Writing the integral in polar coordinates is less taxing:
[tex]\displaystyle\iint_Rf(x,y)\,\mathrm dA=\iiint_Rrf(r\cos\theta,r\sin\theta)\,\mathrm dr\,\mathrm d\theta[/tex]
The region R is captured by the set of points (in polar, of course)
[tex]R=\left\{(r,\theta)\mid1\le r\le3\land\dfrac{5\pi}4\le\theta\le2\pi\right\}[/tex]
so that the integral is
[tex]\displaystyle\int_{5\pi/4}^{2\pi}\int_1^3rf(r\cos\theta,r\sin\theta)\,\mathrm dr\,\mathrm d\theta[/tex]
In rectangular coordinates, we have to split up the region so that
[tex]R=R_1\cup R_2\cup R_3[/tex]
where
[tex]R_1=\left\{(x,y)\mid-\dfrac3{\sqrt2}\le x\le-\dfrac1{\sqrt2}\land-\sqrt{9-x^2}\le y\le x\right\}[/tex]
[tex]R_2=\left\{(x,y)\mid-\dfrac1{\sqrt2}\le x\le1\land-\sqrt{9-x^2}\le y\le -\sqrt{1-x^2}\right\}[/tex]
[tex]R_3=\left\{(x,y)\mid1\le x\le3\land-\sqrt{9-x^2}\le y\le0\right\}[/tex]
and the integral is split into 3 over these regions,
[tex]\displaystyle\iint_Rf(x,y)\,\mathrm dA=\sum_{n=1}^3\iint_{R_n}f(x,y)\,\mathrm dx\,\mathrm dy[/tex]
