Find the polar moment of inertia \(I_0\) for a planar disk with uniform density and radius \(a\).

$$\int _{-a}^a\int _{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} \, \rho \left(x^2+y^2\right) \, dx \, dy$$

$$ \rho \int _{-a}^a \, \left( \frac{x^3}{3}+x y^2 \bigg| _{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} \right) \, dy$$

$$ \rho \int _{-a}^a \, \left( (\frac{\sqrt{a^2-y^2})^3}{3}+ y^2 \sqrt{a^2-y^2} + \frac{(\sqrt{a^2-y^2})^3}{3}+y^2 \sqrt{a^2-y^2} \right) \, dy$$

$$ \rho \int _{-a}^a \, \left( \frac{2(\sqrt{a^2-y^2})^3}{3}+ 2 y^2 \sqrt{a^2-y^2} \right) \, dy$$

$$ \frac{2 \rho}{3} \int _{-a}^a \, \left( (\sqrt{a^2-y^2})^3+ 3 y^2 \sqrt{a^2-y^2} \right) \, dy$$

$$ \frac{2 \rho}{3} \int _{-a}^a \, \sqrt{a^2-y^2} (a^2 + 2y^2) \, dy$$

$$ y = a \sin \theta \hspace{20mm} dy = a \cos \theta \, d\theta$$

$$\frac{2 \rho}{3} \int _{-\pi / 2}^{\pi/2} a \cos \theta \left( a^2 + 2 a^2 \sin^2 \theta \right) a \cos \theta \, d\theta$$

$$\frac{2 a^4 \rho}{3} \int _{-\pi / 2}^{\pi/2} \cos^2 \theta + 2 \sin^2 \theta \cos^2 \theta \, d\theta$$

$$\frac{2 a^4 \rho}{3} \int _{-\pi / 2}^{\pi/2} \cos^2 \theta + 2 \left( \frac{1}{2} – \frac{1}{2} \cos 2 \theta \right) \left( \frac{1}{2} + \frac{1}{2} \cos 2 \theta \right) \, d\theta$$

$$\frac{2 a^4 \rho}{3} \int _{-\pi / 2}^{\pi/2} \cos^2 \theta + \left( \frac{1}{4} – \frac{1}{4} \cos 4 \theta \right) \, d\theta$$

$$\frac{2 a^4 \rho}{3} \left[ \left( \frac{\theta}{2} + \frac{1}{4} \sin (2 \theta) \, \bigg|_{-\pi / 2}^{\pi/2} \right) + \left( \frac{\theta}{4} – \frac{1}{16} \sin (4 \theta) \, \bigg|_{-\pi / 2}^{\pi/2} \right) \right]$$

$$\frac{2 a^4 \rho}{3} \left[ \left( \frac{\pi}{4} + \frac{\pi}{4} \right) + \left( \frac{\pi}{8} + \frac{\pi}{8} \right) \right] = \frac{2 a^4 \rho}{3} \left( \frac{ 3 \pi}{4} \right) = \frac{\rho \pi a^4}{2}$$

Using Polar Coordinates

$$\int _{0}^{2 \pi} \int _{0}^{a} \, \rho r^2 r \, dr \, d\theta$$

$$\rho \int _{0}^{2 \pi} \frac{r^4}{4} \, \bigg|_0^a \, d\theta$$

$$\frac{\rho a^4}{4} \cdot 2 \pi = \frac{\rho \pi a^4}{2}$$