# 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?

• Have you tried making a sketch? Oct 19, 2016 at 18:16

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?

• You can take a look at the sketch in Piquito's answer; although I think it's probably a good idea to try to be able to make these simple sketches yourself, for future (similar) problems :-). Oct 19, 2016 at 18:37

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 $$\frac {\sqrt2}{2}\frac{\sqrt6}{2}=\frac{\sqrt3}{2}$$

• You seem to assume the surface integral is to be calculated of the identity function, but perhaps another function has to be integrated over this surface? Then the result is, in general, not simply the area you're referring to. Oct 19, 2016 at 18:35
• It is just the reason for what I said "If I understand well". Thank you. Oct 19, 2016 at 18:40