# 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?

-
To understand the polar domain, draw a picture of the (x,y) domain of integration. – Did Nov 8 '12 at 6:28

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 $$\pi \frac{1}{2} \ln(r^2+1)|_0^1=\frac{\pi}{2}\ln2$$

-
Thanks for your help. May I ask one more question ?? Why do you consider the left of the y-axis ??? – cwk709394 Nov 8 '12 at 7:02
Draw a picture (third time). – Did Nov 8 '12 at 7:05
since x goes from $-\sqrt{1-y^2}$ to zero we know the domain is contained in the left half of the circle $x^2+y^2 = 1$. Since y goes from -1 to 1 we know it is the entire left half. – Tyler McAtee Nov 8 '12 at 7:34

Hint: change to polar coordinates.

-
I tried to change to polar coordinates!!But I don't know how to change the upper and lower limit – cwk709394 Nov 8 '12 at 6:26
Draw a picture. – Robert Israel Nov 8 '12 at 6:58