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convert the double integral

$$\int\limits_{-1}^1\int\limits_{-\sqrt{1-y^2}}^0\frac{1}{(1+x^2+y^2)^2}dxdy$$ to polar coordinates and then evaluate.

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  • $\begingroup$ That 2 at the end of equation should be^2 not 2. $\endgroup$ – Michael Rametta Mar 21 '13 at 23:06
  • $\begingroup$ What keeps you from doing it? Where are you stuck? $\endgroup$ – Julien Mar 21 '13 at 23:14
  • $\begingroup$ Oops! My answer's ready but I deleted it waiting for the OP to show some self work... $\endgroup$ – DonAntonio Mar 21 '13 at 23:16
  • $\begingroup$ Or, on the other side, wait until someone else posts the complete solution... $\endgroup$ – DonAntonio Mar 22 '13 at 6:42
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Draw a picture of the integration region. You will see that the region is the left half of the unit circle centered at the origin. Then conversion to polars is straightforward; the integral is equal to:

$$\begin{align}\int_{\pi/2}^{3 \pi/2} d\theta \: \int_o^1 dr \: \frac{r}{(1+r^2)^2}&= \pi \frac{1}{2} \int_0^1 \frac{du}{(1+u)^2} = \frac{\pi}{4}\end{align}$$

where the substitution $u=r^2$ was effected in the 2nd integral.

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$$x=r\cos t\;,\;\;y=r\sin t$$

$$\begin{cases}-1\le y\le 1\\{}\\-\sqrt{1-y^2}\le x\le 0\end{cases}\Longrightarrow -\frac{\pi}{2}\le t\le \frac{\pi}{2}\;,\;\;0\le r\le 1$$

Thus,

$$\int\limits_{-1}^1\int\limits_{-\sqrt{1-y^2}}^0\frac{1}{(1+x^2+y^2)^2}dxdy=\int\limits_0^1\int\limits_{-\pi/2}^{\pi/2}\frac{r}{(1+r^2)^2}dt\,dr=$$

$$\frac{\pi}{2}\int\limits_0^1\frac{d(1+r^2)}{(1+r^2)^2}dr=\left.-\frac{\pi}{2}\frac{1}{1+r^2}\right|_0^1=\frac{\pi}{4}$$

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Well, your first step is to determine the region. Start by drawing the boundary curves $x=-\sqrt{1-y^2}$ and $x=0$ between $y=\pm1$. How can we describe this region in polar coordinates?

Next, you'll want to translate the rectangular coordinates to polar coordinates, using $x=r\cos\theta,$ $y=r\sin\theta,$ and $dx\,dy=r\,d\theta\,dr.$

Your new integrand shouldn't have any $\theta$ terms in it, so you can immediately integrate with respect to $\theta$, and a quick $u$-substitution will change the resulting integral with respect to $r$ into a friendlier form.

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