Answer
The correct answer is (b) $4\pi $
Work Step by Step
We have $f\left( {x,y} \right) \le 4$ for all $\left( {x,y} \right) \in {\cal D}$.
By part (a) of Theorem 4:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \le \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} 4{\rm{d}}A = 4\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A$
Since ${\cal D}$ is an unit disk, the radius is $1$. So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A = \pi $. Thus,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \le 4\pi $
Therefore, the largest possible value of $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A$ is $4\pi $. So, the correct answer is (b).
