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I'm attempting to calculate the area of the planar region bounded by the level curve $4$ of the function $f(x,y)=∣x∣+∣y∣$.

Step 1: Finding the Level Curve $4$

I've been working on finding the level curve $4$ of the function $f(x,y)=∣x∣+∣y∣$, and here's what I've done:

I equated $f(x,y)$ to $4$ to find the level curve. I then solved the equation:

$∣x∣+∣y∣=4$

Since both $∣x∣$ and $∣y∣$ involve absolute values, I had to consider four possible cases:

  • When $x≥0$ and $y≥0$, the equation becomes $x+y=4$.
  • When $x≥0$ and $y<0$, the equation becomes $x−y=4$.
  • When $x<0$ and $y≥0$, the equation becomes $−x+y=4$.
  • When $x<0$ and $y<0$, the equation becomes $−x−y=4$.

Each of these cases represents a portion of the level curve $4$ in a specific quadrant of the $xy$ plane. The level curve $4$ is the union of these four portions.

After solving each of the above equations, I obtained the four portions of the level curve $4$. However, this is where I got stuck.

When I reached this part, I got stuck and found it challenging to proceed: I'm having difficulty moving forward with the calculation of the area of the region bounded by this level curve. I'm not sure how to proceed to determine the exact area of the region, and I need guidance on how to effectively tackle this problem.

Can you provide additional instructions or a strategy for calculating the area of this region? Any help would be greatly appreciated!

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2 Answers 2

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I would recommend you to notice first that the proposed region can be graphically described as:

enter image description here

Consequently, its area can be calculated as follows: \begin{align*} A = \int_{-4}^{0}\int_{-x - 4}^{x + 4}1\mathrm{d}y\mathrm{d}x + \int_{0}^{4}\int_{x - 4}^{-x + 4}1\mathrm{d}y\mathrm{d}x \end{align*}

Can you take it from here?

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    $\begingroup$ Yes I could! thank you so much! $\endgroup$
    – Ayesca
    Nov 5, 2023 at 5:28
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    $\begingroup$ @Ayesca you are welcome! I am glad I could help. $\endgroup$ Nov 5, 2023 at 5:29
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Since in each quadrant, you can express the portion of the curve as a graph of function $y=f(x)$, you can use integration to compute the area.

In the specific situation here, it is much simpler to notice that the level curve is a square with vertices $(4,0)$, $(0,4)$, $(-4,0)$ and $(0,-4)$, so its area is $A=(4\sqrt{2})^2=32$.

A general advice in this kind of situation: draw a figure!

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  • $\begingroup$ Thank you very much for the response and advice! $\endgroup$
    – Ayesca
    Nov 5, 2023 at 3:56

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