Verifying Stokes' Theorem

Problem- Given the vector field **F**= <yx, 0, x>, calculate both $\displaystyle \oint_{\partial S}$**F•**d**s **and $\displaystyle \iint_S$curl(**F**)•d**S **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 **n**_{G}(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 $\displaystyle \oint_{\partial S}$**F•**d**s**. I tried defining $\displaystyle \partial$S as the sum of a set of three parameterized line segments **c**_{1} **c**_{2 }**c**_{3} in the xy-plane, however none have a z-component, so when I took **F**(**c**_{1})•**c'**_{1 }, **F**(**c**_{2})•**c'**_{2}, and **F**(**c**_{3})•**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 $\displaystyle \oint_{\partial S}$**F•**d**s** and all of the three scalar line integrals composing it equated to be 0. I feel like I'm doing something wrong.

Any help?

Thanks

Anthony

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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.

Attachment 26132