Double Integral Over Region Common to Two Circles Evaluate $\iint\frac{(x^2+y^2)^2}{x^2y^2} dx dy$ over the region common to the circles $x^2+y^2=7x$ and $x^2+y^2=11y$.
 A: You may use the Cylindrical coordinates to find the value of the integrals. This, as @Ron's Caesarian's approach, will get a bit messy but it works as well. We have two circles on $z=0$ intersected each other, so $$7r\cos\theta=7x=x^2+y^2=11y=11r\sin\theta\longrightarrow\theta_{0}=\tan^{-1}\left(\frac{7}{11}\right)\cong0.56$$ Now we have $$I=\int_{\theta=0}^{\theta_0}\int_{r=0}^{11\sin\theta}\frac{r^4}{r^4\cos^2\theta\sin^2\theta}rdrd\theta+\int_{\theta=\theta_0}^{\pi/2}\int_{r=7\cos\theta}^{0}\frac{r^4}{r^4\cos^2\theta\sin^2\theta}rdrd\theta$$

A: The intersection region is a football-shaped region symmetric about the line $y=7/11 x$.  When you do out the algebra to locate the circles' center and radii, you may derive the integration limits in Cartesians.  In this case, the integral takes the form
$$\int_0^{x_0} \frac{dx}{x^2} \, \int_{y_-(x)}^{y_+(x)} dy \frac{(x^2+y^2)^2}{y^2}$$
where $x_0$ is the rightmost intersection point $x$ coordinate, whch may be found to take the value 
$$x_0=\frac{7 \cdot 11^2}{7^2+11^2} = \frac{847}{170}$$
$y_-$ and $y_+$ define the boundaries of the two circles:
$$y_-(x) = \frac{11}{2}-\sqrt{\frac{121}{4}-x^2}$$
$$y_+(x) = \sqrt{\frac{49}{4}-\left(\frac{7}{2}-x\right)^2}$$
Expanding the integrals, one gets a sum of three single integrals:
$$\int_0^{x_0} dx \, x^2 \left (\frac{1}{y_-(x)} - \frac{1}{y_+(x)}\right ) + 2\int_0^{x_0} dx \, [y_+(x)-y_-(x)] + \\\frac13 \int_0^{x_0} dx \frac{1}{x^2} \left [ y_+(x)^3-y_-(x)^3\right]$$
I will stop here, noting that the evaluation of these integrals, while messy, should be straightforward.  I should note that each of these integrals has been done in Wolfram Alpha (for example, here's one), and are doable through trig substitutions.
