Answer
The integral:
$\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 p\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
represents the probability that both random variables $X$ and $Y$ are between $0$ and $1$.
The integral:
$\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - x} p\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
represents the probability that the sum $X+Y$ is at most $1$.
Work Step by Step
The double integral of $p\left( {x,y} \right)$ over $\left[ {0,1} \right] \times \left[ {0,1} \right]$ is given by
$\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 p\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
This integral represents the probability that both random variables $X$ and $Y$ are between $0$ and $1$. Geometrically, it is the volume of the region bounded below by $z=0$ and bounded above by the surface $z = p\left( {x,y} \right)$ over the domain $\left[ {0,1} \right] \times \left[ {0,1} \right]$.
Consider the integral of $p\left( {x,y} \right)$ over the triangle bounded by $x=0$, $y=0$, and $x+y=1$, given by
$\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{1 - x} p\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
This integral represents the probability that the sum $X+Y$ is at most $1$. Geometrically, it is the volume of the region bounded below by $z=0$ and bounded above by the surface $z = p\left( {x,y} \right)$ over the domain $0 \le x \le 1$, $0 \le y \le 1 - x$.
