Calculating flux over an area of a triangle? Let $T$ be the triangle with the vertices $\{(-1,0) , (1,1), (0,2)\}$ traversed in the anticlockwise direction. let $\hat{a}$ be the outward normal to $T$ in the $xy$ plane
Evaluate 
$$ \oint_{\partial T} \vec{F} \cdot \hat{a} \, ds $$
where
$$\vec{F}(x,y) = (2x^2 + 3x -2 \cos^4(y) \sin^3(y) , 4 e^{2x} \sinh(x) - 3y)$$
 A: Best here to use Stoke's theorem.  This may be done by recognizing that $\hat{a}=\hat{z}$, and rewriting $\vec{F}(x,y) = (2x^2 + 3x -2 \cos^4(y) \sin^3(y) , 4 e^{2x} \sinh(x) - 3y,0)$.  Then
$$\nabla \times \vec{F} \cdot \hat{z} = \left (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right )\cdot \hat{z}$$
The line integral above may be expressed as a double integral over $T$:
$$\iint_T dx dy \, \left (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right ) $$
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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$\Large\tt Hint:$
$$
\nabla\cdot\vec{F}
=
\pars{4x + 3} + \pars{-3} = \nabla\cdot\pars{2x^{2}\hat{x}}
\quad\imp\quad
\nabla\cdot\pars{\vec{F} - 2x^{2}\hat{x}} = 0
$$
Then, $\vec{F} - 2x^{2}\hat{x} = \nabla\times\vec{A}$
and
$$
\int_{S}\vec{F}\cdot\dd\vec{S}
=
2\quad\overbrace{\quad\int_{S}x^{2}\hat{x}\cdot\dd\vec{S}\quad}
^{=\ 0\,\quad\mbox{Why ?}}
+
\int\vec{A}\cdot\dd\vec{r}
$$
It's easier to integrate over three line segments. You have to determine $\vec{A}$:
$$
\partiald{A_{z}}{y} - \partiald{A_{y}}{z} = F_{x} - 2x^{2}\,,
\quad
\partiald{A_{x}}{z} - \partiald{A_{z}}{x} = F_{y}\,, 
\quad
\partiald{A_{y}}{x} - \partiald{A_{x}}{y} = 0 
$$
