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Use green’s theorem to find the surface integral of F(x,y)= (y^2, x^2) where R= the square with vertices at (0,0,) , (1,0), (1,1) and (0,1)
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\[F = y^2 + x ^2\] P = Y^2 Q = X^2 dQ/dx = 2x dP / dy = 2y \[\int\limits_{0}^{1}\int\limits_{0}^{1}\left( \frac{ dQ }{ dX } -\frac{dP }{ dy }\right) \]
\[\int\limits_{0}^{1}\int\limits_{0}^{1} (2x - 2y) dydx\]
Errr, and F(x,y)=(y^2,x^2)
If F is a vector field I have a problem, The vectors on the field is parallel to the x-y plane. So the surface integral of this vector field on the given surface (which computes the sum of the perpendicular components of the surface, in this case 0) will be zero. won't it be?
I got zero too, but the right answer is 11/3, I don't know how....???
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