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Consider the region $C$ of all points in the $x,y$ plane that satisfy $ |x| + |y| < k $. Now I wish the find out the area of $C$ using an integral. It's easy all points in the interior of a square of side $k\sqrt{2}$ centered at the origin and tilted by $\pi/4$ satisfy the equation. The image below illustrates the region for $k=6$

enter image description here

So I guess that the area of $C$ is $2k^2$. I wasn't able to find this by using integral so I'd like an approach by using integrals. Any help would be appreciated. Thanks.

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    $\begingroup$ Take it quadrant by quadrant. In the first quadrant, $f(x) = k - x.$ So, you compute $\int_0^k (k-x)dx.$ $\endgroup$
    – user2661923
    Apr 24, 2021 at 2:12
  • $\begingroup$ As an alternative approach, after dealing with 1st quadrant, you can use a symmetry argument to multiply it by 4, rather than dealing with each quadrant separately. $\endgroup$
    – user2661923
    Apr 24, 2021 at 2:17
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    $\begingroup$ A general formula for the area of the figure $\left|ax\right|+\left|by\right|=c$ is $\frac{2c^{2}}{ab}$ $\endgroup$
    – Vega
    Dec 13, 2021 at 16:17

2 Answers 2

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Area can be calculated by double integral $$S=\int\limits_{-k}^{k}\int\limits_{-k+|x|}^{k-|x|}dydx$$

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You could move to a new coordinate system $(s,t)$ with $s=x+y$, $t=x-y$.

Then $dA=\begin{vmatrix}1&1\\1&-1\end{vmatrix}ds\,dt=2\,ds\,dt$.

In that coordinate system, $$A=\int_{-k/2}^{k/2}\int_{-k/2}^{k/2}kdA=\int_{-k/2}^{k/2}\int_{-k/2}^{k/2}2\,ds\,dt=2k^2$$

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