Integration limits in a triple integral I'm having trouble describing a region in space.
Let $B$ the region bounded by the planes $x=0$, $y=0$, $z=0$, $x+y=1$, $z=x+y$. One way of calculating the volume of B is:
\begin{equation}
V(B)=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{x+y}dzdydx=\frac{1}{3}
\end{equation}
I'm trying to change the order to integration to $\int \int \int dxdydz$, but I can't get my integration limits to work (I get different values for the volume).
I thought that $(0<z<1)$, and $(z-y<x<1-y)$, but I can't find the limits on the Y axis. I believe it sould be something like:
\begin{equation}
\int_0^1\int_?^?\int_{z-y}^{1-y}dxdydz
\end{equation}
 A: Try to work it out from the hints before reading the spoilers.
The first hint is to make a sketch of the cross-section of a plane $z=c$ with the volume of interest and $c$ some constant for which $0 \leq c \leq 1$. What is the shape of this cross-section?

Answer : a trapezoid, bounded by $x \geq 0, x+y\geq z, y \geq 0, 1 \geq x+y$
Can you now find the integration limits?

Did you think of splitting the area in two parts to make it easier?

The simplest way would be to split the trapezoid in a parallelogram and a triangle separated by the line $y=c$. Can you solve the problem for both these areas?

$$ 
\int_0^1 d z \left\{ \int_0^z d y \int_{z-y}^{1-x} d x + \int_z^1 d y \int_0^{1-y} d x \right\} = \frac{1}{3}
$$
        Obviously not the easiest way to evaluate the integral, but not too difficult either.




A: That suggestion I made to 'cheat': it's not necessary - the 'fog' has lifted, and I can see now that $$\int_{x=0}^1\int_{y=0}^{1-x}\int_{z=0}^{x+y}dz.dy.dx$$ does actually comprise the whole of the volume specified. If you see it through, you get $$\int_{x=0}^1\int_{y=0}^{1-x}(x+y)dy.dx$$$$=$$$$\int_{x=0}^1{1\over2}(2x(1-x)+(1-x)^2)dx$$$$=$$$${1\over2}\int_{x=0}^1(1-x^2)dx$$$$=$$$${1\over3} .$$ It is of course necessary to do it this way if you are integrating some function of $x$ $y$ & $z$ over the region - which is generally the case when triple integrals arise in natural course ... the geometric shortcut that I broached in the comments can only be used for finding the volume ... except perhaps in the occasional problem having extraordinary symmetry.
I think my 'cheat' is still relevant though. Because of the shape & orientation, a more 'fitting' set of coordinates for the integral would be $$r=x+y$$&$$s=x-y ,$$& $z$ kept as it is, whence the integral would be$$\int_{r=0}^1\int_{s=-r}^r\int_{z=0}^r \operatorname{f}({r+s\over2}, {r-s\over2},z).dz.{ds\over\sqrt{2}}.{dr\over\sqrt{2}} ,$$ with the factors of $1/\sqrt{2}$ at the differentials $dr$ & $ds$ so that they represent distance in this space. It might also fit the function being integrated over that region better also, as if the region has that shape & orientation, there's an appreciable likelyhood that the function also has some convenient symmetry with respect to that orientation.
