Looks like the integral is
[tex]\displaystyle\iint_R2(x+1)y^2\,\mathrm dA[/tex]
where [tex]R=[0,1]\times[0,3][/tex]. (The inclusion of [tex]y=3[/tex] will have no effect on the value of the integral.)
Let's split up [tex]R[/tex] into [tex]mn[/tex] equally-sized rectangular subintervals, and use the bottom-left vertices of each rectangle to approximate the integral. The intervals will be partitioned as
[tex][0,1]=\left[0,\dfrac1m\right]\cup\left[\dfrac1m,\dfrac2m\right]\cup\cdots\cup\left[\dfrac{m-1}m,1\right][/tex]
and
[tex][0,3]=\left[0,\dfrac3n\right]\cup\left[\dfrac3n,\dfrac6n\right]\cup\cdots\cup\left[\dfrac{3(n-1)}n,3\right][/tex]
where the bottom-left vertices of each rectangle are given by the sequence
[tex]v_{i,j}=\left(\dfrac{i-1}n,\dfrac{3(j-1)}n\right)[/tex]
with [tex]1\le i\le m[/tex] and [tex]1\le j\le n[/tex]. Then the Riemann sum is
[tex]\displaystyle\lim_{m\to\infty,n\to\infty}\sum_{i=1}^m\sum_{j=1}^nf(v_{i,j})\frac{1-0}m\frac{3-0}n[/tex]
[tex]\displaystyle=\lim_{m\to\infty,n\to\infty}\frac3{mn}\sum_{i=1}^m\sum_{j=1}^n\frac{18}{mn^2}(j-1)^2(i-1+m)[/tex]
[tex]\displaystyle=\lim_{m\to\infty,n\to\infty}\frac{54}{m^2n^3}\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}j^2(i+m)[/tex]
[tex]\displaystyle=\frac92\lim_{m\to\infty,n\to\infty}\frac{(3m-1)(2n^3-3n^2+n)}{mn^3}[/tex]
[tex]\displaystyle=\frac92\left(\lim_{m\to\infty}\frac{3m-1}m\right)\left(\lim_{n\to\infty}\frac{2n^3-3n^2+n}{n^3}\right)=\boxed{27}[/tex]
