Did I mess up when plugging in values?

Solution 1:

Your error is not in the values plugged in for the integrand, but in the bounds of the integral. Your surface is the plane $x+y+z=1$ for $x,y,z\geq0$. If we solve this equation at for $z$ and apply $z\geq0$, we get $1-x-y\geq0$, or $y\leq1-x$. Applying $0\leq y$ to this, we get $x\leq1$. Thus our integral should be:

$$\int_0^1\int_0^{1-x}x+y\ {\rm d}y\,{\rm d}x$$

Which evaluates to $\frac{1}{3}$.

Edit:

You said the answer is supposed to be $-\frac{1}{3}$, not $\frac{1}{3}$. So there is some error in the integrand, and I believe it has to do with the "oriented downward" text of the problem statement.

In particular, you used $dS=\langle g_x,g_y,1\rangle$ in your solution. I believe that the "downward-facing" part of the problem indicates that the $z$-component of this vector should be negative, so it should be negative what you found, that is, $dS=\langle -g_x,-g_y,-1\rangle$.

This results in an integrand of $-x-y$, and an answer of $\frac{-1}{3}$.