Answer
$\dfrac{1}{4} $ or, $=0.25$
Work Step by Step
Region of integration
$R=${$ (x,y) | -\sqrt {2-y^2} \lt x \leq -1, -1 \leq y \leq 1$}
conversion of rectangular coordinates into polar coordinates:
$x =r \cos \theta$ and $y=r \sin \theta$
Consider
$I=\dfrac{1}{(x^2+y^2)^2}=\dfrac{1}{[(r \cos \theta)^2+(r \sin \theta)^2]^2}=\dfrac{1}{(r^2)^2}=\dfrac{1}{r^4}$
Now,
$ \iint_{R} \dfrac{1}{(x^2+y^2)^2} \ dA=\int^{-1}_{1} \int_{-\sqrt {2-y^2}}^2{-1} \dfrac{1}{(x^2+y^2)^2} \ dx \ dy $
$=\int_{3\pi/4}^{5\pi/4} \int_{(-1/\cos \theta)}^{\sqrt 2} \dfrac{1}{r^3} dr \ d \theta $
$=-\dfrac{1}{2} \times \int_{3\pi/4}^{5\pi/4} (\dfrac{1}{2}-\cos^2 \theta) \ d \theta $
$=-\dfrac{1}{4} \times \int_{3\pi/4}^{5\pi/4} d \theta +\dfrac{1}{2} \int_{3\pi/4}^{5\pi/4} \dfrac{1+\cos 2 \theta}{2} \ d \theta $
$=-\dfrac{1}{4} \times \int_{3\pi/4}^{5\pi/4} d \theta +\dfrac{1}{4} \times \int_{3\pi/4}^{5\pi/4} d \theta +\dfrac{1}{4} \times \int_{3\pi/4}^{5\pi/4} \cos 2\theta \ d \theta$
$=\dfrac{1}{4} \times (\dfrac{1}{2}) [\sin 2 \theta]_{3\pi/4}^{5\pi/4}$
$=\dfrac{1}{8} \times [\sin (\dfrac{5 \pi}{2})-\sin (\dfrac{5 \pi}{2})] $
$=\dfrac{1}{4} $ or, $=0.25$
