Finding c with a density function defined inside a 3-d object

Question

I have a region $W$ with vertexes in $(0,0,0), (2,0,0), (0,3,0),(0,0,1)$ (a pyramid) i have the random vector $(X,Y,Z)$ with the density $c$ if $(x,y,z) \in W$ and $0$ otherwise.

I found that my regions of integrations could be

1. $0\leq x \leq 2$
2. $0 \leq y \leq \frac{-3}{2}x + 3$
3. $0 \leq z \leq \frac{6-3x-2y}{6}$

How can i now interpret all of this as a triple integral for getting the actual value of $c$?

Answer 1

The total volume of your region would be given by $$ V = \int_{x=0}^{x=2} \int_{y=0}^{3-3x/2} \int_{z=0}^{z=1-x/2-y/3} dzdydx, $$ and the total probability is $cV = 1$, so you need to set $c = 1/V$...