I have to calculate area bounded by curves : $(x^3+y^3)^2=x^2+y^2 $ for $ x,y \ge 0 $. I tried to use polar coordinates, but I have : $r^4(\cos^6\alpha +2\sin^3\alpha\cos^3\alpha + \sin^6\alpha)=1$
2026-03-27 00:54:47.1774572887
Area between curves.
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By your work: $$r=\frac{1}{\sqrt{\sin^3\alpha+\cos^3\alpha}}.$$ Since it's symmetric in respect to $y=x$, we obtain that the needed area it's $$2\int\limits_0^{\frac{\pi}{4}}d\alpha\int\limits_0^{\frac{1}{\sqrt{\sin^3\alpha+\cos^3\alpha}}}rdr=\int_0^{\frac{\pi}{4}}\frac{1}{\sin^3\alpha+\cos^3\alpha}d\alpha.$$ Can you end it now?