Answer
Using Theorem 4, we prove that
$\frac{4}{3} \le \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + {y^2}} \right){\rm{d}}A \le \frac{{20}}{3}$
Work Step by Step
We have the domain ${\cal D}$ bounded by $y = {x^2} + 1$ and $y=2$. We can consider ${\cal D}$ as a vertically simple region (please see the figure attached) whose description is given by
${\cal D} = \left\{ {\left( {x,y} \right)| - 1 \le x \le 1,{x^2} + 1 \le y \le 2} \right\}$
We notice that the integrand ${x^2} + {y^2}$ is the square distance ${d^2}$ from $\left( {x,y} \right) \in {\cal D}$ to the origin. From the figure attached we see that the minimum of ${d^2}$ occurs at $x=0$ and the maximum of ${d^2}$ occurs at $x=-1$ and $x=1$. Therefore, $1 \le {d^2} \le 5$. Thus,
$1 \le {x^2} + {y^2} \le 5$
By part (b) of Theorem 4,
$1\cdot Area\left( {\cal D} \right) \le \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + {y^2}} \right){\rm{d}}A \le 5\cdot Area\left( {\cal D} \right)$
Next, we evaluate the area of ${\cal D}$ as an iterated integral:
$Area\left( {\cal D} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A = \mathop \smallint \limits_{x = - 1}^1 \left( {\mathop \smallint \limits_{y = {x^2} + 1}^2 {\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( {y|_{{x^2} + 1}^2} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( {1 - {x^2}} \right){\rm{d}}x$
$ = \left( {\left( {x - \frac{1}{3}{x^3}} \right)|_{ - 1}^1} \right)$
$ = 1 - \frac{1}{3} + 1 - \frac{1}{3}$
$ = \frac{4}{3}$
Hence, $\frac{4}{3} \le \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + {y^2}} \right){\rm{d}}A \le \frac{{20}}{3}$.
