# Can't understand the solution of the integral

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 $$\int_0^1 \int_{x^2}^{\sqrt{x}}$$?

• Have you sketched the set $S$? Jan 9, 2020 at 14:52

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.
• On the x-axis the region spans from $0$ to $1$, giving the outer integral. : I agree. The inner integral follows from the double inequality $x^2 \leq y \leq \sqrt{x}$ we just derived. :Yes I saw but why we considered theses bounds for $y$ although 'from the graph' the region on the y-axis also spans from $0$ to $1$ ? Jan 9, 2020 at 15:04
• @user8003788 The bounds for $y$ depend on $x$. Jan 9, 2020 at 15:05
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.