Integration limits in a triple integral

Question

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}

Answer 1

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.