## Thomas' Calculus 13th Edition

$\dfrac{1}{4}$ or, $0.25$
The region of integration can be expressed as: $R=${$(x,y) | -\sqrt {2-y^2} \lt x \leq -1, -1 \leq y \leq 1$} To convert rectangular coordinates into polar coordinates, we will use: $x =r \cos \theta$ and $y=r \sin \theta$ The function we want to integrate can be written in polar coordinates as: $\dfrac{1}{(x^2+y^2)^2}=\dfrac{1}{[(r \cos \theta)^2+(r \sin \theta)^2]^2}=\dfrac{1}{(r^2)^2}=\dfrac{1}{r^4}$ $\iint_{R} \dfrac{1}{(x^2+y^2)^2} \ dA=\int^{-1}_{1} \int_{-\sqrt {2-y^2}}^2{-1} \dfrac{1}{(x^2+y^2)^2} \ dx \ dy \\=\int_{3\pi/4}^{5\pi/4} \int_{(-1/\cos \theta)}^{\sqrt 2} \dfrac{1}{r^3} dr \ d \theta \\=-\dfrac{1}{2} \int_{3\pi/4}^{5\pi/4} (\dfrac{1}{2}-\cos^2 \theta) \ d \theta \\=-\dfrac{1}{4} \int_{3\pi/4}^{5\pi/4} d \theta +\dfrac{1}{2} \int_{3\pi/4}^{5\pi/4} \dfrac{1+\cos 2 \theta}{2} \ d \theta \\=-\dfrac{1}{4} \int_{3\pi/4}^{5\pi/4} d \theta +\dfrac{1}{4} \int_{3\pi/4}^{5\pi/4} d \theta +\dfrac{1}{4} \int_{3\pi/4}^{5\pi/4} \cos 2\theta \ d \theta \\=\dfrac{1}{4} [\dfrac{\sin 2 \theta}{2}]_{3\pi/4}^{5\pi/4}\\=\dfrac{1}{8} [\sin (\dfrac{5 \pi}{2})-\sin (\dfrac{5 \pi}{2})] \\=\dfrac{1}{4} \\=0.25$