Calculate $ \int \int_{R^2} \frac{dxdy}{(1+4x^2+9y^2)^2}$. Calculate: $ \int \int_{R^2} \frac{dxdy}{(1+4x^2+9y^2)^2}$.
Solution from book:
We form a set $D_n=\{ (x,y)|4x^2+9y^2\leq n^2 \}$.
Then $4x^2+9y^2=r^2$; $x=\frac{1}{2}rcost$ and $y=\frac{1}{3}rsint$ where $0\leq r \leq n$ and $0\leq t \leq 2\pi$. After that I know how to solve the integral. I just don't understand why is it done this way, why do we form this set $D_n$. The theory behind this is bugging me..
And another example is: $ \int \int_{D} \frac{dxdy}{(1-x^2-y^2)^2}$. Where D is a unit circle.
Here $D_n=\{ (x,y)|x^2+y^2\leq (1-\frac{1}{n})^2 \}$. Why?
Also $ \int \int_{R^2} \frac{dxdy}{1+(x^2+y^2)^2}$ what would $D_n$ be for this one?
 A: A natural way to parametrize the region $\{(x,y)\in\Bbb R^2\mid ax^2+by^2\leqslant R^2\}$ consists in considering $(r,\theta)\mapsto\bigl(ar\cos(\theta),br\sin(\theta)\bigr)$, with $0\leqslant r\leqslant R$. And the integral$$\iint_{\Bbb R^2}\frac{\mathrm dx\,\mathrm dy}{1+4x^2+9y^2},$$which is equal to$$\lim_{R\to\infty}\iint_{\{(x,y)\in\Bbb R^2\mid x^2+y^2\leqslant R\}}\frac{\mathrm dx\,\mathrm dy}{1+4x^2+9y^2},$$becomes, through change of variables$$\lim_{R\to\infty}\int_0^{2\pi}\int_0^R\frac r{1+r^2}\,\mathrm dr\,\mathrm d\theta,$$which is easy to compute.
Also,\begin{align}\int_{\{(x,y)\in\Bbb R^2\mid x^2+y^2<1\}}\frac{\mathrm dx\,\mathrm dy}{1-x^2-y^2}&=\lim_{n\to\infty}\int_{\{(x,y)\in\Bbb R^2\mid x^2+y^2\leqslant1-1/n\}}\frac{\mathrm dx\,\mathrm dy}{1-x^2-y^2}\\&=\lim_{n\to\infty}\int_0^{2\pi}\int_0^{1-1/n}\frac r{1-r^2}\,\mathrm dr\,\mathrm d\theta,\end{align}which, again, is easy to compute.
Can you deal with the other integral now?
A: An alternative approach.
For any $A>0$ we have $\int_{0}^{+\infty}\frac{dx}{A+x^2}=\frac{\pi}{2\sqrt{A}}$, hence by differentiating with respect to $A$ we have
$$ \int_{0}^{+\infty}\frac{dx}{(A+x^2)^2}=\frac{\pi}{4 A^{3/2}},\qquad \int_{-\infty}^{+\infty}\frac{dx}{(1+y^2+x^2)^2}=\frac{\pi}{2(1+y^2)^{3/2}} $$
and by Fubini's theorem
$$ \iint_{\mathbb{R}^2}\frac{dx\,dy}{(1+x^2+y^2)^2}=\frac{\pi}{2}\int_{-\infty}^{+\infty}\frac{dy}{(1+y^2)^{3/2}}=\frac{\pi}{2}\left[\frac{y}{\sqrt{1+y^2}}\right]_{-\infty}^{+\infty}=\pi, $$
so by letting $x=2u$ and $y=3v$ we have
$$ \iint_{\mathbb{R}^2}\frac{du\,dv}{(1+4u^2+9v^2)^2}=\color{red}{\frac{\pi}{6}}.$$
