# Double integral over complicated region

Suppose we wanted to compute $\iint\frac {1}{1 + x^2 + y^2} dxdy$ over the region $\frac {(x-1)^2}4 + \frac {(y+2)^2}9 \leqslant 1$.

It gets quite hairy if we use elliptical polar coordinates i.e. $(x,y) = (1 + 2rcos \theta, -2 + 3r\sin \theta )$, so say we insist on letting $x^2 + y^2 = r^2$ with $(x,y) = (rcos \theta, r \sin \theta)$ in order to make the integrand simple.

How can we then use our insisted change of coordinates $(x,y) = ( r \cos \theta, r \sin \theta )$ to describe the region $\frac {(x-1)^2}4 + \frac {(y+2)^2}9 \leqslant 1$ and actually carry out the integration?

The example function chosen, $\frac {1}{1 + x^2 + y^2}$, may perhaps been a bad one but hopefully the point comes across.

• As you have stated the problem, your integral is equal to 0 since the region of integration is a curve which has measure zero. – Brandon Jul 10 '14 at 4:48
• you are asking how to find a change of coordinate that reduces the amount of work beyond just fitting the region of integration with elliptical polar coordinates? – frogeyedpeas Jul 10 '14 at 4:55
• @Brandon: I forget the less than or equal to sign, fixed now. – thelionkingrafiki Jul 10 '14 at 5:00
• @frogeyedpeas: We insist with polar coordinates instead of $(x,y) = (1 + 2rcos \theta, -2 + 3r\sin \theta )$ in order to make the integrand simpler. But how can we now find the boundaries for $r$ and $theta$ over $(x,y) = (1 + 2rcos \theta, -2 + 3r\sin \theta )$ if we let $(x,y) = (r \cos \theta, r \sin \theta )$? – thelionkingrafiki Jul 10 '14 at 5:01
• why would you want polar coordinates, knowing that it will introduce new problems regarding integral regions? Usually you want change of coordinates so to simplify the region you are integrating, not to simplify integrand itself. I doubt integral region would have any simple form knowing that original region was elliptical.. – user160738 Jul 10 '14 at 5:18

It is worth to calculate the integral regarding the ellipse as a normal domain, from which: $$\begin{eqnarray*} I &=& \int_{-1}^{3}\int_{-2-3\sqrt{1-\left(\frac{x-1}{2}\right)^2}}^{-2+3\sqrt{1-\left(\frac{x-1}{2}\right)^2}}\frac{1}{1+x^2+y^2}\,dy\,dx.\end{eqnarray*}$$ Now exploiting the fact that $\int\frac{dy}{1+x^2+y^2}=\frac{1}{\sqrt{1+x^2}}\arctan\left(\frac{y}{\sqrt{1+x^2}}\right)$ and the identity: $$\arctan(a)+\arctan(b)=\arctan\left(\frac{a+b}{1-ab}\right)$$ we have: $$I = \int_{-1}^{3}\frac{1}{\sqrt{1+x^2}}\arctan\left(\frac{12\sqrt{(1+x^2)(3-x)(x+1)}}{13x^2-18x-7}\right)\,dx$$ and by setting $x=\sinh t$ we arrive at: $$I = \int_{\log(\sqrt{2}-1)}^{\log(3+\sqrt{10})}\arctan\left(\frac{12\cosh t\sqrt{(3-\sinh t)(1+\sinh t)}}{13\sinh^2 t-18\sinh t-7}\right)dt$$ that is hard to evaluate in terms of elementary functions but quite easy to calculate numerically. As an alternative, by setting $x=1+2\rho\cos\theta,y=-2+3\rho\sin\theta$ we have: $$I = \int_{0}^{1}\int_{-\pi}^{\pi}\frac{\rho}{6+4\rho\cos\theta-12\rho\sin\theta+4\rho^2\cos^2\theta+9\sin^2\theta}\,d\theta\,d\rho$$ and by setting $\theta=2\arctan t$ we have: $$I = \int_{0}^{1}\int_{-\infty}^{+\infty}\frac{2\rho(1+t^2)}{6(1+t^2)^2+4\rho(1-t^4)-24\rho t(1+t^2)+4\rho^2(1-t^2)^2+36t^2}\,dt\,d\rho,$$ $$I = \int_{0}^{1}\int_{-\infty}^{+\infty}\frac{2\rho(1+t^2)}{(6-4\rho+4\rho^2)t^4-24\rho t^3+(48-8\rho^2)t^2-24\rho t+(6+4\rho+4\rho^2)}\,dt\,d\rho,$$
$$I = \int_{0}^{1}\int_{-\infty}^{+\infty}\frac{\rho(1+t^2)}{2(1-t^2)^2\rho^2+(2-12t-12t^3-2t^4)\rho+(3+24t^2+3t^4)}\,dt\,d\rho.$$