Solution to specific surface integral involving a projection

Question

I have an integral computation question:

Given the vector field $v = y \hat{z}$ I want to compute the surface integral $\int (\nabla \times v) \cdot da$ of the surface within the triangle with vertices $(a,0,0)$, $(0,2a,0)$ and $(0,0,a)$. The way I did it was to use Stokes Theorem and calculate the corresponding path integral $\int v \cdot dI$. The answer obtained was $a^2$ which as I understand is correct.

I am interested in the direct approach, in the literature I am using the solution is given simply as "$\nabla \times v = \hat{x}$ hence $\int (\nabla \times v)\cdot da$ is the projection of this surface on the $xy$ plane $= \frac{1}{2} \cdot a \cdot 2a = a^2$" Could someone provide some detail regarding this, this explanation does not seem clear to me. Thanks.

Answer 1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Vertices: $\ds{\vec{v}_{1} = a\,\hat{x}}$, $\ds{\vec{v}_{2} = 2a\,\hat{y}}$ and $\ds{\vec{v}_{3} = a\,\hat{z}}$. A unit vector perpendicular to the triangle is $\ds{\propto}$:

\begin{align} \pars{\vec{v}_{1} - v_{2}}\times\pars{\vec{v}_{1} - v_{3}} & = -\vec{v}_{1}\times\vec{v}_{3} - \vec{v}_{2}\times\vec{v}_{1} + \vec{v}_{2}\times\vec{v}_{3} = \pars{-a^{2}\,\hat{y}} - \pars{-2a^{2}\,\hat{z}} + \pars{2a^{2}\,\hat{x}} \end{align} The unit vector is $\ds{\hat{n} = \color{#f00}{{2 \over 3}\,\hat{x}} - {1 \over 3}\,\hat{y} - {2 \over 3}\,\hat{z} \propto \dd\vec{\mathbf{S}}}$. Then, \begin{align} \int\nabla\times\vec{\nu}\cdot\dd\vec{\mathbf{S}} & = \int\hat{x}\cdot\dd\vec{\mathbf{S}} = \int\verts{\dd\vec{\mathbf{S}}} \verts{\cos\pars{\angle\pars{\hat{x},\dd\vec{\mathbf{S}}}}} = \int\verts{\dd\vec{\mathbf{S}}} {2/3 \over \verts{\hat{n}}\verts{\hat{x}}} \\[3mm] & = {2 \over 3}\,\times \overbrace{\mbox{Triangule Surface}} ^{\ds{\mathcal{A = {3 \over 2}\,a^{2}}}} = \color{#f00}{a^{2}} \end{align}

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The triangle has sides of lengths $\ds{{\root{3} \over 2}\,a}$, $\ds{\root{2} a}$ and $\ds{{\root{3} \over 2}\,a}$ such that $\ds{\mathcal{A} = {3 \over 2}\,a^{2}}$.