Need help finding the area inside an implicitly defined curve $x^2 + (y + \sqrt[3]{|x|})^2=1$. (I think it is a heart shape). I've been trying to parameterize it with no luck. I also tried to restrict my attention to $0 \leq x \leq 1$ and do an integral against the $y$ axis from $y=-1$ to $y=1$ (since $x$ is a function of $y$ in that interval) and then add that to the area of the curve in $1 \leq y \leq 1.7$, which I would do with another integral but against $x$ this time.

None of these have worked, because it's difficult to rearrange well enough to get $y$ or $x$ alone on one side.

How would you go about solving this? Is there a neat parameterization I'm missing? Also, I believe the area is $\pi$ but I don't know how they figured that.

enter image description here

  • $\begingroup$ Integration solves it easily. $\endgroup$ – user499203 Nov 19 '17 at 19:08
  • $\begingroup$ @ThePirateBay Really? Ok...the integration of what and between what limits? $\endgroup$ – DonAntonio Nov 19 '17 at 19:12
  • $\begingroup$ First of all, the plot is wrong. It gives upside down image. "What and between what limits" - there are two curves, just combine their integrals $\endgroup$ – user499203 Nov 19 '17 at 19:15

Find the top and bottom of the curve in terms of $x$. $$x^2 + (y + \sqrt[3]{|x|})^2=1\\ (y + \sqrt[3]{|x|})^2=1-x^2\\y + \sqrt[3]{|x|}=\pm\sqrt{1-x^2}\\y=\pm\sqrt{1-x^2}-\sqrt[3]{|x|}$$ Note that your curve is symmetric around the vertical axis, so the area is twice the area of the positive side. Also, from the second equation, you can see that $-1\le x\le1$. So, using vertical rectangles of thickness $dx$, you write the area as $$A=2\int_0^1(\sqrt{1-x^2}-\sqrt[3]{x}+\sqrt{1-x^2}+\sqrt[3]{x})dx=4\int_0^1\sqrt{1-x^2}dx$$ The last integral can be computed using $z=\sin(x)$ substitution, so $$A=4\int_0^{\pi/2}\cos^2x dx=4\frac{\pi}{4}=\pi$$

  • $\begingroup$ Note in fact that you can substitute $\sqrt[3]{|x|}$ with any function of $|x|$ and you get the same result $\endgroup$ – Andrei Nov 19 '17 at 19:17

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