<s>[INST] Applying the Cartesian Coordinate Find the surface area of the portion of $2x+y+z=8$ in the first octant.
I know that $f(x,y)=8-y-2x$, $x=0$ to $4$ and $y= 0$ to $8-2x$, but I'm having trouble solving the problem in Cartesian form. The answer is supposedly $16\sqrt6$ 
 [/INST] This is an analytic proof: 
$$\int_0^4\int_0^{8-2x} \sqrt{(-1)^2+(-2)^2+1} \ dydx=\sqrt{6}\int_0^4 8-2x dx=16\sqrt6$$
This is a geometric proof: the surface is a triangle, whose three vertices lie on the three axes. The three vertices have the following coordinates: $(4,0,0);(0,8,0);(0,0,8)$ which yields the following lengths for the sides: $$a=4\sqrt{5}\quad b=4\sqrt{5}\quad c=8\sqrt{2}$$
Applying Heron's formula yields:$$s=\frac{a+b+c}2=4\sqrt5+4\sqrt2$$
$$A^2=s(s-a)(s-b)(s-c)=(4\sqrt5+4\sqrt2)\times(4\sqrt5+4\sqrt2-4\sqrt5)^2\times(4\sqrt5-4\sqrt2)$$ 
$$A^2=16\times16\times(5-2)\times2$$
$$A=16\sqrt{6}$$
</s>