Performing double integration on a triangle in $R^3$. I have to determine the surface integral $\int\int_S z^2 dS$ on the triangle with vertices $(1,0,0), (0,2,0)$ and $(0,0,3)$. 
I have performed double integration on a triangle in $R^2$, but never in $R^3$. How should I go about this? 
 A: You can use a parametrization of your triangle by two variables - any two of $x$, $y$, $z$ will do. It is easy to check that your triangle lies on the plane with equation $6x + 3y + 2z = 6$, so that you may write $z$ in terms of $x$, $y$ as $z = 3 (1-x - y/2)$.
Now the tricky part is how to write the $dS$ surface element in terms of $dx$, $dy$. The “good” (as in: most general) way to do this would be to find a full change of coordinates in an open subset of $\mathbb R^3$ bringing the triangle to a triangle in the plane, and to use the Jacobian of this. But given that we are working with a triangle (affine functions), you can write $dS$ as a constant times $dx dy$, and simply deduce the constant from the total area of the triangle and its projection on the $(x,y)$ plane. Given that the sides of your triangle have lengths $\sqrt{5}$, $\sqrt{10}$ and $\sqrt{13}$, Heron's formula gives an area of $7/2$ for the triangle. The projection triangle on the other hand, with vertices $(0,0)$, $(1,0)$ and $(0,2)$, has area $1$. So you should find $dS = 7/2 dx dy$. A quick back-of-the-envelope computation, including the change of variables v=y/2, then gives back the integral
$(63/4) \int_T (1-x-v)^2 dxdv$, where $T$ is the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$. You should be able to finish from there.
