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

Question

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.

Answer 1

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.

Answer 2

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 $${x\approx -0.0623934, y \approx 0.998052}, {x \approx 0.0623934, y \approx -0.998052}$$

Answer 3

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.