Calculate the following integral: double integral over
Calculate the following integral: $$ \iint_{|x|+|y| \leq 1} x^2 \, dx \, dy. $$
Explanation of Method
Since $x^2$ is even in both $x$ and $y$ and the diamond is symmetric about both axes, the integral over the whole diamond is $4$ times the integral over just the first-quadrant triangle $x+y\leqslant1$, $x,y\geqslant0$.
- Use symmetry to reduce to the first quadrant.
- Integrate with respect to $y$, then $x$, over the triangle.
Step 1 · Identify the region
$|x|+|y|\leqslant1$ is the diamond with vertices $(\pm1,0)$, $(0,\pm1)$.
Step 2 · Use symmetry
$x^2$ is even in $x$ and the diamond is symmetric about both axes, so:
$$ I = 4\iint_{x\geqslant0,\,y\geqslant0,\,x+y\leqslant1} x^2\,dx\,dy. $$
Step 3 · Set up the integral
$$ I = 4\int_0^1\int_0^{1-x} x^2\,dy\,dx. $$
Step 4 · Integrate with respect to $y$
$$ \int_0^{1-x} x^2\,dy = x^2(1-x). $$
Step 5 · Integrate with respect to $x$
$$ I = 4\int_0^1 (x^2-x^3)\,dx = 4\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1 = 4\left(\frac13-\frac14\right) = 4\cdot\frac{1}{12} = \frac13. $$
Additional Notes
- By the same symmetry argument, $\iint y^2\,dx\,dy$ over the same diamond also equals $\frac13$, since $x$ and $y$ play interchangeable roles in the diamond.
Result
The solution to the integral is: