Projection of a plane on $xy$ plane I am tryi4ng to solve a surface integral, where the given surface is the plane $x + y + z = 1$, where $x,y,z \geq 0$. Now, to do this I need to find the projection of this plane on to the $xy$ plane, to see what domain $D$ this gives me, but I am unsure how to do this?
 A: I could draw it, but it's even better if you draw it yourself!
Sketch a cartesian coordinate system ($xyz$-axes) and notice that the intersection of the plane $x+y+z=1$ with the three coordinate planes gives you the following lines:


*

*in the $xy$-plane (so $z=0$): $x+y=1$;

*in the $xz$-plane (so $y=0$): $x+z=1$;

*in the $yz$-plane (so $x=0$): $y+z=1$.


These lines are easy to draw and you only need the parts where $x,y,z \ge 0$ so e.g. in the $xy$-planes that means you connect $(1,0,0)$ with $(0,1,0)$ and continue like that for the two other coordinate planes, giving you three line segments forming a triangle. Together with the coordinate planes, they bound a region in the first octant ($x,y,z \ge 0$).
This should give you a good idea of the plane $x+y+z=1$ in the first octant. Clearly, the projection onto the $xy$-plane is then bounded by the $x$- and $y$-axis and the line (segment) you drew in the $xy$-plane, the one connecting $(1,0,0)$ and $(0,1,0)$.
Now recall that this line in the $xy$-plane has the equation $x+y=1$. For setting up the integral, you can now let $x:0 \to 1$ and then $y:0 \to 1-x$ or you take $y:0\to 1$ and then $x: 0 \to 1-y$. Does that help?
A: If I understand well your post you want to calculate the area of the green triangle in the figure below which is an equilateral triangle of side $\sqrt 2$ so whatever your integral be your answer is $$\frac {\sqrt2}{2}\frac{\sqrt6}{2}=\frac{\sqrt3}{2}$$

