What is the volume of one of the regions bounded by the planes $x=0,y=0,z=0,x=y,y=z,x=z,x=1,y=1,z=1$? What is the volume of one of the regions bounded by the planes $x=0,y=0,z=0,x=y,y=z,x=z,x=1,y=1,z=1$?
My Try :: As the planes divide the unit cube up into 6 pieces, and they are symetricall, the area of each region is 1/6. However, I am wondering how I would solve the problem with an integral. I tried to do something like $\displaystyle \iiint_{A} dA = \iiint_{A} dx dy dz$ but I am unsure of the bounds.
 A: You claim there are $6$ such parts. Let's verify this formally by checking the location of a point in the unit cube not on one of these planes. For each plane, we can determine on what side of the plane our point is, e.g. for the plane $x = y$, either it is on the side $x > y$ or on the side $x < y$. We indeed have $6$ (valid - see note below) combinations of inequalities for the coordinates $(x,y,z)$ of a point:


*

*$0 < x < y < z < 1,$

*$0 < x < z < y < 1,$

*$0 < y < x < z < 1,$

*$0 < y < z < x < 1,$

*$0 < z < x < y < 1,$

*$0 < z < y < x < 1.$


Let's pick the first case: $0 < x < y < z < 1$. To find bounds for this volume, you have to find the dependencies of the variables $x,y,z$.
We'll start by choosing $z$ to be the independent variable. Within our volume, $z$ can vary from $0$ to $1$. So we write our integral as
$$\int_0^1 \cdots dz.$$ 
Next, let's look at $y$ when $z$ is already fixed. Since $y < z$, we find that $y$ can still vary between $0$ and $z$. Our integral becomes
$$\int_0^1\int_0^z \cdots dydz.$$ 
Finally, $y$ and $z$ are already fixed, and because $x < y$, we obtain that $x$ can vary between $0$ and $y$:
$$\int_0^1\int_0^z\int_0^y dxdydz.$$ 
Solving this integral gives:
$$\begin{align}
\int_0^1\int_0^z\int_0^y dxdydz &= \int_0^1\int_0^z [x]_0^y dy dz\\
&= \int_0^1\int_0^z y dy dz\\
&= \int_0^1 \left[\frac{y^2}{2}\right]^z_0 dz\\
&= \int_0^1 \frac{z^2}{2} dz\\
&= \left[\frac{z^3}{6} \right]^1_0\\
&= \frac{1}{6}.
\end{align}$$
So indeed, the volume is $1/6$. 
You could calculate the same volume by choosing a different independent variable! For example, let $x$ be the independent variable, then $y > x$ and $z > y$ so you would get
$$\int_0^1\int_x^1\int_y^1 dzdydx = \frac{1}{6}.$$
You can even choose $y$ as independent variable here. Then $x < y$ and $z > y$ so you get
$$\int_0^1\int_y^1\int_0^y dxdzdy = \int_0^1\left(\int_y^1 dz \right)\left( \int_0^y dx\right)dy = \frac{1}{6}.$$
In this case you see that the three integrals don't necessarily have to be nested! Once you've fixed $y$, $x$ and $z$ no longer have to depend on eachother, as they are "separated by $y$" in the inequality $x < y < z$.
Or, of course, you could pick any of the other $5$ volumes, which would obviously give the same result. The trick to finding the bounds, is to choose a good independent variable and work out how to other variables depend on it.
Note: a priori we have $8$ combinations of the form "$x \dots y$ and $y \dots z$ and $x \dots z$", where the dots are either $>$ or $<$. However, two of these end up in contradiction. The combination "$x < y$ and $y < z$ and $z < x$" implies that $x < y < z < x$ and of course $x < x$ is impossible. The other combination that doesn't work is "$x > y$ and $y > z$ and $z > x$, which gives $x < z < y < x$".
