Find $\lambda^{2}(A)$ while $A:=\{(x,y)\in \mathbb R^{2}: x<y,y<2,y>\frac{1}{x}\}$

Question

Find $\lambda^{2}(A)$ whereby $A:=\{(x,y)\in \mathbb R^{2}: x<y,y<2,y>\frac{1}{x}\}$

I am struggling to find suitable restrictions.

Possible ideas:

Looking at restrictions:

$\frac{1}{x} < y < 2$ and therfore $1 < x < 4$

$\int_{1}^{4}\int_{\frac{1}{x}}^{2}1dydx=(2-\frac{1}{x})(4-1)$

Is this correct?

Answer 1

Your answer is wrong because it should be a constant which does not depend on $x$ as @BigbearZzz pointed out. Note that we can write $$A=\{(x,y)\in\mathbb{R}^2\;|\;\max\{x,\frac{1}{x}\}<y<2\}.$$ As @Ingix pointed out, if we allow $x$ to be negative, then $(-\infty,0)\times [0,2)$ is contained in $A$ and thus we have $\lambda_2(A)=\infty$. So I'll presume that $x$ is always positive.

To evaluate the volume of $A$, we can write $$\begin{eqnarray} \lambda_2(A)&=&\int_A1\; dxdy\\ &=&\int_{x>0, \max\{x,\frac{1}{x}\}<y<2}1\; dxdy. \end{eqnarray}$$ Note that the range of $x$ for which $\max\{x,\frac{1}{x}\}<2$ is $\frac{1}{2}<x<2$. We can see this by solving $0<x<2$ and $0<\frac{1}{x}<2$. By Fubini-Tonelli's theorem the given integral is $$\begin{eqnarray} \int_{x>0, \max\{x,\frac{1}{x}\}<y<2}1\; dxdy&=&\int_{\frac{1}{2}<x<2}\left[\int_{\max\{x,\frac{1}{x}\}}^2 1\; dy\right]dx\\ &=&\int_{\frac{1}{2}<x<2}\left(2-\max\{x,\frac{1}{x}\}\right)dx\\ &=&\int_\frac{1}{2}^1\left(2-\max\{x,\frac{1}{x}\}\right)dx+\int_1^2\left(2-\max\{x,\frac{1}{x}\}\right)dx\\ &=&\int_\frac{1}{2}^1\left(2-\frac{1}{x}\right)dx+\int_1^2\left(2-x\right)dx\\ &=&1-\left[\log x\right]\Big|^1_{\frac{1}{2}}+2-\left[\frac{1}{2}x^2\right]\Big|^2_1\\ &=&\frac{3}{2}-\log 2. \end{eqnarray}$$ Thus $\lambda_2(A)=\frac{3}{2}-\log 2$.

Answer 2

The left picture shows the integral computed first dy and then dx (this is represented by the direction of the segments coloring the domain). It is necessary to split the integral in two parts $\int\limits_{1/2}^1\;\int\limits_{1/x}^2 dy\;dx + \int\limits_{1}^2\int\limits_{x}^2 dy\;dx.$

The right one presents the same area but the integral is easier to manage, $$\int_{1}^2\int\limits_{1/y}^y dx\;dy=\int_{1}^2 \left(y-\frac1y\right) dy =\left[\frac{y^2}{2}-\log y\right]_{1}^{2}=\frac32 - \log 2. $$