Answer
The average value of the square distance from the origin to a point in the domain ${\cal D}$ is $5.5143$.
Work Step by Step
Referring to Figure 31, the square distance from the origin to a point in the domain ${\cal D}$ is given by ${d^2} = {x^2} + {y^2}$, and we see that $1 \le d \le 3$.
We can consider the domain ${\cal D}$ as a horizontally simple region. From Figure 31, we see that the left and right boundaries are $x = {y^2} + 1$ and $x=3$, respectively. For the lower and upper boundaries, we find the intersection of the curve $x = {y^2} + 1$ and $x=3$ by solving the following equation:
$x = {y^2} + 1 = 3$
$y = \pm \sqrt 2 $
So, the lower and upper boundaries are $y = - \sqrt 2 $ and $y = \sqrt 2 $, respectively.
Thus, the domain description of ${\cal D}$ is
${\cal D} = \left\{ {\left( {x,y} \right)| - \sqrt 2 \le y \le \sqrt 2 ,{y^2} + 1 \le x \le 3} \right\}$
The average value of $f\left( {x,y} \right)$ on a domain ${\cal D}$ is given by Eq. (8):
$\bar f = \frac{{\mathop \smallint \nolimits_{}^{} \mathop \smallint \nolimits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A}}{{\mathop \smallint \nolimits_{}^{} \mathop \smallint \nolimits_{\cal D}^{} 1{\rm{d}}A}}$
Write $f\left( {x,y} \right) = {x^2} + {y^2}$. So, the average value of the square distance from the origin to a point in the domain ${\cal D}$ is
$\bar f = \frac{{\mathop \smallint \nolimits_{}^{} \mathop \smallint \nolimits_{\cal D}^{} \left( {{x^2} + {y^2}} \right){\rm{d}}A}}{{\mathop \smallint \nolimits_{}^{} \mathop \smallint \nolimits_{\cal D}^{} {\rm{d}}A}}$
1. Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + {y^2}} \right){\rm{d}}A$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} + {y^2}} \right){\rm{d}}A = \mathop \smallint \limits_{y = - \sqrt 2 }^{\sqrt 2 } \mathop \smallint \limits_{x = {y^2} + 1}^3 \left( {{x^2} + {y^2}} \right){\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - \sqrt 2 }^{\sqrt 2 } \left( {\left( {\frac{1}{3}{x^3} + {y^2}x} \right)|_{{y^2} + 1}^3} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = - \sqrt 2 }^{\sqrt 2 } \left( {9 + 3{y^2} - \frac{1}{3}{{\left( {{y^2} + 1} \right)}^3} - {y^2}\left( {{y^2} + 1} \right)} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = - \sqrt 2 }^{\sqrt 2 } \left( { - \frac{1}{3}{y^6} - 2{y^4} + {y^2} + \frac{{26}}{3}} \right){\rm{d}}y$
$ = \left( {\left( { - \frac{1}{{21}}{y^7} - \frac{2}{5}{y^5} + \frac{1}{3}{y^3} + \frac{{26}}{3}y} \right)|_{ - \sqrt 2 }^{\sqrt 2 }} \right)$
$ = - \frac{8}{{21}}\sqrt 2 - \frac{8}{5}\sqrt 2 + \frac{2}{3}\sqrt 2 + \frac{{26}}{3}\sqrt 2 - \frac{{16}}{{21}}\sqrt 2 - \frac{8}{5}\sqrt 2 + \frac{2}{3}\sqrt 2 + \frac{{26}}{3}\sqrt 2 $
$ = \frac{{1544\sqrt 2 }}{{105}}$
2. Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A = \mathop \smallint \limits_{y = - \sqrt 2 }^{\sqrt 2 } \mathop \smallint \limits_{x = {y^2} + 1}^3 {\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - \sqrt 2 }^{\sqrt 2 } \left( {x|_{{y^2} + 1}^3} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = - \sqrt 2 }^{\sqrt 2 } \left( {2 - {y^2}} \right){\rm{d}}y$
$ = \left( {\left( {2y - \frac{1}{3}{y^3}} \right)|_{ - \sqrt 2 }^{\sqrt 2 }} \right)$
$ = 2\sqrt 2 - \frac{2}{3}\sqrt 2 + 2\sqrt 2 - \frac{2}{3}\sqrt 2 $
$ = \frac{8}{3}\sqrt 2 $
Thus,
$\bar f = \frac{{\mathop \smallint \nolimits_{}^{} \mathop \smallint \nolimits_{\cal D}^{} \left( {{x^2} + {y^2}} \right){\rm{d}}A}}{{\mathop \smallint \nolimits_{}^{} \mathop \smallint \nolimits_{\cal D}^{} {\rm{d}}A}} = \frac{{1544\sqrt 2 }}{{105}}\cdot\frac{3}{{8\sqrt 2 }} = \frac{{193}}{{35}} \simeq 5.5143$
So, the average value of the square distance from the origin to a point in the domain ${\cal D}$ is $5.5143$.
