Can the calculation of the surface integral of a specific vector field be simplified?

Question

Suppose the two vector fields are $F(x,y,z)=(x^2,0,0)$ and $G(x,y,z)=(0,0,x z)$ respectively.

The surface $S$ is a triangle determined by three points $A:(a_1,a_2,a_3)$, $B:(b_1,b_2,b_3)$ and $C: (c_1,c_2,c_3)$ as below:

Can the surface integrals of $F$ and $G$ on $S$:

$$\iint\limits_S\; F(x,y,z)\cdot {\bf n}{\rm d}S=\iint\limits_S\;x^2\;{\rm d}y\wedge{\rm d}z$$ and

$$\iint\limits_S\; G(x,y,z)\cdot {\bf n}{\rm d}S=\iint\limits_S\;x z\;{\rm d}x\wedge{\rm d}y$$

be simplified so that a double integral is unnecessary?

I guess there should be a very simple result such that both the integral can be represented as $\sum\limits_{k=1}^3\; f(a_k,b_k,c_k)$, but don't know how to prove it.

Answer 1

$\renewcommand{\tr}[1][ABC]{\,\bigtriangleup\!\s{#1}} \renewcommand{\s}[1]{\mathsf{#1}} \renewcommand{\b}{\begin} \newcommand{\e}{\end} \newcommand{\n}{\vec{\mathbf{n}}} \newcommand{\u}{\vec{\mathbf{u}}} \newcommand{\w}{\vec{\mathbf{v}}} \newcommand{\ab}{\overrightarrow{AB}} \newcommand{\ac}{\overrightarrow{AC}} \newcommand{\bc}{\overrightarrow{BC}} \renewcommand{\v}[1]{\b{sm} {#1}_{1} \\ {#1}_{2} \\ {#1}_{3} \\ \e{sm}} \renewcommand{\vv}[2]{\b{sm} {#1}_{1} - {#2}_{1} \\ {#1}_{2} - {#2}_{2} \\ {#1}_{3} - {#2}_{3} \\ \e{sm}} \def\A{\s{A}} \def\B{\s{B}} \def\C{\s{C}} \def\i{\hat{\mathbf{i}}} \def\j{\hat{\mathbf{j}}} \def\k{\hat{\mathbf{k}}} \newenvironment{sm}{\left[\begin{smallmatrix}}{\end{smallmatrix}\right]} \renewenvironment{m}{\begin{bmatrix}}{\end{bmatrix}} \newcommand{\V}[1]{\b{m} {#1}_{1} \\ {#1}_{2} \\ {#1}_{3} \\ \e{m}} \newcommand{\VV}[2]{\b{m} {#1}_{1} - {#2}_{1} \\ {#1}_{2} - {#2}_{2} \\ {#1}_{3} - {#2}_{3} \\ \e{m}} $

This is a complement to my previous answer. Here I provide an alternative way to estimate surface integral. In contrast with the previous approach, (which, as was pointed out in comments, may fail under certain circumstance) this one should be applicable for any triangle $S$.

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Given triangle $S = \tr$, where $\A = \v{a}$, $\B = \v{b}$, and $\C =\v{c}$, we compute its sides $\ab, \ac, \bc$ and normal $\n= \ab \times \ac$ as

$$ \ab = \V{b} - \V{a} = \VV{b}{a}, \quad \ac = \VV{c}{a}, \quad \bc = \VV{c}{b}, \label{*} \tag{*} $$ $$ \n = \b{m} \i & \j & \k \\ b_1 - a_1 & b_2 - a_2 & b_3 - a_3 \\ c_1 - a_1 & c_2 - a_2 & c_3 - a_3 \\ \e{m} = \b{m} \left(b_2-a_2\right)\left(c_3-a_3\right)-\left(b_3-a_3\right)\left(c_2 - a_2\right) \\ \left(b_3-a_3 \right)\left(c_1-a_1\right)-\left(b_1-a_1\right)\left(c_3 - a_3\right) \\ \left(b_1-a_1\right)\left(c_2 - a_2\right) -\left(b_2-a_2\right)\left(c_1 - a_1\right) \e{m} = \V{n} $$

Then $$ \b{aligned} I_G = \iint_S G\cdot \mathbf{dS} & = \iint_S G\cdot \n \,dS = \iint_S \left \langle \b{sm} 0 \\ 0 \\ xz \e{sm} , \v{n} \right \rangle \, dS = \iint_S xz n_3 \, dS = \\ & = \big(\left(b_1-a_1\right)\left(c_2 - a_2\right) -\left(b_2-a_2\right)\left(c_1 - a_1\right) \big) \iint_S xz \, dS = \\ & = \big(b_1 c_2 - a_2 b_1 - a_1 c_2 - b_2 c_1 + a_1 b_2 + a_2 c_1 \big) \iint_S xz \, dS = \\ & = \big(a_1\left(b_2-c_2\right) -a_2\left(b_1 - c_1\right) + b_1 c_1 - b_2 c_2 \big) \iint_S xz \, dS \e{aligned} \label{**} \tag{**} $$

Let us now parametrize sides of triangle by parameters $r,s$, and $t$: $$ \b{aligned} &\A\B: & \ab \,t + \A \implies \b{cases} x(r) = \left(b_1 - a_1\right) t + a_1 \\ y(r) = \left(b_2 - a_2\right) t + a_2 \\ z(r) = \left(b_3 - a_3\right) t + a_3 \e{cases} \\ &\A\C: & \ac \,s + \A \implies \b{cases} x(s) = \left(c_1 - a_1\right) s + a_1 \\ y(s) = \left(c_2 - a_2\right) s + a_2 \\ z(s) = \left(c_3 - a_3\right) s + a_3 \e{cases} \\ &\B\C: & \bc \,r + \B \implies \b{cases} x(t) = \left(c_1 - b_1\right) r + b_1 \\ y(t) = \left(c_2 - b_2\right) r + b_2 \\ z(t) = \left(c_3 - b_3\right) r + b_3 \e{cases} \\ \e{aligned} $$

Take $\u = \frac{\ab}{\left\|\ab\right\|}$ and $\w = \frac{\ac}{\left\|\ac\right\|}$ as new basis vectors. Then equations above will be rewritten as $$ \b{aligned} &\A\B: &\b{sm} x \\ y \\ z \e{sm} &= \left\|\ab\right\| \u \,t + \A \\ &\A\C: &\b{sm} x \\ y \\ z \e{sm} &= \left\|\ac\right\| \w \,s + \A \\ &\B\C: &\b{sm} x \\ y \\ z \e{sm} &= \left\|\bc\right\| \left(\w - \u\right)\,s + \B \e{aligned} \label{***}\tag{***} $$

Next, express $x$, $y$, $z$, and $dS$ in terms of $\u$, and $\w$ from $\eqref{***}$. Keep in mind that $ \bc = \ac - \ab $.

Finally, integrate expression $xz$ in new coordinates along $\ab$ and $ \ac$, and multiply by $\big(a_1\left(b_2-c_2\right) -a_2\left(b_1 - c_1\right) + b_1 c_1 - b_2 c_2 \big)$, just like in $\eqref{**} $. Resulting integral will still be double, but using definitions $\eqref{*}$ and the fact that $ \bc = \ac - \ab $, it might be possible to simplify it, even in general case.

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I hope I was able to make my chain of conclusions understandable, and that working out remaining analytics will not be too tedious.

Answer 2

I would make (linear) change of coordinates $\Phi:\mathbb{R}^3 \to \mathbb{R}^3$, so that the image $Q = \Phi(S)$ of the triangle $S$ lies on "axis plane" (i.e. plane defined by the setting a coordinate variable $x$, $y$, or $z$ to $0$). In that case the normal vector $\mathbf{n}$ will have only one non-zero component, and each of two original surface integrals will take form of a single integral.

For Example, given $S$ as defined in the question, let us compute surface integral $$ I_G = \iint_S G(x,y,z) \cdot \mathbf{dS} = \iint_S \big\langle G(x,y,z) , \mathbf{n}\big\rangle dS = \iint_S \left\langle \left[ \begin{smallmatrix} 0\\ 0\\ xz \end{smallmatrix} \right], \left[ \begin{smallmatrix} n_1\\ n_2\\ n_3 \end{smallmatrix} \right] \right\rangle dS. $$ First, we map the original triangle $S$ to $xy$ plane. For convenience we will look for a map $\Phi: \mathbb{R}^3 \to \mathbb{R}^3 $ such that $\Phi(A ) = O, \Phi(B) = \hat{i}, \Phi(A ) = \hat{j} $, where $O$ is the origin, and $\hat{i}, \hat{j}, \hat{k} $ are standard basis unit vectors.

Since $\Phi$ is linear, it can be represented as $3\times 3$ matrix: $ \Phi = \left[\begin{smallmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33}\end{smallmatrix}\right]$. Therefore we can write $$ \Phi(A) = O \iff \begin{bmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{bmatrix} \cdot \begin{bmatrix} a_{1} \\ a_{2} \\ a_{3} \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix} \\ \Phi(B) = \hat{i} \iff \begin{bmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{bmatrix} \cdot \begin{bmatrix} b_{1} \\ b_{2} \\ b_{3} \end{bmatrix} = \begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix} \\ \Phi(C) = \hat{j} \iff \begin{bmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{bmatrix} \cdot \begin{bmatrix} c_{1} \\ c_{2} \\ c_{3} \end{bmatrix} = \begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix} $$ Combining equations above, we get $\Phi \cdot \left[\begin{smallmatrix} A & B & C \end{smallmatrix}\right] =\left[\begin{smallmatrix} O & \hat{i} & \hat{j} \end{smallmatrix}\right]$: $$ \begin{bmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{bmatrix} \cdot \begin{bmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} $$ Note that the matrix $\left[\begin{smallmatrix} A & B & C \end{smallmatrix}\right]$ is invertible, since its rows are linearly independent because $A$, $B$, and $C$ do not lie on one line. Therefore we can compute $\Phi$ explicitly: $$ \begin{bmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{bmatrix}^{-1} $$ Denote new coordinates $x_\Phi$, $y_\Phi$, and $z_\Phi$, so that $\Phi(x,y,z) = (x_\Phi, y_\Phi,z_\Phi)$, and thus $$ \begin{bmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x_\Phi \\ y_\Phi \\ z_\Phi \end{bmatrix} \implies \begin{cases} x_\Phi = p_{11} x + p_{12} y + p_{13} z \\ y_\Phi = p_{21} x + p_{22} y + p_{23} z\\ z_\Phi = p_{21} x + p_{22} y + p_{23} z \end{cases} $$

The inverse transformation $\Phi^{-1}: =\Theta: \mathbb{R}^3 \to \mathbb{R}^3$ s.t. $\Theta(x_\Phi, y_\Phi,z_\Phi) = (x,y,z)$ is represented by reciprocal matrix of $\Phi$: $$ \begin{bmatrix} t_{11} & t_{12} & t_{13} \\ t_{21} & t_{22} & t_{23} \\ t_{31} & t_{32} & t_{33} \end{bmatrix} \cdot \begin{bmatrix} x_\Phi \\ y_\Phi \\ z_\Phi\end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \implies \begin{cases} x = t_{11} x_\Phi + t_{12} y_\Phi + t_{13} z_\Phi \\ y = t_{21} x_\Phi + t_{22} y_\Phi + t_{23} z_\Phi\\ z = t_{21} x_\Phi + t_{22} y_\Phi + t_{23} z_\Phi \end{cases} $$

Denote $S_\Phi = \Phi(S)$ the new triangle with vertices $O, \hat{i},\hat{j}$. Then its unit normal $\mathbf{n}_\Phi$ is $\hat{k} =\left[\begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix}\right] $. Then, after the linear transformation $\Phi$ of coordinates, our surface integral $I_G$ will take form $$ \begin{aligned} I_G & = \iint_S G(x,y,z) \cdot \mathbf{dS} = \iint_{S_\Phi} \left| \Theta \right| G\big(\Theta\left(x_\Phi,y_\Phi,z_\Phi\right)\big) \cdot \mathbf{dS_\Phi} \\ & = \left| \Theta \right| \iint_{S_\Phi} \big\langle G\left( t_{11} x_\Phi + t_{12} y_\Phi + t_{13} z_\Phi, t_{21} x_\Phi + t_{22} y_\Phi + t_{23} z_\Phi, t_{21} x_\Phi + t_{22} y_\Phi + t_{23} z_\Phi \right) , \hat{k}\big\rangle \,dS_\Phi \\ & = \left| \Theta \right| \iint_{S_\Phi} \left\langle \left[ \begin{smallmatrix} 0\\ 0\\ x\cdot z \end{smallmatrix} \right], \left[ \begin{smallmatrix} 0\\ 0\\ 1 \end{smallmatrix} \right] \right\rangle \,dS_\Phi = \left| \Theta \right| \iint_{S_\Phi} x \cdot z\, dS_\Phi = \\ & =\left| \Theta \right| \iint_{S_\Phi} \big( t_{11} x_\Phi + t_{12} y_\Phi + t_{13} z_\Phi \big) \cdot \big( t_{21} x_\Phi + t_{22} y_\Phi + t_{23} z_\Phi \big) \, dS_\Phi . \end{aligned} $$

Since triangle $S_\Phi$ has vertices $\big( 1, 0, 0\big)$, $\big( 0, 0, 0\big)$, and $\big( 0, 1, 0\big)$, we know that the boundaries are $$ x_\Phi \in [0,1], \qquad y_\Phi = \implies y \in [0,1], \qquad z_\Phi = 0 \\ $$ Thus, the integral can be written as $$ \begin{aligned} I_G & = \big| \Theta \big| \, \int_0^1 \big( t_{11} x_\Phi + t_{12} \left( 1-x_\Phi\right) + t_{13} 0 \big) \cdot \big( t_{31} x_\Phi + t_{32} \left( 1-x_\Phi\right) + t_{33} 0 \big) \, dx_\Phi = \\ & = \big| \Theta \big| \, \int_0^1 \big(\left( t_{11} - t_{12}\right) x_\Phi + t_{12} \big) \cdot \big(\left( t_{31} - t_{32}\right) x_\Phi + t_{32} \big) \, dx_\Phi, \end{aligned} $$ where $\big| \Theta \big| $ is determinant of inverse linear transformation $\Theta = \left[ t_{ij} \right]$. Note that $x_\Phi$ becomes a dummy variable of integration in the last formula, so we can replace it with any other letter we want, for example with $\xi$: $$ \begin{aligned} \iint_S G\left(x,y,z \right) \cdot \mathbf{dS} = \big| \Theta \big| \, \int_0^1 \big(\left( t_{11} - t_{12}\right) \xi + t_{12} \big) \cdot \big(\left( t_{31} - t_{32}\right) \xi + t_{32} \big) \, d\xi = \\ = \big| \Theta \big| \, \int_0^1 \Big( \big(t_{11} - t_{12}\big)\big(t_{31} - t_{32}\big) \xi^2 + \big(t_{12}\left(t_{31}-t_{32}\right)+t_{32}\left(t_{11}-t_{12}\right) \big)\xi + t_{12}t_{32} \Big) \, d\xi \end{aligned} $$

Thus, we finally expressed the surface integral $ I_G = \iint_S G\left(x,y,z \right) \cdot \mathbf{dS}$ as one-dimensional integral $$ \boxed{ \ I_G =\big| \Theta \big| \, \int_0^1 \Big[ \big(t_{11} - t_{12}\big)\big(t_{31} - t_{32}\big) \xi^2 + \big(t_{12}\left(t_{31}-t_{32}\right)+t_{32}\left(t_{11}-t_{12}\right) \big)\xi + t_{12}t_{32} \Big] \, d\xi \ \ } $$

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Similarly, you can derive explicit formulas for vector field $F$. I believe you can pick it up from here.