Surface integral of a plane I have this function
$$ f(x,y,z) = x + y $$
and I want to find
$$ \iint f(x,y,z) dS $$
Where S is the first octant part of the plane 
$$ x + y + z =1 $$
Now, I know the method, and have paramatrized like this 
$$
z = 1 - x - y,\, x = x,\, y = y.
$$
However, I thought the bounds of integration for x and y were between 0 and 1, but doing this gives me an answer of $3^{0.5}$, when the answer is $(3^{1/2})/3$, implying that the bounds that my textbook is using are different.
What are they?  
 A: Here is the solution,
$$ \iint f(x,y,z) dS=\int_{0}^{1}\int_{0}^{1-x} (x+y) \sqrt{1+(\frac{dz}{dx})^{2}+(\frac{dz}{dy})^2}dy\,dx$$
$$ = \sqrt{3} \int_{0}^{1}\int_{0}^{1-x} (x+y)dydx= \frac{\sqrt{3}}{3}. $$
Note that, S is the first octant part of the plane, then you will have the region in xy-plane bounded by the $x$-axis, $y$-axis and the $y=1-x$, $x\geq 0, y\geq 0.$   
A: The region of integration is the part of the plane $x + y + z = 1$ that is cut out by the inequalities $x, y, z \geq 0$.  The resulting surface is a triangle with vertices $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.  You can project the triangle onto the $x$-$y$ plane for your calculation, in which case your region of integration is given by $x, y \geq 0$ and $1 - x - y \geq 0$.  If you integrate $y$ first, then the limits of integration are from $y = 0$ to $y = 1 - x$.  Then $x$ has to go from $x = 0$ to $x = 1$.  Then $dS = \sqrt{3} \, dx \,dy$, so this yields
$$
\iint f \,\, dS = \int_{x = 0}^{x = 1} \int_{y = 0}^{y = 1 - x} (x + y) \, \sqrt{3} \,\, dy \,\, dx.
$$


*

*Edited because the original had the wrong $dS$ term.

