Area of $\{(x,y)\in \mathbb{R}^2:\ 1\leq x y\leq 2, x \leq y \leq 2x\}$

Question

Trying to evaluate the area spanned by the set $\{(x,y)\in \mathbb{R}^2:\ 1\leq x y\leq 2, x \leq y \leq 2x\}$, I use the following reasoning:

\begin{equation} 1\leq x y \leq 2\ \Longrightarrow\ \frac{1}{y}\leq x\leq \frac{2}{y} \end{equation}

But $x\leq y\leq 2x$, so \begin{equation} \frac{1}{y}\leq y \leq \frac{4}{y}\ \Longrightarrow\ 1 \leq y^2 \leq 4 \Longrightarrow y \in [-2,-1] \cup [1,2] \end{equation}

To calculate the total area, we integrate the area element $dx\,dy$ over these boundaries. Now, given that the area spanned by a set of points $(x,y)$ satisfying $\frac{1}{y}\leq x\leq \frac{2}{y}$ is the same for either $y\in [-2,-1]$ or $y \in [1,2]$, I conclude that the total area is twice the area spanned by $\{(x,y):\ \frac{1}{y}\leq x\leq \frac{2}{y}, y\in [1,2] \}$.

Hence, the total area is \begin{equation} 2\int_1^2\int_{1/y}^{2/y}dx\,dy = 2 \log{2} \end{equation}

However I am said the correct value of the area is $\frac{1}{2}\log{2}$. I wonder therefore what is wrong with my reasoning and how do we get the correct answer.

Answer 1

Besides the simple and nice way Carlo described, you can do it directly...but observe you have an irregular domain, so you must find intersection points and subdivide the domain. We thus can evaluate for $\;x,y\ge0\;$ (as this follows from the domain's definition) and then multiply by two by symmetry:

$$\begin{cases}y=2x=\frac1x\implies x=\frac1{\sqrt2}\\{}\\ y=2x=\frac1{2x}\implies x=\frac12\\{}\\ y=x=\frac1x\implies x=1\\{}\\ y=x=\frac1{2x}\implies x=\frac1{\sqrt2}\end{cases}$$

and thus

$$A=\int_{1/2}^{1/\sqrt2}\int_{1/x}^{2x} dy\,dx+\int_{1/\sqrt2}^1\int_x^{2/x} dy\,dx=\int_{1/2}^{1/\sqrt2}\left(2x-\frac1x\right)dx+\int_{1/\sqrt2}^1\left(\frac2x-x\right)dx=$$

$$=\left.\left(x^2-\log x\right)\right|_{1/2}^{1/\sqrt2}+\left.\left(2\log x-\frac12x^2\right)\right|_{1/\sqrt2}^1=\frac12-\log\frac1{\sqrt2}-\frac14+\log\frac12+0-\frac12-2\log\frac1{\sqrt2}+\frac14=$$

$$=\log\frac12-3\log\frac1{\sqrt2}=-\log2+3\log\sqrt2=-\log2+\frac32\log2=\color{red}{\frac12\log2}$$

Of course, much easier and faster with variables change...

Answer 2

Consider the assignment of variables $u(x, y) = xy$ and $v(x, y) = y/x.$ Observe that $1 \leq u \leq 2$ and $1 \leq v \leq 2,$ hence the transformation $G(x, y) = (u(x, y), v(x, y))$ maps the curvilinear rectangle $\mathcal R = \{(x, y) \,|\, 1 \leq xy \leq 2 \text{ and } x \leq y \leq 2x \}$ in the $xy$-plane to the translation of the unit square $\mathcal S = \{(u, v) \,|\, 1 \leq u \leq 2 \text{ and } 1 \leq v \leq 2 \}$ in the $uv$-plane.

Observe that the Jacobian of the transformation $G$ is given by $$\operatorname{Jac}(G) = \det \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix} = \det \begin{pmatrix} y & x \\ -\frac y {x^2} & \frac 1 x \end{pmatrix} = \frac y x + \frac y x = \frac{2y} x = 2v.$$ By the Change of Variables Formula, we have that $$\iint_\mathcal R 1 \, dA = \iint_\mathcal S \frac{1}{\operatorname{Jac}(G)} \, dA = \int_1^2 \int_1^2 \frac 1 {2v} \, dv \, du = \frac 1 2 \ln 2. \text{ QED.}$$