is to any of them.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook