Calculate $\int_{S} xyz~d{\sigma}$ where $S$ :
is the portion $x+y+z=1$ in the first octant $ 0 \leq x , 0 \leq y , 0 \leq z$ .
should i calculate
$\sqrt{3}\int_0^1\int_0^{1-x} (xy-x^2y-xy^2) dy~dx$ =
= $\sqrt{3}~\int_0^1\int_0^1 (u-1)(u-v)(v)~du~dv$
where :$ ~~~~~u = x+y ~,~~~~ v=x ~~~, ~~~Jacobian = -1$
and so the full answer is $\boxed{\frac{\sqrt{3}}{12}}$
The integrated region in the $xy$ plane is $\{(x,y):x,y\ge0,x+y\le1\}$. $$0\le x\le1\implies0\le v\le1\\0\le y\le1\implies0\le u-v\le1\\0\le x+y\le1\implies0\le u\le1$$
The integrated region in $uv$ plane is the intersection of the solution regions of the above three inequalities in $u,v$;$$\{(u,v):0\le v\le u\le 1,v\le1\}$$which is not very different from the original integral. Also note that when applying change of coordinates, you are supposed to multiply by the absolute value of the jacobian. The integral should look like$$\sqrt{3}\int_0^1\int_v^1v(1-u)(u-v)~du~dv$$which gives the answer as $\dfrac{\sqrt3}{120}$.
Remark. You could just as well avoid the change of coordinates and solve it directly.$$\sqrt3\int_0^1\int_0^{1-x} xy(1-x-y)dy~dx=\sqrt3\int_0^1\int_0^{1-x}x(1-x)y-xy^2dy~dx\\=\sqrt3\int_0^1\frac{x(1-x)^3}6dx=\sqrt3\int_0^1\frac{x^3(1-x)}6dx=\frac{\sqrt3}{120}$$