Triple integral - how to make a projection on the $xy$ plane? I'm starting to study triple integrals. In general, I have been doing problems which require me to sketch the projection on the $xy$ plane so I can figure out the boundaries for $x$ and $y$. For example, I had an exercise where I had to calculate the volume bound between the planes $x=0$, $y=0$, $z=0$, $x+y+z=1$ which was easy. For the projection on the $xy$ plane, I set that $z=0$, then I got $x+y=1$ which is a line.
However, now I have the following problem:
Calculate the volume bound between:
$$z=xy$$
$$x+y+z=1$$
$$z=0$$
now I know that if I put $z=0$ into the second equation I get the equation $y=1-x$ which is a line, but I also know that $z=xy$ has to play a role in the projection. If I put $xy=0$ I don't get anything useful. Can someone help me understand how these projections work and how I can apply it here?
 A: As you wrote plane $x+y+z=1$ cut plane $OXY$ in line $x+y=1$, which together with $x,y,z \geqslant 0$ gives triangle
$$\left\lbrace\begin{array}{l}0\leqslant x \leqslant 1 \\
0 \leqslant y \leqslant 1-x\end{array}\right\rbrace$$
Projection of intersection of $x+y+z=1$  and $z=xy$ gives $y=\frac{1-x}{1+x}$ hiperbola on plane $OXY$, which divides triangle in 2 parts. In left part plane $x+y+z=1$ is above "seddle"(Hyperbolic paraboloid) $z=xy$ and reverse on right.
Volume between plane and "seddle" from above and plane $z=0$ from below, for $x,y,z \geqslant 0$ can be calculated as
$$\int\limits_{0}^{1}\int\limits_{0}^{\frac{1-x}{1+x}}\int\limits_{0}^{xy}+\int\limits_{0}^{1}\int\limits_{\frac{1-x}{1+x}}^{1-x}\int\limits_{0}^{1-x-y}$$
A: 
The diagram that I have added is the projection in XY-plane. This curve will be the reference as we set up our integral.
As the curve in red is the projection of intersection of both surfaces in XY-plane, we equate $z$ and the equation of projection is,
$xy = 1 - x - y \implies xy + x + y = 1$
If you set up the integral in the order $dz$ first then you have to split the integral into two.
i) For $xy + x + y \leq 1, x, y \geq 0$,
$0 \leq z \leq xy$ as we are bound above by the hyperboloid.
ii) For $xy + x + y \geq 1, x + y \leq 1$,
$0 \leq z \leq 1-x-y$ as we are bound above by the plane.
Now after $dz$, you can choose either order $dx \ dy$ or $dy \ dx$. The bounds are straightforward.
But the part that I wanted to bring to your notice is that if we decide to integrate in the order $dx$ followed by $dz$ (and then $dy$ last) or in the order $dy$ followed by $dz$ (and then $dx$ last), we can set this up in one integral.
As we found, at the intersection of the plane and the hyperboloid,
$xy + x + y = 1 \implies y = \dfrac{1-x}{1+x}$
$z = xy = 1-x-y = \dfrac{x-x^2}{1+x}$
So summarizing it, for any given $x$ and $z$, you can see that $y$ is bound by hyperboloid $(\frac{z}{x})$ and the plane $(1-x-z)$ and for any given $x$, $z$ is bound by xy-plane $(z = 0)$ and the $z$ at the intersection of surfaces $\left(\frac{x-x^2}{1+x}\right)$. That leads us to the integral,
$\displaystyle \int_0^1 \int_0^{(x-x^2)/(1+x)} \int_{z/x}^{1-x-z} \ dy \ dz \ dx$
