Area defined by $x^2+y^2 \leq 1$ and $y\geq x(x^2-16)$

One very obvious way would be to find the points of intersection which would be messy and subject to many conditions. I was trying to solve this using polar coordinates $(r\cos \theta,r \sin \theta)$ substituted it into the other curve to get $$\frac{\tan \theta + 16}{\cos \theta}\geq r$$

where $r \subset [0,1]$

I don't know how to proceed further.


Say $A$ is the set where $x^2+y^2\le 1$ and $y>x(x^2-16)$ and $B$ is the set where $x^2+y^2\le 1$ and $y<x(x^2-16)$. The mapping $(x,y)\mapsto(-x,-y)$ shows that $A$ and $B$ have the same area. It's easy to see the set where $y=x(x^2-16)$ has area zero. So your set has area $\pi/2$, half the area of the unit disk.

  • $\begingroup$ I did not understand how does the mapping show that $A,B$ have the same areas $\endgroup$ – GenericQuantumPh6Term May 17 at 16:08
  • $\begingroup$ @GenericQuantumPh6Term Say $\phi(x,y)=(-x,-y)$. Then $\phi(A)=B$, and the Jacobian of $\phi$ is $1$, so $\phi$ preserves area, $\endgroup$ – David C. Ullrich May 17 at 16:12
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    $\begingroup$ @GenericQuantumPh6Term Or: $B$ is the same as $A$, rotated $180$ degrees. $\endgroup$ – David C. Ullrich May 17 at 16:16

I would solve the System $$x^2+y^2=1$$ and $$y=x(x^2-16)$$ The solutions are looking terrible! Taking a calculator we obtain $${x\approx -0.0623934, y \approx 0.998052}, {x \approx 0.0623934, y \approx -0.998052}$$


Your "online tools" appear to be approximating the cubic by a linear function. That line being a diameter of the circle, yes, the area of each part is half the area of the circle. But I have no confidence that is correct for the true area.

  • $\begingroup$ The derivative being $3x^2-16$ means the graph is going to look somewhat like a line near the origin. Regarding the true area, see the other answers. $\endgroup$ – David C. Ullrich May 17 at 19:57

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