$\int_E \frac{y^2}{x^2}d\lambda^2(x,y)$ for $E=\{ (x,y)\in\mathbb R^2 : x\le y, xy\ge 1, y\le2 \}$

Question

$\int_E \frac{y^2}{x^2}d\lambda^2(x,y)$ for $E=\{ (x,y)\in\mathbb R^2 : x\le y, xy\ge 1, y\le2 \}$

I often struggle to find the boundaries of integration from a given area.

My approach was to use Fubini as the function is positive and compute the iterated integral

$\int_{-\infty}^{2}\int_{1/y}^{y} \frac{y^2}{x^2} dx dy=...=\infty$

My questions is whether the boundaries $\frac{1}{y} \le x\le y$ and $-\infty \le y\le 2$ are correct and if there is a general/elegant approach to find them from a given arbitrary area.

Answer 1

Wolfram Alpha produces this plot for x <= y, xy >= 1, y <= 2:

So it makes sense that you would need to break the integration up into at least two pieces. Since $xy > 0$, $x$ and $y$ are always of the same sign.

When $x > 0, y> 0$, then $y = 2$ is the top, $y = x$ is bottom and left, and $y = 1/x$ is bottom and right. $y = 1/x$ and $y = x$ cross at $(1,1)$, so if you integrate by $y$ first, you need $y \in [1/x, 2]$ for $x \le 1$ and $y \in [x, 2]$ for $x \ge 1$. $y = 1/x$ crosses $y = 2$ at $x = 1/2$, so you get $\int_{1/2}^1\int_{1/x}^2 \, dydx$ for the first and $\int_1^2\int_{x}^2 \, dydx$ for the second.

But if you integrate $x$ first, then this piece can be done in a single integration: $\int_1^2\int_{1/y}^y \, dxdy$

The lower left piece where $x < 0, y < 0$ is simpler, as $y \le 1/x$ implies $y \le 2$ here. The integral is $\int_{-\infty}^{-1}\int_{x}^{1/x}\, dydx$

So the simplest form is $$\int_E \frac{y^2}{x^2}d\lambda^2(x,y) = \int_{-\infty}^{-1}\int_{x}^{1/x}\frac{y^2}{x^2}\,dydx + \int_1^2\int_{1/y}^y \frac{y^2}{x^2}\, dxdy$$