Thread: Verifying Stokes' Theorem

Problem- Given the vector field F= <yx, 0, x>, calculate both F•ds ​and curl(F)•dS over the portion of the plane + + z = 1 where x, y, z > 0, oriented with an upward normal vector.

I took x F to get <0, y-1, z> which I dotted with n_G(x,y)= < , ,1> of my parameterization of the plane G(x,y) = (x, y, 1-( + ) to get +z. I then computed +z dy dz to get which I believe is correct.

My issue lies in finding F•ds. I tried defining 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 S in the yz-plane, but I met the same <0, 0, 0> dot products, so F•ds​ 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

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.