The area of the region R=(x,y):xy 8,1 y x^2,x 0 is

Question

The area of the region $R={(x,y):xy\le 8,1\le y\le {x}^{2},x\ge 0}$ is

Accepted Answer

The region is bounded by three curves in the first quadrant ( $x\ge 0$ ) :

1. $y=1$ : A horizontal line.

2. $y={x}^{2}$ : A parabola opening upwards.

3. $y=\frac{8}{x}$ : A rectangular hyperbola.

Find the Intersection Points

We need to know where these curves meet to set our integration limits:

Parabola and Line $(y={x}^{2}$ and $y=1)$ : ${x}^{2}=1\implies x=1$ (since $x\ge 0$ ). Point: $(1,1)$.

Parabola and Hyperbola $(y={x}^{2}$ and $y=\frac{8}{x}):{x}^{2}=\frac{8}{x}\implies {x}^{3}=8\implies x=2$. Point: $(2,4)$.

Hyperbola and Line $(y=\frac{8}{x}$ and $y=1):\frac{8}{x}=1\implies x=8$. Point: $(8,1)$.

Understanding the Inequalities

$x\ge 0$ : The region is located in the first and fourth quadrants (but limited to the first quadrant by $1\le y$ ).

$1\le y\le {x}^{2}$ : This gives us a lower bound of the horizontal line $y=1$ and an upper bound of the parabola $y={x}^{2}$.

$xy\le 8$ : This can be rewritten as $y\le \frac{8}{x}$. This represents the area below the rectangular hyperbola.

Setting up the Integrals

The region starts at $x=1$ and ends at $x=8$. Because the "upper" boundary changes at $x=2$, we must split the area into two parts:

Part 1 (from $x=1$ to $x=2$ ): Bounded above by the parabola $y={x}^{2}$ and below by $y=1$.

Part 2 (from $x=2$ to $x=8$ ): Bounded above by the hyperbola $y=\frac{8}{x}$ and below by $y=1$.

Area $=\int _{1}^{2}({x}^{2}-1)dx+\int _{2}^{8}(\frac{8}{x}-1)dx$

$={[\frac{{x}^{3}}{3}-x]}_{2}^{1}+[8\ln x-x{]}_{8}^{2}$

$=[\frac{8}{3}-2-{\frac{1}{3}-1}]+[(8\ln 8-8)-{8\ln 2-2}]$

$=\frac{7}{3}-1+8\ln {2}^{3}-8\ln 2-6$

$=\frac{7}{3}-7+24\ln 2-8\ln 2\, {\ln {a}^{m}=m\ln a}$

$=-\frac{14}{3}+16\ln 2$

$=\frac{2}{3}[24\ln 2-7]$

Solutions · 1