Answer
Please see the figure attached.
$\mathop \smallint \limits_1^2 \mathop \smallint \limits_0^{\sqrt {2x - {x^2}} } \frac{1}{{\sqrt {{x^2} + {y^2}} }}{\rm{d}}y{\rm{d}}x \simeq 0.533$
Work Step by Step
We have $\mathop \smallint \limits_1^2 \mathop \smallint \limits_0^{\sqrt {2x - {x^2}} } \frac{1}{{\sqrt {{x^2} + {y^2}} }}{\rm{d}}y{\rm{d}}x$.
From the order of the integration, we obtain the region of integration:
$1 \le x \le 2$, ${\ \ \ \ }$ $0 \le y \le \sqrt {2x - {x^2}} $
Write $f\left( {x,y} \right) = \frac{1}{{\sqrt {{x^2} + {y^2}} }}$. Using $x = r\cos \theta $ and $y = r\sin \theta $, in polar coordinates $f\left( {r\cos \theta ,r\sin \theta } \right) = \frac{1}{r}$.
We sketch the region and see that the region is part of a unit disk centered at $\left( {1,0} \right)$. Notice that this is the region depicted in Figure 8 in Example 2. To verify this, write:
$y = \sqrt {2x - {x^2}} $
${x^2} - 2x + {y^2} = 0$
${\left( {x - 1} \right)^2} + {y^2} - 1 = 0$
${\left( {x - 1} \right)^2} + {y^2} = 1$
Hence, the region is part of a unit disk centered at $\left( {1,0} \right)$.
Therefore, we can use the results obtained in Example 2, that is, the domain ${\cal D}$ has polar description:
${\cal D} = \left\{ {\left( {r,\theta } \right)|\sec \theta \le r \le 2\cos \theta ,0 \le \theta \le \frac{\pi }{4}} \right\}$
Using the Change of Variables Formula for ${\cal D}$, we evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_1^2 \mathop \smallint \limits_0^{\sqrt {2x - {x^2}} } \frac{1}{{\sqrt {{x^2} + {y^2}} }}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = 0}^{\pi /4} \mathop \smallint \limits_{r = \sec \theta }^{2\cos \theta } \left( {\frac{1}{r}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{\pi /4} \left( {r|_{\sec \theta }^{2\cos \theta }} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{\pi /4} \left( {2\cos \theta - \sec \theta } \right){\rm{d}}\theta $
Using the result in Example 6 of Section 8.2, that is
$\smallint \sec x{\rm{d}}x = \ln \left| {\sec x + \tan x} \right| + C$
we obtain
$\mathop \smallint \limits_{\theta = 0}^{\pi /4} \left( {2\cos \theta - \sec \theta } \right){\rm{d}}\theta = 2\sin \theta |_0^{\pi /4} - \left( {\ln \left| {\sec \theta + \tan \theta } \right|} \right)|_0^{\pi /4}$
$ = \sqrt 2 - \ln \left( {\sqrt 2 + 1} \right) \simeq 0.533$
Thus, $\mathop \smallint \limits_1^2 \mathop \smallint \limits_0^{\sqrt {2x - {x^2}} } \frac{1}{{\sqrt {{x^2} + {y^2}} }}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = 0}^{\pi /4} \mathop \smallint \limits_{r = \sec \theta }^{2\cos \theta } \left( {\frac{1}{r}} \right)r{\rm{d}}r{\rm{d}}\theta \simeq 0.533$.
