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!)
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.