Volumes of Regions in Cube This is a multivariable calculus question about regions bounded by certain planes.  I have several planes that divide the cube $[-1/2, 1/2] \times [-1/2, 1/2] \times [-1/2, 1/2]$ into 24 total regions.  (There is a lot of symmetry.)
Here are the seven planes: 
$x = 0, y = 0, z = 0$;
$x + y = 0, x + z = 0, y + z = 0$;
$x + y + z = 0$.  

My overarching aim is to determine the volumes of the regions enclosed by the planes and inside the cube.
Taking $x = 0, y = 0, z = 0$, we divide the cube into eight regions of volume $1/8$.  In the final set-up there are two regions defined only by these planes. 
Now, taking all of the planes except for $x + y + z = 0$, we end up dividing some of those regions with volume $1/8$ into four regions.  Using symmetry it appears that the volumes of these regions should be $1/48$ and $1/24$.  (I would like to be able to check this, too).  
I would like to determine the volume of the regions that are also sliced by the plane $x + y + z = 0$.  
Problem: Could you help me determine the volume contained the cube $[-1/2, 1/2]^3$ and satisfying:
$$x + y > 0, x + z < 0 \text{ and } x + y + z < 0.$$ 
(My calculus is very rusty! Thank you!)
 A: The region that you describe is a union of regions in your picture (since you don't have choices on all of your inequalities).
Here's our plan: 


*

*Pick an integration order.

*Determine in which interval the outermost variable can take its values.

*Determine the limits of the other variables.
Let's try the integration order $dydxdz$.  We should be able to do any order, but this order is the most obvious to me.
$$
\int_\ast^\ast\int_\ast^\ast\int_\ast^\ast1dydxdz.
$$
It's easiest to start with the variable for the outermost integral; in this case $z$.  In other words, we need to find the biggest and smallest possible values for $z$.  The first inequality can be rewritten as
$$
-x-y<0.
$$
Adding this to the last inequality gives $z<0$.  Therefore, $z$ is negative; since we're in the given cube, the smallest that $z$ could be is $-\frac{1}{2}$.  If we need to further restrict the value for $z$, we will when it is required.  Therefore, we currently have:
$$
\int_{-\frac{1}{2}}^0\int_\ast^\ast\int_\ast^\ast1dydxdz.
$$
This gives the reason why I chose the integration order with $z$ on the outside - it was obvious to me that I could get $z$ being negative from the first and third inequalities.  Now, taking the second inequality, we have that $x<-z$.  Therefore, we use this to determine our limits for $x$.  In this case
$$
\int_{-\frac{1}{2}}^0\int_{-\frac{1}{2}}^{-z}\int_\ast^\ast1dydxdz.
$$
Finally, we can use the first inequality to get that $y>-x$ and the last inequality to get that $y<-x-z$.  We would like to use these two as the limits for the integral with respect to $y$, but we have to make sure that these don't describe an impossible situation, i.e., that $-x>-x-z$, but $-x<y<-x-z$.  However, $-x>-x-z$ implies that $z>0$, which is not possible, so the inequalities always define a region.  Therefore, our limits of integration are
$$
\int_{-\frac{1}{2}}^0\int_{-\frac{1}{2}}^{-z}\int_{-x}^{-x-z}1dydxdz.
$$
Now, I haven't done out all the details (and I skipped over a few checks that one might do - you can't always skip over the inequalities that have variables other than what you're looking for), but this should give the general idea.
