Double integral of floor function and exponential function over rectangular region I am not getting any idea how to deal with this two sums of double integral where functions and rectangular domain is given as 


*

*$\lfloor x+y\rfloor$ and $1\leq x\leq 3$, $2\leq y\leq 5$

*$e^{\max(x^2,y^2)}$ and $0\leq x\leq 1,0\leq y\leq 1$.
For first one I use inequality $x+y-1<\lfloor x+y\rfloor\leq x+y$, and I just get hint that my answer will in between $27$ and $33$. But how I will get exact answer.

For a given value of $x$ and integer $k$, $$\lfloor x+y\rfloor = \left\{ \begin{matrix} \lfloor x \rfloor + k  & \text{for $k \le y   < k + 1 - x + \lfloor x \rfloor $,} \\ \lfloor x \rfloor + k+1  & \text{for $k + 1 - x + \lfloor x \rfloor \le y   < k + 1$.}\end{matrix}\right.$$ 
Consequently, for an integer $k$,
$$\int_{k}^{k+1} \lfloor x+y\rfloor \,dy = \int_{k}^{k + 1 - x + \lfloor x \rfloor} (\lfloor x \rfloor+ k) \,dy + \int_{k + 1 - x + \lfloor x \rfloor}^{k+1} (\lfloor x \rfloor +k +1 )\,dy = k +x.$$
Therefore,
$$\int_{1}^{3}\int_{2}^{5} \lfloor x+y\rfloor \, dy \, dx = \int_{1}^{3}\left(\int_{2}^{3} \lfloor x+y\rfloor \, dy+\int_{3}^{4} \lfloor x+y\rfloor \, dy+\int_{4}^{5} \lfloor x+y\rfloor \, dy\right)\,dx$$ implies
$$\int_{1}^{3}\int_{2}^{5} \lfloor x+y\rfloor \, dy \, dx = \int_{1}^{3} [(2+x)+(3+x)+(4+x) ]\,dx = \color{red}{30}.$$
