Finding and area given (integration in polar coordinates)

Question

There is an area given: $$S = \big\{ (x,y):(x^2+y^2)^2 \le 2(x^2-y^2)\big\}$$ I know that I have to use polar coordinate system and I know that: $$r \le\sqrt{2\cos2\phi}$$ However I don't know how to find the range of $\phi$.
Is there a method to draw that kind of areas?

Answer 1

You have $$ P:\{(r,\phi)\in[0,\infty)\times[0,2\pi]~:~r^2\leq 2\cos(2\phi)\}\to S,~P(r,\phi)=(r\cos(\phi),r\sin(\phi)). $$ So you have to restrict your domain to get $S$ as the image of $P$. It looks like here:

To get the range for $\phi$ you have to consider $r^2\leq 2\cos(2\phi)$. The LSH is positive, so the inequality can just hold if and only if $\cos(2\phi)\geq 0$. Since $\phi\in[0,2\pi]$ you get $2\phi\in[0,4\pi]$ and $$ \cos(2\phi)\geq 0 \Leftrightarrow 2\phi\in\left[0,\frac12\pi\right]\cup\left[\frac32\pi,\frac52\pi\right]\cup\left[\frac72\pi,4\pi\right]\\ \Leftrightarrow \phi\in\left[0,\frac14\pi\right]\cup\left[\frac34\pi,\frac54\pi\right]\cup\left[\frac74\pi,2\pi\right]. $$ So you get $$ M:=\{(r,\phi)\in[0,\infty)\times[0,2\pi]~:~r^2\leq 2\cos(2\phi)\}\\ =\left\{(r,\phi)\in[0,\infty)\times\left(\left[0,\frac14\pi\right]\cup\left[\frac34\pi,\frac54\pi\right]\cup\left[\frac74\pi,\frac2\pi\right]\right)~:~r^2\leq 2\cos(2\phi)\right\}. $$

Now you get $$ \int_S1~d(x,y)=\int_M r~d(r,\phi)\\ =\int_0^{\frac14\pi}\int_0^\sqrt{2\cos(2\phi)}r~dr~d\phi+\int_{\frac34\pi}^{\frac54\pi}\int_0^\sqrt{2\cos(2\phi)}r~dr~d\phi+\int_{\frac74\pi}^{2\pi}\int_0^\sqrt{2\cos(2\phi)}r~dr~d\phi $$

Because of the symmetry you can simplify $$ \int_S1~d(x,y)=4\int_0^{\frac14\pi}\int_0^\sqrt{2\cos(2\phi)}r~dr~d\phi. $$

Result:

The area should be $2$.