Find the area inside the polar curve $r^2=\cos(2\theta)$ I feel a bit silly asking this question as it is no doubt relatively simple, but it has been bugging me.
Given the polar curve described by $r^2 = cos(2\theta)$, find the area inside the curve. 
My intuition is to integrate from $0$ to $2 \pi$, but this clearly just yields 0. I have found that the area is 1, but this is just from dividing the curve into 4 pices and integrating from $0$ to $\pi/4$, then multiplying the resultant are by 4.
Why is it that the usual method of integrating $\frac{1}{2}\int_{0}^{2 \pi} r(\theta)^2 d\theta$ fails here?

It's the same reason why integrating $f(x)=sin(x)$ from $T: 0 \rightarrow 2\pi$ yields a value of $0$; you're integrating over the negative portion of the $x$ and $y$ axes and because your function displays radial symmetry, when you take the integral over half your period it is equal in magnitude to the other half, but the signs are opposite. This results in the values canceling one another out.
