Answer
Using Theorem 4, we prove that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{{{\rm{d}}A}}{{4 + {x^2} + {y^2}}} \le \pi $
Work Step by Step
We have the double integral $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{{{\rm{d}}A}}{{4 + {x^2} + {y^2}}}$, where ${\cal D}$ is the disk ${x^2} + {y^2} \le 4$.
Since ${x^2} \ge 0$ and ${y^2} \ge 0$, for all $\left( {x,y} \right) \in {\cal D}$, the following inequality holds:
$4 \le 4 + {x^2} + {y^2} \le 8$
Taking reciprocals, we get
$\frac{1}{8} \le \frac{1}{{4 + {x^2} + {y^2}}} \le \frac{1}{4}$
By part (b) of Theorem 4,
$\frac{1}{8}\cdot Area\left( {\cal D} \right) \le \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{{{\rm{d}}A}}{{4 + {x^2} + {y^2}}} \le \frac{1}{4}\cdot Area\left( {\cal D} \right)$
Since the domain ${\cal D}$ is a disk of radius $2$, its area is $\pi \cdot{2^2}$. So, $Area\left( {\cal D} \right) = 4\pi $.
Hence, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{{{\rm{d}}A}}{{4 + {x^2} + {y^2}}} \le \pi $.
