Calculating flux over an area of a triangle?

Question

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

$$\vec{F}(x,y) = (2x^2 + 3x -2 \cos^4(y) \sin^3(y) , 4 e^{2x} \sinh(x) - 3y)$$

Answer 1

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 $T$:

$$\iint_T dx dy \, \left (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right ) $$

Answer 2

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\Large\tt Hint:$ $$ \nabla\cdot\vec{F} = \pars{4x + 3} + \pars{-3} = \nabla\cdot\pars{2x^{2}\hat{x}} \quad\imp\quad \nabla\cdot\pars{\vec{F} - 2x^{2}\hat{x}} = 0 $$ 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 $\vec{A}$:

$$ \partiald{A_{z}}{y} - \partiald{A_{y}}{z} = F_{x} - 2x^{2}\,, \quad \partiald{A_{x}}{z} - \partiald{A_{z}}{x} = F_{y}\,, \quad \partiald{A_{y}}{x} - \partiald{A_{x}}{y} = 0 $$