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Help to find the area bounded by the curve: $$ (x^2+2y^2)^3 = xy^4 $$ From wolframalpha

Update:

$ x = r\cos(\phi) $

$ v = r\sin(\phi) $

$ (r^2)^3=\frac{1}{4}r\cos(\phi)(r\sin(\phi))^4 $

$ r=\frac{1}{4}\cos(\phi)\sin^4(\phi) $

And the area is equal to? And where do the inverse change $v = \sqrt{2}y$?

$$ S = 2 \int_0^{PI/2} \, d\phi \int_0^{\frac{1}{4}\cos(\phi){\sin^4(\phi)}} r\, dr $$

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    $\begingroup$ have you an idea? what have you tried so far and where did you get stuck? $\endgroup$ – flonk Jan 17 '14 at 13:14
  • $\begingroup$ Have you ever tried to find its parametrization? $\endgroup$ – mrs Jan 17 '14 at 13:54
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A hint:

Writing $y:={\displaystyle{v\over\sqrt{2}}}$ transports the figure to the $(x,v)$-plane, and it then has the equation $$(x^2+v^2)^3={1\over4} x v^4\ .$$ At the same time the area has been multiplied by $\sqrt{2}$. In order to compute the $(x,v)$-area, introduce polar coordinates in the $(x,v)$-plane.

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