<s>[INST] Finding the limits of integration for $\iint\limits_A \frac{1}{(1 + x^2 + y^2)^2} dxdy$ where A is one loop of lemniscate $(x^2 + y^2)^2 = x^2 - y^2$ Work so far: Switch to polar coordinates
$A \rightarrow r^2 = \cos^2\theta - \sin^2\theta \rightarrow r^2 = \cos2\theta$
So $A$ is $r^2 = \cos 2 \theta$
And the integral becomes 
$$\iint\limits_A \frac{r}{(1 + r^2)}drd\theta$$
Now I want to find the limits of integration without actually graphing the curve.
So
$$2rdr = -2\sin(2\theta)$$
$$\frac{dr}{d\theta} = \frac{-\sin(2\theta)}{r}$$
Critical points for $\theta$ come out to be $0$ and $\frac{\pi}{2}$
Now I know how to solve the integral using u substitution but I'm not sure about the limits.
 [/INST] In polar coordinates the curve is determined by 
$$r^2=\cos2\theta,
$$
which has real solutions for $r $ if and only if:
$$
\cos2\theta\ge0 \implies -\frac\pi4\le\theta\le\frac\pi4\text { or }\frac{3\pi}4\le\theta\le\frac{5\pi}4.$$
As the problem asks to find the area of a single loop it suffices to consider only the first interval. We have:
$$\iint\limits_A \frac{r}{(1 + r^2)^2}drd\theta=
\int\limits_{-\frac\pi4}^{\frac\pi4}d\theta
\int\limits_0^\sqrt {\cos2\theta}\frac{rdr}{(1 + r^2)^2}.
$$
Can you take it from here?
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