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Stokes Theorem

  1. Nov 28, 2006 #1
    1. The problem statement, all variables and given/known data
    Let M be the capped cylindrical surface which is the union of two surfaces, a cylinder given by x^2 + y^2 = 9, 0 ≤ z ≤ 1, and a hemispherical cap defined by x^2 + y^2 + (z−1)^2 = 9, z ≥ 1. For the vector field F = (zx + z^2y + 2 y, z^3yx+ 8 x, z^4x^2), compute doubleintM (∇×F) ·dS in any way you like.

    doubleintM (∇×F) ·dS = ?

    2. Relevant equations

    I thought the line integral would make the most sense in solving this - I wanted to take the line integral with respect to the bottom circle. parameterized, it is (3costheta, 3sintheta, 1).

    Line int = F(r) (rprime)

    3. The attempt at a solution

    I have no idea what's going on here - I tried to take the line integral by saying r=(3sint,3cost,0) Since z=0 for circle the only parts that matter in the F is the 2y and the 8x, but I don't think that's right.

    Integral w/ limits 0 to 2pi

    (2(3sintheta),8(3costheta),0) dot product (-3sintheta, 3costheta, 0)

    Can anyone help me with this one?
  2. jcsd
  3. Nov 29, 2006 #2


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    Science Advisor

    Not quite. That is the bottom of the hemispherical cap. The bottom of M is the circle in the xy-plane: [itex](3 cos\theta , 3 sin\theta ,0)[/itex]

    Oh! Okay, now you moved down to the correct circle!

    [itex]F= (zx+ z^2y+ 2y, z^3yx+8x, z^4x)[/itex] which, on z= 0 becomes
    [itex]F= (2y, 8x, 0)[/itex]. Good. That's what you have. But [itex]s= (3 cos(\theta), 3 sin(\theta), 0)[/itex] so [itex]ds= (-3 sin(\theta), 3 cos(\theta), 0)d\theta[/itex]. Your integral should be:
    [tex]\int_{\theta= 0}^{2\pi}(6sin(\theta ), 24cos(\theta ),0)\cdot (-3sin(\theta ),3cos(\theta ),0)d\theta[/tex].
    Never forget the "d" in an integral!
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