1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook