Your tetrahedron is bordered by the coordinate planes, and the plane $x+y+z=1$. Sketch a picture! So $x\ge0,$ $y\ge0$, $z\ge0$ and $x+y+z\le1$.
[Edit: If you have trouble with this step, then go and review how you determine
the equation of a plane given that you know three points on it. Here the coordinate planes $x=0$, $y=0$ and $z=0$ stand out. The fourth plane passes via the points $(1,0,0), (0,1,0)$ and $(0,0,1)$. If everything else fails, you can go through that process. Here with a bit of experience you should notice that the coordinates of these three points sum up to $1$. That gives you the equation of the last plane./Edit]
As a next step you should ask yourself the questions:
- If I know the values of $x$ and $y$, what is the allowable range for $z$?
- Ignoring $z$, if I know the value of $x$, what is the allowable range for $y$?
- Ignoring $z$ and $y$, what is the allowable range for $x$?
This will give you the limits, if you first integrate w.r.t. $z$, then $y$, last $x$.
If you prefer (sometimes it will be to your advantage), you can process the variables in a different order.
If you have a volume $A$, and want to express $\int_A f(x,y,z)\,d(x,y,z)$ as an iterated integral, i.e. as $$
\int_A f(x,y,z) \,d(x,y,z) = \int_{A_x} \int_{A_y(x)} \int_{A_z(x,y)} f(x,y,z) \,dz\,dy\,dx
$$
you have to find $$\begin{eqnarray}
A_x &\subset& \mathbb{R} \\
A_y &\,:\,& A_x \to \mathcal{P}({\mathbb{R}}) \\
A_z &\,:\,& \{(x,y) \,:\, x \in A_x, y \in A_y(x)\} \to \mathcal{P}(\mathbb{R})
\end{eqnarray}$$
such that $$
A = \left\{(x,y,z) \,:\, x \in A_x,\, y \in A_y(x),\, z \in A_z(x,y,z) \right\}.
$$
Integration order $dz\,dx\,dy$
Let's look at the case of a pyramid with base $[-2,2]\times [-2,2]$ in the $x,y$-plane, tip at $(0,0,5)$, and integration order $dz\,dx\,dy$. We have $$
A = \left\{(x,y,z) \,:\, x \in [-2,2], y \in [-2,2], 0 \leq z \leq \tfrac{5}{2}\min \{2-|x|,2-|y|\}\right\} \text{.}
$$
So $A_y$ doesn't actually depend on $x$, and $A_z(x,y) = [0, \tfrac{5}{2}\min \{2-|x|,2-|y|\}]$. Therefore, $$
\int_A 1 \,d(x,y,z) = \int_{-2}^2 \int_{-2}^2 \int_0^{\tfrac{5}{2}\min \{2-|x|,2-|y|\}} 1 \,dz\,dy\,dx \text{.}
$$
Integration order $dx\,dy\,dz$
For integration order $dx\,dy\,dz$ we have to find $$\begin{eqnarray}
A_z &\subset& \mathbb{R} \\
A_y &\,:\,& A_z \to \mathcal{P}({\mathbb{R}}) \\
A_x &\,:\,& \{(z,y) \,:\, z \in A_z, y \in A_y(z)\} \to \mathcal{P}(\mathbb{R})
\end{eqnarray}$$
such that $$
A = \left\{(x,y,z) \,:\, x \in A_x(z,y),\, y \in A_y(z),\, z \in A_z \right\}.
$$
Obviously, $A_z = [0,5]$. For a fixed $z \in A_z$, $(x,y,z) \in A$ exactly if $$\begin{eqnarray}
z &\leq& \tfrac{5}{2}(2 - |x|) &\Leftrightarrow& |x| \leq \tfrac{2}{5}z - 2 &\text{ and } \\
z &\leq& \tfrac{5}{2}(2 - |y|) &\Leftrightarrow& |y| \leq \tfrac{2}{5}z - 2 \text{,}\\
\end{eqnarray}$$
which means $$
\int_A 1 \,d(x,y,z) = \int_0^5 \int_{-\frac{2}{5}z + 2}^{\frac{2}{5}z - 2} \int_{-\frac{2}{5}z + 2}^{\frac{2}{5}z - 2} 1 \,dx\,dy\,dz \text{.}
$$
Best Answer
The equation of plane $ADC$ is $x+z=1$.
The equation of plane $BDC$ is $y+z=1$.
$$\iiint_D (1-z^2)dxdydz= \int_0^1\int_0^x\int_0^{1-x}(1-z^2)\,dz\,dy\,dx+\int_0^1\int_x^1\int_0^{1-y}(1-z^2)\,dz\,dy\,dx$$