Question: For $D=[0,+\infty)×[0,+\infty)$ and $f(x,y)=(1+x+y)^{-1}$, why does the integral over $D$ of $f(x,y)$ not converge? 

It's not like $\rho/(1+\rho|\sin\theta+\cos\theta|)$ which to infinite is like $1/(|\sin\theta+\cos\theta|)$ and, close to $0$ is like $0$?

Answer: 
Consider 

$$D'=[0,+\infty)\times [0,1]$$

and note that

$$\int\_D \frac1{1+x+y} \,dx\,dy \ge \int\_{D'} \frac1{1+x+y} \,dx\,dy\ge \int\_{D'} \frac1{2+x} \, dx \, dy$$
First, we see that $f$ is positive on all of $D$. This means that we can consider a smaller region, and if it diverges there it still diverges on ll of $D$. Consider the region $E\subseteq D$ given by $E = [0,\infty)\times [0,1]$. On $E$, we have $f(x, y)\geq \frac{1}{2+x}$, and we have
$$
\iint\_Df(x, y)\,dx\,dy \geq \iint\_Ef(x, y)\,dx\,dy = \int\_0^\infty \int\_0^1f(x, y)\,dy\,dx\\
\geq \int\_0^\infty\int\_0^1\frac{1}{2+x}dy\,dx = \int\_0^\infty\frac{1}{2+x}dx
$$
which diverges.