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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?

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    $\begingroup$ First, don't use the same letter $D$ for the original domain and the transformed domain in polars. What domain you don't know how to determine? Start drawing. $\endgroup$ Commented Feb 13, 2014 at 9:09
  • $\begingroup$ @Martín-BlasPérezPinilla: good advice to start drawing. But what on earth do you mean about the transformed domain? It is the same domain, just expressed in polars. $\endgroup$
    – Ron Gordon
    Commented Feb 13, 2014 at 9:16
  • $\begingroup$ The set of pairs $(r,\phi)$ s.t. $(r\cos\phi,r\sin\phi)\in D$ is $\ne D$. $\endgroup$ Commented Feb 13, 2014 at 9:19
  • $\begingroup$ You can also view $x$, $y$, $r$, $\theta$ as functions defined over $D$, in which case the notation is fine. $\endgroup$
    – apt1002
    Commented Feb 13, 2014 at 9:21
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    $\begingroup$ @Martín-BlasPérezPinilla: $D$ is the geometrical region defined by those points. It doesn't matter how you describe them. You are mapping those points to a rectangular region in another cartesian system. But we do not do this when, for example, we perform polar plots. Along these lines, the plot of $r=\sin{\phi}$ is the same as that of $x^2+(y-1)^2=1$. $\endgroup$
    – Ron Gordon
    Commented Feb 13, 2014 at 9:22

1 Answer 1

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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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  • $\begingroup$ I think i'm starting to "get" it :D [wolframalpha.com/input/… is what you meant with the drawing? $\endgroup$
    – ELEC
    Commented Feb 13, 2014 at 9:36
  • $\begingroup$ so $\phi =[\pi/4, 3\pi/4]$ ? $\endgroup$
    – ELEC
    Commented Feb 13, 2014 at 9:42
  • $\begingroup$ Yes, that's correct. =) $\endgroup$ Commented Feb 13, 2014 at 9:45
  • $\begingroup$ The area is basicly $1/4$ of a circle? could I use $\phi = [0,\pi/2]$? $\endgroup$
    – ELEC
    Commented Feb 13, 2014 at 9:50

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