How do you find area of the loop in the graph of $x(x^2+y^2)=(x^2-y^2)$

Question

The graph of the given equation is

$x(x^2+y^2)=(x^2-y^2)$">

I believe I have to use (r,θ) coordinates but I do not know how to integrate this in (r,θ).

Answer 1

Let $x=r\cos\theta$, $y=r\sin\theta$

$dxdy = r \ dr \ d \theta$

$x^2 + y^2 = r^2$

As, $x(x^2+y^2) = x^2-y^2$

$r(\cos\theta)r^2 = r^2(\cos^2\theta - \sin^2\theta) = r^2(\cos2\theta)$

$r^2(r\cos\theta - \cos2\theta) = 0$

$r = 0$ to $r = \frac{\cos2\theta}{\cos\theta}$

and $\theta = -\pi/4$ to $\pi/4$

So, $A = \int\int dxdy = \int^{\pi/4}_{-\pi/4} \ \int^{\cos2\theta/\cos\theta}_{0} \ \ (rdr\ d\theta)$