# Area under $|x| + |y| < k$

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 $$k=6$$

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.

• Take it quadrant by quadrant. In the first quadrant, $f(x) = k - x.$ So, you compute $\int_0^k (k-x)dx.$ Apr 24, 2021 at 2:12
• 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. Apr 24, 2021 at 2:17
• A general formula for the area of the figure $\left|ax\right|+\left|by\right|=c$ is $\frac{2c^{2}}{ab}$
– Vega
Dec 13, 2021 at 16:17

Area can be calculated by double integral $$S=\int\limits_{-k}^{k}\int\limits_{-k+|x|}^{k-|x|}dydx$$
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, $$A=\int_{-k/2}^{k/2}\int_{-k/2}^{k/2}kdA=\int_{-k/2}^{k/2}\int_{-k/2}^{k/2}2\,ds\,dt=2k^2$$