Calculate the following integral: double integral over
Calculate the following integral: $$ \iint_{|x|+|y|\leqslant1} (|x|+|y|)\,dx\,dy. $$
Explanation of Method
The region $|x|+|y|\leqslant1$ is a diamond symmetric about both axes, and the integrand $|x|+|y|$ takes the same value under reflection across either axis. So the integral equals $4$ times the integral of $x+y$ over just the first-quadrant triangle.
- Use symmetry to reduce to $4$ times the first-quadrant integral of $x+y$.
- Integrate with respect to $y$, then $x$, over the triangle $x+y\leqslant1$.
Step 1 · Use symmetry
$$ I = 4\iint_{D_1} (x+y)\,dx\,dy, $$
where $D_1$ is the triangle $x,y\geqslant0$, $x+y\leqslant1$.
Step 2 · Set up the integral
$$ I = 4\int_0^1\int_0^{1-x} (x+y)\,dy\,dx. $$
Step 3 · Integrate with respect to $y$
$$ \int_0^{1-x}(x+y)\,dy = \left[xy+\frac{y^2}{2}\right]_0^{1-x} = x(1-x)+\frac{(1-x)^2}{2} = \frac12-\frac{x^2}{2}. $$
Step 4 · Integrate with respect to $x$
$$ I = 4\int_0^1\left(\frac12-\frac{x^2}{2}\right)dx = 2\int_0^1(1-x^2)\,dx = 2\left[x-\frac{x^3}{3}\right]_0^1 = 2\cdot\frac23=\frac43. $$
Additional Notes
- Verified numerically: direct double integration over the diamond gives $I\approx1.3333$, matching $\frac43$.
Result
The solution to the integral is: