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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: $(3sqrt{2}+5)/24$
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*}

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, $xy = scdot0 = 0$, 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 $D = {(x,y,0)mid 0le x le 1, 0 le y le x }$
And the integral over D is $$int_0^1int_0^x xy dydx = frac12int_0^1x^3dx =frac18 $$

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.

Here is a sketch of the tetrahedron.

The slanted red surface is $x+z=1$, so along this surface, $vec r=langle x,y,zrangle=langle x,y,1-xrangle$. Then $d vec r=langle1,0,-1rangle,dx+langle0,1,0rangle,dy$ and
$$d^2vec A=pmlangle1,0,-1rangle,dxtimeslangle0,1,0rangle,dy=pmlangle1,0,1rangle,dx,dy$$
$$d^2A=left|left|d^2vec Aright|right|=sqrt2,dx,dy$$
$$I_1=int_0^1int_0^xxysqrt2,dy,dx=int_0^1xsqrt2frac{x^2}2dx=frac{sqrt2}8$$
The blue surface in back that's kind of hard to see is $x=y$, so along this surface, $vec r=langle x,x,zrangle$, $dvec r=langle1,1,0rangle,dx+langle0,0,1rangle,dz$ and
$$d^2vec A=pmlangle1,1,0rangle,dxtimeslangle0,0,1rangle,dz=pmlangle1,-1,0rangle,dx,dz$$
$$d^2 A=left|left|d^2vec Aright|right|=sqrt2,dx,dz$$
$$I_2=int_0^1int_0^{1-x}x^2sqrt2,dz,dx=int_0^1x^2sqrt2(1-x)dx=frac{sqrt2}{12}$$
The yellow surface on the bottom is $z=0$ so along the surface, $vec r=langle x,y,0rangle$. Then $dvec r=langle1,0,0rangle,dx+langle0,1,0rangle,dy$ and
$$d^2vec A=pmlangle1,0,0rangle,dxtimeslangle0,1,0rangle,dy=pmlangle0,0,1rangle,dx,dy$$
$$d^2A=left|left|d^2vec Aright|right|=dx,dy$$
$$I_3=int_0^1int_0^xxy,dy,dx=int_0^1frac12x^3,dx=frac18$$
The green surface in front is $y=0$, but here the integrand is indentically $0$, so
$$I_4=0$$
Adding up,
$$I=I_1+I_2+I_3+I_4=frac{sqrt2}8+frac{sqrt2}{12}+frac18+0=frac{5sqrt2+3}{24}$$
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.