I'm going through some old calculus, and I'm struggling a bit with surface integrals.
Here's the problem:
Compute the integral
$\iint\limits_{\sigma} (x-y-z)d\sigma$
where $\sigma$ is the plane $x+y=1$ in the first octant, limited by $z=0$ and $z=1$.
So, what I've done so far is to convert the equation for sigma as a function of $y$, i.e. $y = 1-x$
$\therefore\quad\frac{\partial y}{\partial x} = -1,\quad\frac{\partial y}{\partial z} = 0\\ \therefore\quad\sqrt{(\frac{\partial y}{\partial x})^2 + (\frac{\partial y}{\partial z})^2+1} = \sqrt{2}\\ \therefore\quad\displaystyle{\iint\limits_{\sigma}} (x-y-z)d\sigma = \sqrt{2} \displaystyle{\iint\limits_{R}} (x-(1-x)-z)dxdz $
where $R$ is the projection of the given region $\sigma$ on the $xz$ plane.
Simplifying:
$\sqrt{2}\iint\limits_{R}(2x-1-z)dxdz$
So, it seems that the projection is a right triangle, with vertices at $(0,0,0), (0,0,1), (1,0,0)$.
Am I on the right track, and, how do I proceed from here? I full worked out example would help me a lot.