Given the following curve:
${(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2=\frac{x^2}{a^2}-\frac{y^2}{b^2}$, $a,b>0$
Find the space enclosed by it.
Now, obviously (I think) the idea is to subsitute $y=brsin\theta$ and $x=arcos\theta$, and you get the equation:
$r^2=cos(2\theta)$
From here we should perform double integration, I assume. I thought of:
$\int_0^{2\pi}\int_0^{cos2\theta}r^2 drd\theta$
However that obviously ends up as 0. The problem is, even if I try to take into consideration the symmetry with:
$4\times \int_0^{\frac{\pi}{2}}\int_0^{cos2\theta}r^2 drd\theta$
It still ends up as 0. Am I computing it wrong, or is my double-integral wrong? What am I doing wrong?
First the integral in polar coordinates is:
$$A=\iint_T r\;\mathrm{d}r\;\mathrm{d}\theta$$
Second is more clear if you see the graph of the area:
And observe the next:
In the first quadrant $\theta$ goes from $0$ (point $(1,0)$) to $\frac{\pi}{4}$ (point $(0,0)$).
Second when $\theta$ goes from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ $\cos{2\theta}$ is negative therefore the square of the distance to the origin is not well defined.
The same with $\theta$ from $\frac{\pi}{2}$ to $\frac{3\pi}{4}$.
Third from $\frac{3\pi}{4}$ to $\pi$ you have the part of the curve which in the second quadrant.
I think you can continue and define the rest of the curve when is well defined.
But you can take advantage of the simmetry and just take the first quadrant and calculate:
$$A=4\int_{0}^{\frac{\pi}{4}} \int_0^{\sqrt{\cos{2\theta} }} r\;\mathrm{d}r\;\mathrm{d}\theta$$