Double integrals in polar coordinates Determine the domain of
$$\!\!\!\!\!\!\!{\small
D \equiv
\left\{\left(x,y\right) \in \mathbb{R}^{2}\
{\large\mid}\
x \in \left[\,{-\,\frac{1}{\,\sqrt{\,{2}\,}\,},
\frac{1}{\,\sqrt{\,{2}\,}\,}}\,\right],\
y \in \left[\,{\left\vert\,{x}\,\right\vert,
\,\sqrt{\,{1 - x^{2}}\,}\,}\,\right]\right\}}
$$
in polar coordinates and draw it.
Also how would you integrate $$\int\int_D \frac{1}{1+x^2 + y^2}dA$$ which is i guess
$$\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{|x|}^{\sqrt{1-x^2}} \frac{1}{1+x^2 + y^2}dydx$$
I guess in the integral you can use the polar coordinates
$$\int\int_D \frac{1}{1+r^2\cos^2(\phi) + r^2\sin^2(\phi)}rdrd\phi$$
$$\int\int_D \frac{r}{1+r^2}drd\phi$$
$$\int_{\frac{1}{4\pi}}^{\frac{3}{4\pi}}\int_0^1 \frac{r}{1+r^2}drd\phi=\int_{\frac{1}{4\pi}}^{\frac{3}{4\pi}}\left(\frac 12 \ln(1+1^2)-\frac 12\ln(1+0^2) \right)d\phi$$
$$\int_{\frac{1}{4\pi}}^{\frac{3}{4\pi}}\frac{\ln{2}}{2}d\phi=\left(\frac{3}{4\pi}\frac{\ln{2}}{2}-\frac{1}{4\pi}\frac{\ln{2}}{2} \right)=\frac{\pi}{4}\ln{2}$$
Did I get it right?

Draw a picture.  A simple plot reveals that the domain $D$ is simply the sector of the circle $r=1$ between two values of $\theta$.  A little thought provides those values of $\theta$ (i.e., what purpose does the absolute value serve?).
The integrand you show is also wrong, as $1+r^2 \ne 2$.
The answer I get is $(\pi/4) \log{2}$.
