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Stokes theorem help

  1. May 7, 2006 #1
    Hi, i cant seem to figure out how stokes theorem works. I've run through a lot of examples but i still am not having any luck. Anyway, some advice on a particular problem would be greatly appreciated.

    The problem is: F(x,y,z) = <2y,3z,-2x>. The surface is the part of the unit sphere in the first octant; the normal vector n is directed upward.

    I get that the curl of F is <-3,2,-2>. What I tried next was writing the equation for the sphere as z = f(x,y) = sqrt(1-x^2-y^2) and finding from that fx(x,y) and fy(x,y). I then tried evaluating the double integral in cylindrical coordinates over R of (3fx-2fy-2)dA, where R is the region from 0 to pi/2 and r = 0 to r=1. I changed all the x's and y's to their polar equivalents and didn't forget the r in the dA or anything. I got nothing close to the right answer.

    I understand that the way i approached it might be flawed, so if you can help in either helping me understand why what i was doing is wrong or if you have a different way of approaching it i would greatly appreciate the help. Thanks
     
  2. jcsd
  3. May 7, 2006 #2
    Work you way through the problem systematically. What is Stoke's Theorem?

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

    What does [tex] d\vec S [/tex] mean?

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

    So what do you have now?

    [tex] \iint_S \nabla \times \vec F \cdot \hat n \,\,dS [/tex]

    Now finding the curl is straightfoward, thus:

    [tex] \iint_S \nabla \times \vec F \cdot d\vec S= \iint_S (-3,2,-2) \cdot \hat n \,\, dS [/tex]

    What's a general expression for solving a surface integral?

    This one works right?
    [tex] \iint_S f(x,y,z)\,\,dS = \iint_D f(x,y,g(x,y))\sqrt{\left(\frac{\partial z}{\partial x}\right)^2 + \left( \frac{\partial z}{\partial y} \right)^2}\,\,dA [/tex]

    Why does this work?
    Well when you take the dot product of two vectors, what do you get? Yup... a scalar. And f(x,y,z) doesn't return a vector right?

    So what is the unit vector? And then what happens when you take the dot product? What would f(x,y,z) equal?
     
    Last edited: May 8, 2006
  4. Jun 20, 2006 #3
    so, what exactly does d(vector)S represent? How do I get a vector to dot with curl(vector)F? My particular problem has a circle in 3-space, with z=1. The formula above is for when one is given an equation in form of z=g(x,y).
     
  5. Jun 20, 2006 #4

    arildno

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    [tex]d\vec{S}[/tex] is an infinitesemal vector normal to the surface with magnitude equal to the area of the parallellogram spanned by two linearly independent tangent vectors of infinitesemal magnitudes.
     
  6. Jun 21, 2006 #5

    HallsofIvy

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    Use the "fundamental vector product": the surface of the unit sphere can be written in terms of 2 parameters, the two angles in spherical coordinates: [itex]x= cos(\theta)sin(\phi)[/itex], [itex]y= sin(\theta)sin(\phi)[/itex], [itex]z= cos(\phi)[/itex].
    The derivatives of <x, y, z> with respect to [itex]\theta[/itex] and [itex]\phi[/itex] are the vectors
    [tex]<-sin(\theta)sin(\phi), cos(\theta)sin(\phi), 0>[/tex]
    and
    [tex]<cos(\theta)cos(\phi), sin(\theta)cos(\phi),-sin(\phi)>[/tex]

    The "fundamental vector product" is the cross product of those:
    [tex]<cos(\theta)sin^2(\phi),sin(\theta)sin^2(\phi),sin(\phi)cos(\phi)>[/tex]
    (positive since it is oriented upward).

    Finally, the vector differential of surface area is
    [tex]<cos(\theta)sin^2(\phi),sin(\theta)sin^2(\phi),sin(\phi)cos(\phi)>d\theta d\phi[/tex]
     
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