Answer
The volume of the mound $V \simeq 0.19376$.
Work Step by Step
The Riemann sum ${S_{4,3}}$ implies that $N=4$ and $M=3$. From the table we see that $\Delta x = \Delta y = \frac{1}{4}$. And also, $x$ and $y$ start from the origin. Denote the $x$-interval by $\left[ {0,b} \right]$ and $y$-interval by $\left[ {0,d} \right]$. So,
$\Delta x = \frac{1}{4} = \frac{{b - 0}}{N}$
$b=1$
$\Delta y = \frac{1}{4} = \frac{{d - 0}}{M}$
$d = \frac{3}{4}$
So, the domain of the Riemann sum is ${\cal R} = \left[ {0,1} \right] \times \left[ {0,\frac{3}{4}} \right]$.
The Riemann sum ${S_{4,3}}$ to estimate the volume of the mound is given by
${S_{4,3}} = \mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^3 f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = \frac{1}{{16}}\mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^3 f\left( {{P_{ij}}} \right)$,
where $y = f\left( x \right)$.
Step 1. Evaluate ${S_{4,3}}$ using lower-left vertices as sample points
$S_{4,3}^{lower}$
$= \frac{1}{{16}}(f\left( {0,0} \right) + f\left( {0.25,0} \right) + f\left( {0.5,0} \right) + f\left( {0.75,0} \right)$
$ + f\left( {0,0.25} \right) + f\left( {0.25,0.25} \right) + f\left( {0.5,0.25} \right) + f\left( {0.75,0.25} \right)$
$ + f\left( {0,0.5} \right) + f\left( {0.25,0.5} \right) + f\left( {0.5,0.5} \right) + f\left( {0.75,0.5} \right))$
$ = \frac{1}{{16}}(0.1 + 0.15 + 0.2 + 0.15$
$\ \ \ $ $ + 0.15 + 0.2 + 0.4 + 0.3$
$\ \ \ $ $ + 0.2 + 0.3 + 0.5 + 0.4)$
$S_{4,3}^{lower} \simeq 0.19063$
Step 2. Evaluate ${S_{4,3}}$ using upper-right vertices as sample points
$S_{4,3}^{upper}$
$= \frac{1}{{16}}(f\left( {0.25,0.25} \right) + f\left( {0.5,0.25} \right) + f\left( {0.75,0.25} \right) + f\left( {1,0.25} \right)$
$ + f\left( {0.25,0.5} \right) + f\left( {0.5,0.5} \right) + f\left( {0.75,0.5} \right) + f\left( {1,0.5} \right)$
$ + f\left( {0.25,0.75} \right) + f\left( {0.5,0.75} \right) + f\left( {0.75,0.75} \right) + f\left( {1,0.75} \right))$
$ = \frac{1}{{16}}(0.2 + 0.4 + 0.3 + 0.2$
$\ \ \ $ $ + 0.3 + 0.5 + 0.4 + 0.2$
$\ \ \ $ $ + 0.2 + 0.2 + 0.15 + 0.1)$
$S_{4,3}^{upper} \simeq 0.19688$
From these results, we compute the average of the two Riemann sums ${S_{4,3}}$ with lower-left and upper-right vertices:
$\overline {{S_{4,3}}} = \frac{{S_{4,3}^{lower} + S_{4,3}^{upper}}}{2} \simeq \frac{{0.19063 + 0.19688}}{2} \simeq 0.19376$
So, the volume of the mound $V \simeq 0.19376$.
