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Homework Help: Stokes' Theorem Question

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Note: the bullets in the equations are dot products, the X are cross products

    Evaluate: [over curve c][tex]\oint[/tex]( F [tex]\bullet[/tex] dr ) where F = < exp(x^2), x + sin(y^2) , z> and C is the curve formed by the intersection of the cone: z = [tex]\sqrt{(x^2 + y^2)}[/tex] and the cylinder: x^2 + (y-1)^2 = 1 oriented CCW looking down from + z axis

    2. Relevant equations

    I'm assuming this is a Stokes' theorem question, so:

    [over curve C][tex]\oint[/tex] F [tex]\bullet[/tex] dr = [over surface S][tex]\int[/tex][tex]\int[/tex]( curl(F) [tex]\bullet[/tex] dS)

    3. The attempt at a solution

    First, the surface looks like it's on the cone, so paramaterizing the cone in polar coordinates:

    (S) r(t,r) = < rcost , rsint , r >

    now for finding curl(F) i did: gradient X F = <0,0,1>

    So according to the equation mentioned above,

    [over curve C][tex]\oint[/tex] F [tex]\bullet[/tex] dr = [over surface S][tex]\int[/tex][tex]\int[/tex]( curl(F) [tex]\bullet[/tex] dS)

    and: [over surface S][tex]\int[/tex][tex]\int[/tex]( curl(F) [tex]\bullet[/tex] dS) = [over domain D][tex]\int[/tex][tex]\int[/tex] curl(F) [tex]\bullet[/tex] < r(partial derivative with t) X r(partial derivative with r) > dA

    so i found curl(F) already, < r(partial with t) X r(partial with r) > = <rcost, rsint, -r> , and dA = rdrdt
    the domain is the domain of the parameters, (t,r), so plugging in the cylinder equation i get:

    r = 2sint, so: 0 < r < 2sint and 0 < t < pi , and plugging all that in i get:

    [tex]\int[/tex][0<t<pi][tex]\int[/tex][0<r<2sint] <0,0,1> [tex]\bullet[/tex] <rcost,rsint,-t> r dr dt

    i get this down to (-8/2)[tex]\int[/tex][0 to pi] (sint)^3 dt which is = -32/9.

    I know the answer is supposed to = pi , but I have no idea what I'm doing wrong. Any help would be greatly appreciated, thanks for your time!
    Last edited: Nov 27, 2009
  2. jcsd
  3. Nov 27, 2009 #2


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    You don't really need to go to polar coordinates. Since the curl(F)=(0,0,1) you only need to find the z component of the vector dS. If you parameterize the cone as r=(x,y,sqrt(x^2+y^2)), just find the z component of dr/dx X dr/dy (both derivatives partial, of course). I think you'll find the integrand is constant over your domain.
  4. Nov 27, 2009 #3
    Thanks for the quick reply. I see your point, and I can solve it with your method. However, I can't quite see what's wrong with using polar coords, it feels like I'm doing the same thing.
  5. Nov 28, 2009 #4


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    It's mostly right. What's wrong is the 'r' in dA=r*dr*dt. The cross product of the two tangent vectors already gives you that 'r'. If Tr and Tt are the two tangents, you should just have dS=Tr X Tt dr dt, not Tr X Tt r*dr*dt.
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