Calculating $\iint xy \,\mathrm{d}S$ where $S$ is the surface of the tetrahedron with sides $z=0$, $y = 0$, $x + z = 1$, $x = y$ 
Calculate $\iint xy \,\mathrm{d}S$ where $S$ is the surface of the tetrahedron with sides $z=0$, $y=0$, $x + z = 1$ and $x=y$. 

The answer is given as: $(3\sqrt{2}+5)/24$
\begin{align*}
   &\, \iint xy \,\mathrm{d}S \\
  =&\, \iint xy \sqrt{1 + (z_x)^2 + (z_y)^2} \,\mathrm{d}A \\
  =&\, \int_{x=0}^1 \int_{y=0}^x xy \sqrt{1 + (-1)^2 + 0^2}
       \,\mathrm{d}y \,\mathrm{d}x \\
  =&\, \sqrt{2}
       \int_{x=0}^1
         \left[ \frac{xy^2}{2} \right]_{y=0}^x
       \,\mathrm{d}x \\
  =&\, \sqrt{2}
         \int_{x=0}^1 \frac{x^3}{2}
       \,\mathrm{d}x \\
  =&\, \frac{\sqrt{2}}{8}.
\end{align*}
 A: Here is a sketch of the tetrahedron.

The slanted red surface is $x+z=1$, so along this surface, $\vec r=\langle x,y,z\rangle=\langle x,y,1-x\rangle$. Then $d \vec r=\langle1,0,-1\rangle\,dx+\langle0,1,0\rangle\,dy$ and
$$d^2\vec A=\pm\langle1,0,-1\rangle\,dx\times\langle0,1,0\rangle\,dy=\pm\langle1,0,1\rangle\,dx\,dy$$
$$d^2A=\left|\left|d^2\vec A\right|\right|=\sqrt2\,dx\,dy$$
$$I_1=\int_0^1\int_0^xxy\sqrt2\,dy\,dx=\int_0^1x\sqrt2\frac{x^2}2dx=\frac{\sqrt2}8$$
The blue surface in back that's kind of hard to see is $x=y$, so along this surface, $\vec r=\langle x,x,z\rangle$, $d\vec r=\langle1,1,0\rangle\,dx+\langle0,0,1\rangle\,dz$ and
$$d^2\vec A=\pm\langle1,1,0\rangle\,dx\times\langle0,0,1\rangle\,dz=\pm\langle1,-1,0\rangle\,dx\,dz$$
$$d^2 A=\left|\left|d^2\vec A\right|\right|=\sqrt2\,dx\,dz$$
$$I_2=\int_0^1\int_0^{1-x}x^2\sqrt2\,dz\,dx=\int_0^1x^2\sqrt2(1-x)dx=\frac{\sqrt2}{12}$$
The yellow surface on the bottom is $z=0$ so along the surface, $\vec r=\langle x,y,0\rangle$. Then $d\vec r=\langle1,0,0\rangle\,dx+\langle0,1,0\rangle\,dy$ and
$$d^2\vec A=\pm\langle1,0,0\rangle\,dx\times\langle0,1,0\rangle\,dy=\pm\langle0,0,1\rangle\,dx\,dy$$
$$d^2A=\left|\left|d^2\vec A\right|\right|=dx\,dy$$
$$I_3=\int_0^1\int_0^xxy\,dy\,dx=\int_0^1\frac12x^3\,dx=\frac18$$
The green surface in front is $y=0$, but here the integrand is indentically $0$, so
$$I_4=0$$
Adding up,
$$I=I_1+I_2+I_3+I_4=\frac{\sqrt2}8+\frac{\sqrt2}{12}+\frac18+0=\frac{5\sqrt2+3}{24}$$
This disagrees with the given answer; maybe the composer of the answer key forgot about the $\sqrt2$ in $I_2$. Either that or I might have some mistake. Sometimes you just can't see your own mistakes when they are obvious to the first person who looks at them.
A: Well you have four surfaces to deal with, so you'll have to parametrize each one of them. For example start with the triangle in the $ \{y = 0\} $ plane. This is easy because the entire plane is parametrized by $\sigma(s,t) = (s,0,t)$. We'd have restraints on $s$ and $t$ which would be easy to calculate, but notice that in this plane, $xy = s\cdot0 = 0$, so the integral becomes $\iint0 dS  = 0$. But that was sort of cheating so we'll actually do the next integral, for example the triangle in the $\{z=0\}$ plane. This is going to be simple as well because, being in the xy-plane, we can describe it as y-simple region and use Fubini's Theorem to evaluate. As seen top down:
So the region is $D = \{(x,y,0)\mid 0\le x \le 1, 0 \le y \le x \}$
And the integral over D is $$\int_0^1\int_0^x xy \ dydx  = \frac12\int_0^1x^3dx =\frac18 $$
For the last two the procedure I'd do is to parametrize the entire plane in consideration as $ \Pi :p + \lambda u + \mu v$ where $p$ is a point in the plane and $u$ and $v$ "lie" on the plane. In other words $\sigma(\lambda,\mu) = (a,b,c) + \lambda(u_1,u_2,u_3) + \mu(v_1,v_2,v_3)$. Afterwards impose the restrictions you have on $(x,y,z) = (a+\lambda u_1+ \mu v_1,b+\lambda u_2+ \mu v_2,c+\lambda u_3+ \mu v_3)$ (for example $0 \le y \le x, 0 \le x \le 1, 0 \le z \le 1$, for the "inclined" triangle). From there derive restrictions on $\lambda$ and $\mu$, and these will determine your integral limits. Also dont forget to calculate the norm of the normal vector to the parametrization you choose, as this needed in calculating the integral.
