What goes wrong in my computation of $\iiint _S 1 dz dy dx$ where $S=\{(x,y,z)\mid ax+by+cz=0, 1\leq x,y,z \leq N\}$? I am trying to calculate this triple integral but I don't really know if it is correct. I don't have much experience with triple integrals. Here it is:
$\iiint _S 1 dz dy dx$ where $S$ is the set $\{(x,y,z)\mid ax+by+cz=0, 1\leq x,y,z \leq N\}$. Here $a$, $b$ and $c$ are fixed non-zero reals, $N$ is a fixed positive real. It seems that $z$ should run from $1$ to $-(ax+by)/c$ and we can let $x$ and $y$ run from $1$ to $N$, my reasoning being that if $x$ and $y$ are fixed, $z$ gets automatically fixed.
Then our integral transforms into $$\iiint _S 1 dz dy dx=\int_1^N \int_1^N \int_1^{-(ax+by)/c} 1 dz dy dx=\int_1^N \int_1^N -(ax+by+c)/c  \mathrm{d}x dy$$ which is equal to $$\int_1^N-(ax^2/2+bxy+cx)|^{N}_1 dy=\int_1^N \left(-(aN^2/2+bNy+cN)+a/2+by+c\right)dy=-aN^3/2-bN^3/2-cN^2+aN/2+bN^2/2+cN+aN^2+bN/2+cN-a/2-b/2-c.$$
I feel like i am making an error somewhere, but I am not sure. Thanks!
 A: The integration region of your problem isn't well-defined. My guess is that the equation $ax+by+cz=0$ should be an inequation, $ax+by+cz\leq0$ or $ax+by+cz\geq0$, because if the equation $ax+by+cz=0$ does define your set, you will actually have a surface and not a volume.
Anyway, it is always easier to solve multiple integrals by visualizing their respective integration regions. Doing so, it is possible to conclude that your question is definitely harder than it looks.
$1 \leq x,y,z,\leq N$ defines a cube of edge length $N-1$ and $ax+by+cz=0$ defines the plane $z=-(ax+by)/c$. From here, obvious questions arise: Does the plane cut the cube? If does, how exactly? You have to study the all the possibilities.
For now, your solution is only valid if $ 1 \leq z=-(ax+by)/c \leq N$ for all points $(x,y)$ in the square $[1,N] \, \text{x} \, [1,N]$ (and if the inequality $ax+by+cz\leq0$ takes the place of $ax+by+cz=0$).
It could be interesting to highlight that the upper bound value of the integral is known: since the integrand is $f(x,y,z)=1$, it is the volume of the cube, $(N-1)^3$.
