Integral of a floor function $\iint_D [x + y] \, dx\, dy$

Question

Given a domain $$D=[0,2]×[0,2]$$ How can I calculate the integral $$\bbox[5px,border:2px solid #C0A000]{\iint_D [x + y] \, dx\, dy}\quad$$

Answer 1

Hint.

Observe that $0\le x+y\le 4$ on $D$. Consider the following cases $$ m\le x+y<m+1,\quad m=0,1,2,3 $$ Then split your integral into four pieces.

Answer 2

How about breaking the domain into $$ D=D_0\cup D_1\cup D_2 $$ where $$ { D_0=\{(x,y): 0<x+y< 0.5\} \\ D_1=\{(x,y): 0.5<x+y< 1.5\} \\ D_2=\{(x,y): 1.5<x+y< 2\} } $$?

Answer 3

We can divide in four areas $A,B,C,D$ inside I indicated the value of $\lfloor x+y\rfloor$

Thus $I=0A+1B+2C+3D$ but notice that $A=D$ so $$I=(A+B+C+D)+(C+D)=Q+\frac 12Q$$

where $Q=4$ is the area of the square.

Answer 4

According to the question, we have$$[x+y]=k-1((x,y)\in D_k,k=1,2,3,4).$$ So$$I=\sum\limits_{k=1}^{4}\iint\limits_{D_k}[x+y]\mathrm{d}x\mathrm{d}y=\sum\limits_{k=1}^{4}(k-1)|D_k|=|D_2|+2|D_3|+3|D_4|.$$ and the $|D_k|$ is the area of $D_k$.Hence $$I=3+\frac{3}{2}.$$