Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D \sqrt{x^2+y^2}\,dx\,dy, $$ where $D$ is the disk $x^2+y^2\leqslant a^2$.

Step 1 · Convert to polar coordinates

With $x=r\cos\theta$, $y=r\sin\theta$, the disk becomes $0\leqslant r\leqslant a$, $0\leqslant\theta\leqslant2\pi$, and $\sqrt{x^2+y^2}=r$:

$$ I = \int_0^{2\pi}\int_0^a r\cdot r\,dr\,d\theta = \int_0^{2\pi}\int_0^a r^2\,dr\,d\theta. $$

Step 2 · Integrate with respect to $r$

$$ \int_0^a r^2\,dr = \left[\frac{r^3}{3}\right]_0^a = \frac{a^3}{3}. $$

Step 3 · Integrate with respect to $\theta$

$$ I = \int_0^{2\pi} \frac{a^3}{3}\,d\theta = 2\pi\cdot\frac{a^3}{3} = \frac{2\pi a^3}{3}. $$

Additional Notes

• Geometrically, $I$ is the volume of the solid bounded below by the plane $z=0$ and above by the cone $z=\sqrt{x^2+y^2}$, over the disk $D$. At height $z_0$, the cross-section is the annulus $z_0\leqslant r\leqslant a$, so this solid is exactly a cylinder of radius $a$ and height $a$ (volume $\pi a^3$) with the solid cone $r\leqslant z\leqslant a$ (volume $\frac13\pi a^3$) drilled out of its core — giving $\pi a^3-\frac13\pi a^3=\frac{2\pi a^3}{3}$, matching the result above.