Answer to Question #153362 in Differential Geometry | Topology for Tony

Question

Shade the region of the xy-plane for which

a) (𝑥2+𝑦2−16)(𝑥2−4)≤0

b) (𝑦−𝑥)(𝑦2+𝑥3)>0

Answer 1

Before shading this region, let's study geometrically to which region it corres[onds. The first inequality is achieved, when two expressions ($x^2+y^2-16, x^2-4$) have opposite signs. Now there is two cases : $x^2+y^2-16\leq 0, x^2-4\geq0$, in this case the point should be in the circle centered at 0 of radius 4 (due to the first inequality) and it should have $|x|\geq 2$ due to the second case. The second case is when the point is outside the circle centered at 0 of radius 4, but it has $|x|\leq 2$. Thus we obtain a region like this :

We will study this region similarly to the previous one : this expression is positive if both the epressions $y-x, y^2+x^3$ are of the same sign. There is two cases: $y>x, y^2>-x^3$ or $y<x, y^2<-x^3$. The first case corresponds to points that are higher than the straight line $y=x$, and for the points $x<0$ we have additional condition $|y|>\sqrt{-x}^3$. In the other case we have points that are lower than $y=x$, we can't have points $x\geq0$ (as we can't have $y^2<0$) and for $x<0$ we have an additional condition $|y|< \sqrt{-x}^3$. Thus we obtain a region like this :