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I had to evaluate $\iint_{S} F.\eta dA$ where $F = [3x^2 , y^2 , 0]$ and $S : r(u,v) = [u,v,2u+3v]$ where $0 \leq u \leq 2 , -1 \leq v \leq 1$.

I solved like this -

$\iint_{S} F(r(u,v)) . (r_{u} \times r_{v}) du dv$ where $r_{u},r_{v}$ are the partial derivatives of $r$ wrt $u$ and $v$ respectively.

I got $F(r(u,v)) . (r_{u} \times r_{v}) = (-6u^2 - 3v^2)$

And $\int_{v=0}^{1} \int_{u=0}^{2} F(r(u,v)).(r_{u} \times r_{v}) du dv$ And I got $\int_{v = -1}^{1} \int_{u=0}^{1} (-6u^2 -3v^2) du dv = -6$

But it is not correct.I could not figure out the mistake?

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    $\begingroup$ The limits on $u$ are $0$ to $2$, but in your final integral, you suddenly changed to $0$ to $1$. Your answer is consistent with this, so I assume that is your error. $\endgroup$ – Paul Sinclair Jan 13 '18 at 17:07

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