0
$\begingroup$

Surface $S_1$ is formed by $|x-1|≤ 1$ and $|y-2|≤ 3$. Surface $S_2$ is formed by $x ≥3$, $y ≤1$ and $x+2y ≤3$. S is union of $S_1$ and $ S_2$.

Find the area of surface $S_1$, $S_2$ and $S$ ?

1 Answers 1

1

$S_1$ is a rectangle centered at $(1,2)$ with sides of length $2$ and $6$ respectively. Therefore $A = 12$

$S_2$ is not a closed polygon. Therefore its area is infinite. Check that $x \geq 3$ and $y \leq 1$ makes any value from $x\in (3,\infty)$ and $y \in (-\infty,1)$ part of it. Then $y \leq \dfrac{3}{2} - \dfrac{x}{2}$ excludes only some upper values but all below the line $y = \dfrac{3}{2} - \dfrac{x}{2}$ will still be valid.

The union of both areas will be an infinite area as well.

enter image description here

  • 1
    I think You also started with solving inequalities and then plotting for $x = 0$, $x = 2$ $y= -1$ and $y = 5$ which makes you say this. But plot of $|x -1 | \le 1$ is $|x| $ shifted by $1$ towards +ve x axis. – 2017-02-08
  • 0
    You are right. I will correct it later on – 2017-02-09
  • 0
    Well, re-reading that in the computer I don't think so, you are talking about $y = |x|$, but what he seems to be writing is the domain of $x$ so it is correct – 2017-02-09