Can't understand the solution of the integral

Question

I prepare for my exam and I have this exercise and its solution.

$\forall (x, y) \in \mathbb{R} \times \mathbb{R}$, define the function $f(x,y) = xy \mathbb{1}_S$, where $\mathbb{1}_S$ is an indicator for the set $S = \{ (x,y) \in (0,\infty) \times (0,\infty): x^2 \leq y, y^2 \leq x\}$.

I know this is an easy exercise but seeing $x^2 \leq y, y^2 \leq x\}$ is confusing. The solution is

We have $0 \leq x \leq 1$, $0 \leq y \leq 1$ and $x^2 \leq y \leq \sqrt{x}$ then

$\int \int f(x,y) dxdy = \int_0^1 \int_{x^2}^{\sqrt{x}} ydy~ xdx= \dots= 1/12.$

I don't understand from where we got $0 \leq x \leq 1$, $0 \leq y \leq 1$ and $x^2 \leq y \leq \sqrt{x}$ ?

and after why we used for the integrals $\int_0^1 \int_{x^2}^{\sqrt{x}}$?

Answer 1

It helps to draw a graph. The curve $y=x^2$ and its reflection on $y=x$ ($x=y^2$) intersect at $(0,0)$ and $(1,1)$, enclosing a region where both $x^2\le y$ and $y^2\le x$. Taking square roots on the second inequality (since the variables are non-negative) gives $y\le\sqrt x$, which then combines with the first inequality.

On the $x$-axis the region spans from $0$ to $1$, giving the outer integral. The inner integral follows from the double inequality $x^2\le y\le\sqrt x$ we just derived.

Answer 2

We have $y^4 \leq x^2 \leq y$, hence $y \in [0,1]$. Same for $x$. Now $y^2 \leq x$ is equivalent to $y \leq \sqrt{x}$, so we get $0 \leq x \leq 1$, $x^2 \leq y \leq \sqrt{x}$. Now you just integrate over that domain.