Can the calculation of the surface integral of a specific vector field be simplified? 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.
 A: $\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 $S$.

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.

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