Calculate area of $r^2 = \cos(2 \theta)$ without breaking into individual petals?

Question

In the area integral, I am integrating first $r$ from: $-sqrt\cos(2\theta)$ to $sqrt\cos(2\theta)$ and theta from $-\pi/4$ to $\pi/4$. Integrand is $r*dr*d\theta$. In r integral it comes 0 after placing limits. Why? There is something wrong in elemental area additions in these limits. Please demonstrate in detail what I'm doing wrong.

Answer 1

$$ A=\iint r dr d\theta = \int \frac{r^2}{2} d \theta = \int_{- \pi /4}^{ \pi/4} \frac{ \cos ( 2 \theta)}{2} d \theta = \int_0^{ \pi/4} \cos(2\theta) d\theta = \frac{1}{2}$$

Answer 2

The idea is to calculate the integral with respect to r first.

$\int\int \ r \ drd\theta$

=$\int \frac{1}{2} \ r^2 \ d\theta$

=$\int \frac{1}{2} \cos({2\theta}) d\theta $

= $[\frac{1}{4}\sin(2\theta)]^{\pi/4}_{-\pi/4}$

= $\frac{1}{2}$