Answer
The area of each subdomain is $Area\left( {{{\cal D}_j}} \right) = \frac{\pi }{{12}}$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \approx 3.2201$
Work Step by Step
Using Figure 33, we list the values of $f\left( {{P_j}} \right)$ in the following table:
$\begin{array}{*{20}{c}}
j&1&2&3&4&5&6\\
{f\left( {{P_j}} \right)}&{2.5}&{2.4}&{2.2}&2&{1.7}&{1.5}
\end{array}$
Since the radius of the circle is $5$, so the the length of a subdomain is $l = 5\cdot\Delta \theta = \frac{{5\pi }}{{12}}$. From Figure 33 we see that the width of a subdomain is $0.2$. So, the area of each subdomain is $Area\left( {{{\cal D}_j}} \right) = \frac{{5\pi }}{{12}}\cdot0.2 = \frac{\pi }{{12}}$.
Using Eq. (11) and the values in the given table, we estimate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \approx \mathop \sum \limits_6^{j = 1} f\left( {{P_j}} \right)Area\left( {{{\cal D}_j}} \right)$
$ = \frac{\pi }{{12}}\left( {2.5 + 2.4 + 2.2 + 2 + 1.7 + 1.5} \right)$
$ = 3.2201$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \approx 3.2201$.
