two variables inequality $y \leq x^2+y^2\leq x$

Question

I want to graph and calculate the Area $ \mathcal D $ , how do i solve this inequality :

$ $ $ \mathcal D $ = $ ${ ${(x,y)\in \Bbb R^2 }$ : $y$ $\geq$ $0 $ , $y \leq x^2+y^2\leq x$ } ?

Answer 1

Hint: You can change the $\leq$ and $\geq$ sections into circles. $$x^2+y^2\leq x \implies x^2-x+y^2\leq 0\implies \left(x-\frac{1}{2}\right)^2+y^2\leq\frac{1}{4}$$ $$x^2+y^2\geq y\implies x^2+y^2-y\geq 0\implies x^2+\left(y-\frac{1} {2}\right)^2\geq\frac{1}{4}$$ Which are two circles. You can then find the region you want after plotting the two circles and determining regions that satisfy the inequalities.

Answer 2

Look at each inequality separately, and take their intersection.

---

$$y \leq x^2+y^2$$

$$0 \leq x^2+y^2-y$$

$$0 \leq x^2+(y-\frac{1}{2})^2-\frac{1}{4}$$

$$\frac{1}{4} \leq x^2+(y-\frac{1}{2})^2$$

---

$$x^2+y^2 \leq x$$

$$(x-\frac{1}{2})^2+y^2 \leq \frac{1}{4}$$

The dark blue region above the $x$ axis is the region described by your set. A suggestion to find the area: switch to polar.