1
$\begingroup$

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 : $r^4(\cos^6\alpha +2\sin^3\alpha\cos^3\alpha + \sin^6\alpha)=1$

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Yes, thank you. $\endgroup$ – pawelK Mar 19 at 20:05
  • $\begingroup$ If the down-voter has some question, I am ready to explain. $\endgroup$ – Michael Rozenberg Mar 22 at 17:02

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.