# 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.

• Why do you need the critical points? Look for the range of $\theta$ where the inequality $\cos2\theta\ge0$ holds. – user Nov 13 '19 at 10:17
• @user $\frac{\pi}{4} \rightarrow 0$ seems right but where does $\cos 2\theta \geq 0$ come from? – atis Nov 13 '19 at 10:27
• The equation $r^2=\cos2\theta$ has real solutions for $r$ if and only if $\cos2\theta\ge0$. – user Nov 13 '19 at 15:24
• Ah so to get every possible value, $\theta$ should go from $\frac{\pi}{4}$ to $0$. So does that mean $r$ should go from $0$ to its maximum possible bound? How should I go about finding that? – atis Nov 13 '19 at 16:19

In polar coordinates the curve is determined by $$r^2=\cos2\theta,$$ which has real solutions for $$r$$ if and only if: $$\cos2\theta\ge0 \implies -\frac\pi4\le\theta\le\frac\pi4\text { or }\frac{3\pi}4\le\theta\le\frac{5\pi}4.$$
$$\iint\limits_A \frac{r}{(1 + r^2)^2}drd\theta= \int\limits_{-\frac\pi4}^{\frac\pi4}d\theta \int\limits_0^\sqrt {\cos2\theta}\frac{rdr}{(1 + r^2)^2}.$$
• One question if you don't mind, why are the limits of $r$ are from $0 \rightarrow \sqrt{\cos 2 \theta}$? – atis Nov 14 '19 at 7:46
• You are welcome. The only solutions of the equation $r^2=\cos2\theta$ are $r=\pm\sqrt{\cos2\theta}$. The negative square root belongs to the other loop of the curve ($\frac{3\pi}4\le\theta\le\frac{5\pi}4$). – user Nov 14 '19 at 8:46