Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A = \frac{1}{4}$
Work Step by Step
We sketch the graph of the region in the first quadrant bounded by the spiral $r = \theta $, the circle $r=1$, and the $x$-axis.
From the figure attached, we see that the ray from the origin passes through $r = \theta $ and ends at $r=1$. So, the range of $r$ is $\theta \le r \le 1$.
To find the range of $\theta $, we find the ray where the spiral $r = \theta $ intersects the circle $r=1$:
$r = 1 = \theta $
The intersection is at $\theta = 1$. So, $\theta $ varies from $0$ to $1$.
Thus, the description of ${\cal D}$:
${\cal D} = \left\{ {\left( {r,\theta } \right)|\theta \le r \le 1,0 \le \theta \le 1} \right\}$
Using $\sqrt {{x^2} + {y^2}} = r$, we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A$ in polar coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^1 \mathop \smallint \limits_{r = \theta }^1 {r^2}{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^1 \left( {{r^3}|_\theta ^1} \right){\rm{d}}\theta $
$ = \frac{1}{3}\mathop \smallint \limits_{\theta = 0}^1 \left( {1 - {\theta ^3}} \right){\rm{d}}\theta $
$ = \frac{1}{3}\left( {\left( {\theta - \frac{1}{4}{\theta ^4}} \right)|_0^1} \right) = \frac{1}{3}\left( {1 - \frac{1}{4}} \right) = \frac{1}{4}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}A = \frac{1}{4}$.
