Determine the domain of $D=\{(x,y) \in \Bbb{R}^2 |x\in [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}],y\in[|x|,\sqrt{1-x^2}]\}$ 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?
Answer
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 $(\pi/4) \log{2}$.
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