Answer
Using Theorem 4, we prove that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \le \frac{3}{2}$.
Work Step by Step
We have $f\left( {x,y} \right) = \sqrt {{x^3} + 1} $ and the domain ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 2, - \frac{1}{8} \le y \le \frac{1}{8}} \right\}$.
Notice that $f\left( {x,y} \right) \le 3$ for all $\left( {x,y} \right) \in {\cal D}$.
By part (a) of Theorem 4,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \le 3\cdot Area\left( {\cal D} \right)$
Since ${\cal D}$ is the rectangle $0 \le x \le 2$, $ - \frac{1}{8} \le y \le \frac{1}{8}$, the area of ${\cal D}$ is $Area\left( {\cal D} \right) = 2\cdot\frac{1}{4} = \frac{1}{2}$.
Hence, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \le \frac{3}{2}$.
