Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D |xy|\,dx\,dy, $$ where $D$ is the region $x^2+y^2\leqslant a^2$.
Explanation of Method
The disk is symmetric under reflection across either axis, and $|xy|$ takes the same value in each of the four quadrants (it equals $xy$ in the first and third, and $-xy$ in the second and fourth — both non-negative). So the integral over the full disk is $4$ times the integral of $xy$ over just the first-quadrant quarter-disk.
- Use symmetry to reduce to $4$ times the first-quadrant integral of $xy$.
- Convert to polar coordinates.
- Integrate with respect to $r$, then $\theta$.
Step 1 · Use symmetry
$$ I = 4\iint_{D_1} xy\,dx\,dy, $$
where $D_1$ is the first-quadrant quarter-disk $x,y\geqslant0$, $x^2+y^2\leqslant a^2$.
Step 2 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$: $0\leqslant r\leqslant a$, $0\leqslant\theta\leqslant\frac\pi2$, and $xy=\frac{r^2\sin2\theta}{2}$:
$$ I = 4\int_0^{\pi/2}\int_0^a \frac{r^2\sin2\theta}{2}\cdot r\,dr\,d\theta = 2\int_0^{\pi/2}\sin2\theta\,d\theta\int_0^a r^3\,dr. $$
Step 3 · Evaluate each factor
$$ \int_0^a r^3\,dr = \frac{a^4}{4}, \qquad \int_0^{\pi/2}\sin2\theta\,d\theta = \left[-\frac{\cos2\theta}{2}\right]_0^{\pi/2} = \frac12+\frac12=1. $$
Step 4 · Combine the results
$$ I = 2\cdot1\cdot\frac{a^4}{4} = \frac{a^4}{2}. $$
Additional Notes
- Verified numerically (with $a=2$): direct double integration gives $I=8$, matching $\frac{a^4}{2}=8$.
Result
The solution to the integral is: