Area bounded by $2 \leq|x+3 y|+|x-y| \leq 4$

Question

Find the area of the region bounded by $$2 \leq|x+3 y|+|x-y| \leq 4$$

I tried taking four cases which are: $$x+3y \geq 0, x-y \geq 0$$ $$x+3y \geq 0, x-y \leq 0$$ $$x+3y \leq 0, x-y \geq 0$$ $$x+3y \leq 0, x-y \leq 0$$ But it becomes so confusing to draw the region. Any better way ?

Answer 1

The transformation $$ \begin{align} u &= x + y \\ v &= 2y \end{align} $$ transforms the region $$ 2 \leq|x+3 y|+|x-y| \leq 4 $$ to $$ 2 \leq|u+v|+|u-v| \leq 4 $$ and increases the area by a factor of two (why?). The new region is limited by two squares with side length $2$ and $4$, respectively (compare How to solve $|a+b|+|a-b|=c$?), so that its area can be computed easily.

Answer 2

Well, if you're looking for a plot:

Now, if we want to find the area of the part where $y\ge0$ and $x\ge0$, we can take a look at the following picture:

We can find the shaded area by finding:

$$3\cdot\left(\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}\right)+\frac{1}{2}\cdot\frac{1}{2}+\int_1^2\left(2-x\right)\space\text{d}x=\frac{9}{8}\tag1$$

And the place where $y\ge0$ and $x\le0$, we can take a look at the following picture:

We can find the shaded area by finding:

$$3\cdot\frac{1}{2}\cdot1+\frac{1}{2}\cdot\frac{1}{2}+\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}=\frac{15}{8}\tag 2$$

So, the total area is given by:

$$\mathscr{A}=2\cdot\frac{9}{8}+2\cdot\frac{15}{8}=6\tag3$$

Answer 3

Here is how I would resolve the absolute equations.

$2 \leq|x+3 y|+|x-y| \leq 4$

If $x \geq y,$

$2 \leq|x+3 y|+ x-y \leq 4$

For $x+3y \geq 0, x+3y \leq 0,$ we get

$1 \leq x+y \leq 2, - 1 \leq y \leq -\frac{1}{2}$

Similarly for $x \leq y$, we have

$-2 \leq x+y \leq -1, \frac{1}{2} \leq y \leq 1$

That gives you the below region to find area of,

Finding area may be easier with a transformation and the transformation choice is obvious but given the focus of the question, I am not getting there.