The domain of integration is a circular sector subtended by an angle of 3π/4 radians, belonging to a circle of radius 3.
The first integral is taken over the left half of the upper semicircle. [tex]\left(\frac\pi2 \le \theta \le \pi\right)[/tex]
The second integral is taken over a sector belonging to the right half of the same semicircle. Notice that the line [tex]y=x[/tex] meets the semicircle [tex]y=\sqrt{9-x^2}[/tex] when
[tex]x = \sqrt{9-x^2} \implies x^2 = 9-x^2 \implies x = \dfrac3{\sqrt2}[/tex]
and the line [tex]y=x[/tex] makes an angle of π/4 with the positive [tex]x[/tex]-axis. [tex]\left(\frac\pi4 \le \theta \le \frac\pi2\right)[/tex]
Then the two integrals combine into one integral in polar coordinates, and
[tex]\displaystyle \int_{-3}^0 \int_0^{\sqrt{9-x^2}} \sqrt{x^2+y^2} \, dy \, dx + \int_0^{3/\sqrt2} \int_0^{\sqrt{9-x^2}} \sqrt{x^2+y^2} \, dy \, dx \\\\ ~~~~~~~~ = \int_{\pi/4}^\pi \int_0^3 \sqrt{r^2}\,r\,dr\,d\theta \\\\ ~~~~~~~~ = \frac{3\pi}4 \int_0^3 r^2 \, dr = \boxed{\frac{27\pi}4}[/tex]
