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**DeMath** Use the polar coordinates

$\displaystyle \displaystyle{(x^2+y^2)^3=4a^2x^2y^2}$

$\displaystyle \displaystyle{x=\rho\cos\varphi,~y=\rho\sin\varphi }$

$\displaystyle \displaystyle{\rho^6=4a^2\rho^4\cos^2\varphi\sin^2 \varphi}$

$\displaystyle \displaystyle{\rho^2=4a^2\cos^2\varphi\sin^2\varph i}$

$\displaystyle \displaystyle{\rho^2=a^2\sin^22\varphi}$

$\displaystyle \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet a\rho^2\,d\varphi=\frac{a^2}{2}\int\limits_0^{2\pi }\sin^22\varphi\,d\varphi=\ldots=\frac{a^2\pi}{2}}$

P.S. $\displaystyle \rho=a\sin2\varphi$ is the equation of the *polar rose* with four * petals* (see wikipedia).