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

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

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.
