a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Riemann sum with m = n = 2 and choose the sample points to be lower right corners.

In the [tex]x[/tex] direction we consider the [tex]m=2[/tex] subintervals [0, 1] and [1, 2] (each with length 1), while in the [tex]y[/tex] direction we consider the [tex]n=2[/tex] subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of [tex]R[/tex] are (1, 0), (2, 0), (1, 2), (2, 2).Let [tex]f(x,y)=7x+5y^2[/tex]. The volume of the solid is approximately[tex]\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}[/tex]###More generally, the lower-right-corner Riemann sum over [tex]m=\mu[/tex] and [tex]n=\nu[/tex] subintervals would be[tex]\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)[/tex]Then taking the limits as [tex]\mu\to\infty[/tex] and [tex]\nu\to\infty[/tex] leaves us with an exact volume of [tex]\dfrac{808}3[/tex].