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

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 $$y. The mapping $$(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$$, half the area of the unit disk.
• I did not understand how does the mapping show that $A,B$ have the same areas – GenericQuantumPh6Term May 17 at 16:08
• @GenericQuantumPh6Term Say $\phi(x,y)=(-x,-y)$. Then $\phi(A)=B$, and the Jacobian of $\phi$ is $1$, so $\phi$ preserves area, – David C. Ullrich May 17 at 16:12
• @GenericQuantumPh6Term Or: $B$ is the same as $A$, rotated $180$ degrees. – David C. Ullrich May 17 at 16:16
I would solve the System $$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}$$
• 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. – David C. Ullrich May 17 at 19:57