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


$ 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 $v = \sqrt{2}y$?

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

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.