Answer
$$\pi $$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$\int^1_{-1} \int^{\sqrt{1-x^2}}_{-\sqrt{1-x^2}}\dfrac{2}{(1+x^2+y^2)} \space dy \space dx =\int^{\pi}_0 \int^1_0 \dfrac{2r}{1+(r \cos \theta )^2+(r \sin \theta )^2} \space dr \space d\theta $
or, $= \int^{\pi}_0 \int^1_0 \dfrac{2r}{(1+r^2)^2}dr d\theta $
or, =4 \times $\int^{\pi/2}_0 [\dfrac{-1}{1+r^2}]^1_0d\theta $
or, $=2 \times \int^{\pi/2}_0 d\theta $
or, $=\pi $
