In order to evaluate $\displaystyle\int_{0}^{2}\int_{0}^{2}\lfloor x+y \rfloor dx \,dy$, I transformed co-ordinates using $x= u-v$ and $y=v$. Then the Jacobian becomes $1$ and $0 \leq u \leq 4$ and $0 \leq v \leq 2$ implying that
$\displaystyle\int_{0}^{2}\int_{0}^{2}\lfloor x+y \rfloor dx \,dy = \displaystyle\int_{0}^{2}\int_{0}^{4}\lfloor u \rfloor du\,dv = \displaystyle\int_{0}^{2}6\, dv = 12$
However the answer on Wolfram alpha is $6$. Can someone point out where I went wrong?
Since $u=x+y$ and $y=v$, the square $[0,2]\times [0,2]$ in the $xy$-plane is transformed in the parallelogram of vertices $(0,0)$, $(2,0)$, $(4,2)$ and $(2,2)$ in the $uv$-plane. Moreover the Jacobian is $1$. Hence \begin{align*}\int_{0}^{2}\int_{0}^{2}\lfloor x+y \rfloor dx \,dy &= \int_{u=0}^{2}\lfloor u\rfloor\left(\int_{v=0}^u dv\right)\,du+\int_{u=2}^{4}\lfloor u \rfloor\left(\int_{v=u-2}^2 dv\right)\,du\\ &= \int_{u=0}^{2}\lfloor u\rfloor u\,du+\int_{u=2}^{4}\lfloor u\rfloor (4-u)\,du\\ &= \int_{u=1}^{2}u\,du+2\int_{u=2}^{3}(4-u)\,du+3\int_{u=3}^{4}(4-u)\,du\\ &=\frac{3}{2}+3+\frac{3}{2}=6. \end{align*}