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Thread: Verifying Stokes' Theorem

  1. #1
    Junior Member
    Sep 2010

    Verifying Stokes' Theorem

    Problem- Given the vector field F= <yx, 0, x>, calculate both $\displaystyle \oint_{\partial S}$Fds ​and $\displaystyle \iint_S$curl(F)dS over the portion of the plane $\displaystyle \frac{x}{2}$+$\displaystyle \frac{y}{3}$+ z = 1 where x, y, z > 0, oriented with an upward normal vector.

    I took $\displaystyle \nabla$ x F to get <0, y-1, z> which I dotted with nG(x,y)= <$\displaystyle \frac{1}{2}$,$\displaystyle \frac{1}{3}$,1> of my parameterization of the plane G(x,y) = (x, y, 1-($\displaystyle \frac{x}{2}$+$\displaystyle \frac{y}{3}$ ) to get $\displaystyle \frac{y-1}{3}$ +z. I then computed $\displaystyle \int_0^1$$\displaystyle \int_0^{3-3z}$$\displaystyle \frac{y-1}{3}$ +z dy dz to get $\displaystyle \frac{1}{2}$ which I believe is correct.

    My issue lies in finding $\displaystyle \oint_{\partial S}$Fds. I tried defining $\displaystyle \partial$S as the sum of a set of three parameterized line segments c1 c2 c3 in the xy-plane, however none have a z-component, so when I took F(c1)c'1 , F(c2)c'2, and F(c3)c'3 they all ended up as <0, 0, 0>. I tried defining $\displaystyle \partial$S in the yz-plane, but I met the same <0, 0, 0> dot products, so $\displaystyle \oint_{\partial S}$Fds​ and all of the three scalar line integrals composing it equated to be 0. I feel like I'm doing something wrong.

    Any help?
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  2. #2
    Junior Member
    Nov 2012
    Jacksonville, FL

    Re: Verifying Stokes' Theorem

    Three parametrized line segments are indeed how you would break up the boundary of S. But one each lies in the xy-, xz-, and yz-planes. Try drawing a picture like the one I've added below.

    Verifying Stokes' Theorem-image.jpg
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