Answer
Choice 1. sample points $ \bullet $ (there are 6):
${S_{3,4}} = - 3$
Choice 2. sample points $ \circ $ (there are 8)
${S_{3,4}} = - 4$
The sample points $ \circ $ is a better approximation to the integral of $f$ over ${\cal D}$ because there are more points in the domain.
Work Step by Step
We have $f\left( {x,y} \right) = x - y$.
From Figure 21, we see that $\Delta {x_j} = \Delta {y_j} = 1$.
We calculate the Riemann sum for $f\left( {x,y} \right) = x - y$ using two choices:
Choice 1. sample points $ \bullet $ (there are 6)
Using Figure 21, we list the values in the following table:
$\begin{array}{*{20}{c}}
{{P_j} = \left( {{x_j},{y_j}} \right)}&{\left( {1,1} \right)}&{\left( {2,1} \right)}&{\left( {1,2} \right)}&{\left( {2,2} \right)}&{\left( {1,3} \right)}&{\left( {2,3} \right)}\\
{f\left( {{x_j},{y_j}} \right)}&0&1&{ - 1}&0&{ - 2}&{ - 1}
\end{array}$
From Figure 21, we see that $\Delta {A_j} = 1$. So, the Riemann sum for $f\left( {x,y} \right)$ is
${S_{3,4}} = \mathop \sum \limits_{j = 1}^6 f\left( {{P_j}} \right)\Delta {A_j} = 0 + 1 - 1 + 0 - 2 - 1 = - 3$
Choice 2. sample points $ \circ $ (there are 8)
Using Figure 21, we list the values in the following table:
$\begin{array}{*{20}{c}}
{{P_j} = \left( {{x_j},{y_j}} \right)}&{\left( {\frac{3}{2},\frac{1}{2}} \right)}&{\left( {\frac{1}{2},\frac{3}{2}} \right)}&{\left( {\frac{3}{2},\frac{3}{2}} \right)}&{\left( {\frac{5}{2},\frac{3}{2}} \right)}&{\left( {\frac{1}{2},\frac{5}{2}} \right)}&{\left( {\frac{3}{2},\frac{5}{2}} \right)}&{\left( {\frac{5}{2},\frac{5}{2}} \right)}&{\left( {\frac{3}{2},\frac{7}{2}} \right)}\\
{f\left( {{x_j},{y_j}} \right)}&1&{ - 1}&0&1&{ - 2}&{ - 1}&0&{ - 2}
\end{array}$
Since $\Delta {A_j} = 1$. So, the Riemann sum for $f\left( {x,y} \right)$ is
${S_{3,4}} = \mathop \sum \limits_{j = 1}^8 f\left( {{P_j}} \right)\Delta {A_j} = 1 - 1 + 0 + 1 - 2 - 1 + 0 - 2 = - 4$
The sample points $ \circ $ is a better approximation to the integral of $f$ over ${\cal D}$ because there are more points in the domain.
