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I have $$\iint_A \frac{1}{(x^2+y^2)^2}\,dx\,dy.$$

$A$ is bounded by the conditions $x^2 + y^2 \leq 1$ and $x+y \geq 1$.

I initially thought to make the switch the polar coordinates, but the line $x+y=1$ is making it hard to find the limits of integration.

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maybe you could rotate your domain (since the function is invariant with respect to rotations) and compute the integral on the normal region $x^2+y^2\le 1$, $y\ge\sqrt 2 / 2$ – Emanuele Paolini Feb 20 '13 at 17:36
maybe you want something like this ... but not sure though – Santosh Linkha Feb 20 '13 at 17:46
You should plot the region to see what's going on. – Mhenni Benghorbal Feb 20 '13 at 19:17
up vote 1 down vote accepted

The integral $Q$ remains unchanged when we replace $A$ by $$A':=\left\{(x,y)\ \bigg|\ x\geq{\sqrt{2}\over 2}, \ x^2+y^2\leq 1\right\}\ .$$ Changing to polar coordinates we get $$Q=\int_{A'}{1\over (x^2+y^2)^2}\ {\rm d}(x,y)=\int_{-\pi/4}^{\pi/4} \int_{1/(\sqrt{2}\cos\phi)}^1 {1\over r^4}\> r\> dr\ d\phi\ .$$ The inner integral computes to $${-1\over 2r^2}\biggr|_{1/(\sqrt{2}\cos\phi)}^1=\cos^2\phi-{1\over2}={1\over2}\cos(2\phi)\ .$$ It follows that $$Q={1\over2}\int_{-\pi/4}^{\pi/4} \cos(2\phi)\ d\phi={1\over4}\sin(2\phi)\biggr|_{-\pi/4}^{\pi/4}={1\over2}\ .$$

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$$ x+y = r(\cos\theta+\sin\theta) = r\sqrt{2}\sin\left(\theta+\frac\pi2\right). $$ So $$ \iint_A \frac{dx\,dy}{(x^2+y^2)^2} = \int_0^{\pi/2} \int_{(\csc(\theta+\pi/4))/\sqrt{2}}^1 \frac{r\,dr \, d\theta}{r^4}. $$ $$ = \int_0^{\pi/2} \left(\int_{(\csc(\theta+\pi/4))/\sqrt{2}}^1\ \frac{dr}{r^3} \right) \, d\theta $$

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If you rotate your domain by $45$ degrees your line becomes $y=\sqrt{2} / 2$ and the integrand is invariant, so you get (also use the simmetry with respect to $x=0$): $$ \iint_A \frac{1}{(x^2+y^2)^2} = 2\int_0^{\sqrt 2/2}\,dx\int_{\sqrt 2/2}^{\sqrt{1-x^2}} \frac{dy}{(x^2+y^2)} $$

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Here is another answer

$$\iint_A \frac{1}{(x^2+y^2)^2}\,dx\,dy=\int_{0}^{1}\int_{1-x}^{\sqrt{1-x^2}} \frac{1}{(x^2+y^2)^2}\,dy\,dx .$$

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