Can anyone help me calculate this area? I have to use double integrals, and the question sounds like this: "Calculate the area bounded by the curve $(x^2+y^2)^2=a^2(x^2-y^2)$, where $a$ is a real constant. I have searched online and found that this type of curve is a lemniscate, but I do not know how to convert $x,y$ to polar coordinates. Do you have any suggestions?

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    $\begingroup$ You can find more info here (in the notation there: $a^2 = 2c^2$) including how to calculate the area. To go to polar coordinates just take $x=r\cos\theta$ and $y=r\sin\theta$ in the equation. $\endgroup$ – Winther Feb 7 '15 at 23:33
  • $\begingroup$ $r^2=a^2\cos(2\theta)$ $\endgroup$ – jimbo Feb 8 '15 at 0:05

Since the lemniscate encloses 4 equal subregions, one in each quadrant, you could use

$\displaystyle A=4\int_0^{\frac{\pi}{4}}\int_0^{a\sqrt{\cos2\theta}}r\; dr d\theta$.


$$A=2(\dfrac{1}{2}\int r^2d\theta)=a^2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(2\theta)d {\theta}=a^2$$

  • $\begingroup$ I have to calculate the area using double integrals, so if I substitute x=rcosθ and y=rsinθ it will give me the equation r^2=a^2cos(2θ) , but r and θ go are in what intervals? $\endgroup$ – andrei985 Feb 8 '15 at 0:16

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