Surface Integral for Curl of Vector Field V over a Parametrized Triangle

[theneedtoknow]

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?

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!

 

[LCKurtz]

I haven't checked all your arithmetic, but almost certainly your limits for s and t are wrong. Why use such a non-intuitive parameterization? Your surface is just a plane:

\frac x 1 + \frac y 3 + \frac z 2 = 1

or

z = 2 - 2x -\frac 2 3 y

so it would be very natural to use, for example, x and y as parameters:

\vec R(x,y) = \langle x, y, 2 - 2x -\frac 2 3 y\rangle

Use that to calculate your dS in terms of dx and dy and your limits will be over the triangular region in the xy plane. The inner upper limit won't be constant.

 

[theneedtoknow]

thank you very much :)