I asked for some help on how to do surface integrals, and this is what I understood from that applied to a question, which i am getting the wrong answer to, so can someone please let me know which part of this I am doing wrong?(adsbygoogle = window.adsbygoogle || []).push({});

I need to find the integral of the curl of V, over the triangle with vertices (1,0,0), (0,3,0), (0,0,2), where V = (zx^2, 2yz, 0)

So the curl of V is (-2y, x^2, 0)

the surface parametrized is x = 1-s-t, y=3s, z=2t

so I need to dot the curl of V with d(1-s-t,3s,2t)/ds cross d(1-s-t,3s,2t)/dt = (-1,3,0)cross(-1,0,2) = (6, 2, 3)

So I have to integrate (-2(3s) , (1-t-s)^2, 0) dot (6, 2, 3) dsdt from s=0 to s=1 and from z=0 to z = 1

This is the double integral of (-36s + 2 -4t -4s + 4ts + 2t^2 + 2s^2) = (2-40s-4t+4ts+2t^2+2s^2) for s,t going from 0 to 1

but when i do this integral, i get -53/3

I have done the integral of V over the closed path around the triangle, and got -35/6

By stokes theorem these should be equal, and I am pretty sure my path integrals are correct at -35/6, so I must be doing something wrong with the surface integral. Please help!

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# Homework Help: Help with Surface Integral

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