<s>[INST] Evaluate : $\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{0}\frac{dxdy}{1+x^2+y^2}$ How to evaluate  $\displaystyle\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{0}\frac{dxdy}{1+x^2+y^2}$
This is my steps :
$\displaystyle\begin{align*}\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{0}\frac{dxdy}{1+x^2+y^2}&=\frac{1}{2}\int_{0}^{2\pi}\int_{-1}^{0}\frac{r}{1+r^2}drd\theta\end{align*}$
Is that right?
 [/INST] Notice that the domain is the half of the circle $x^2+y^2=1$ to the left of the y-axis and do a transformation to polar coordinates
$$\int_{-1}^1 \int_{-\sqrt{1-y^2}}^0 \frac{dxdy}{1+x^2+y^2}=
\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_0^1 \frac{rdrd\theta}{1 + r^2}=
\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} d\theta \int_0^1 \frac{rdr}{1+r^2}=
\pi \int_0^1 \frac{rdr}{1+r^2} 
$$
From trig substitution
$$
\pi \frac{1}{2} \ln(r^2+1)|_0^1=\frac{\pi}{2}\ln2
$$
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