Calculate the surface integral:
$$\iint_\sigma f(x,y,z)\ \mathrm{d}S$$
Where: $f(x,y,z) = x-y-z$ and $\sigma$ is the portion of the plane $x+y=1$ on the first octant between $z=0$ e $z=1$
I don't know how to determinate the limits of the integral. I guess it must be $\mathrm{d}x\:\mathrm{d}z$, but x variates in function of y, and not of z.
How start it?
On
It actually does not matter, you can use $dx\,dz$ or $dy\,dz$ and express the third coordinate with these two from $x+y=1$. Then just be careful to get the area correctly (there is a $\sqrt{2}$ factor).
The $z$ coordinate is independent anyway, it just goes from $0$ to $1$. The rest you can work out in the $xy$ intersection sketch that you also need a range from $0$ to $1$ (in x or in y, it doesn't matter).
EDIT:
A "divine inspiration" method: $x-y$ is symmetric on this domain, it is negative on one half of the plane and positive on the other. So $\iint(x-y)dS=0$. What's left is $-\iint z\,d S$. The double integral goes over $z$ and over the "width" of the area, on which $z$ does not depend, so it leaves you with $$-\sqrt{2}\int_0^1 z\,dz=-\frac{\sqrt2}{2}$$
Choose $x$ and $z$ as parameters, for example.
First, show that $dS=\sqrt2dxdz$, and that $f(x,y,z)=2x-z-1$ on $\sigma$.
Then, since $0<x<1$ and $0<z<1$, transform it into a double integral.
Finally, compute the double integral.