Answer
$$\dfrac{8\pi a^5}{15}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{2a}_0 \int^{a}_0 \int^{\sqrt{a^2-r^2}}_{-\sqrt{a^2-r^2}} r^3 \space dz \space dr \space d\theta \\=\int^{2\pi}_0 \int^a_0 2r^3\sqrt{a^2-r^2} \space dr \space d\theta \\=2 \times \int^{2\pi}_0 [(-\dfrac{r^2}{5}-\dfrac{2a^2}{15})(a^2-r^2)^{3/2}]^a_0 \space d\theta \\=(2) \times \dfrac{2}{15} \times \int^{2\pi}_0 a^5d\theta \\=\dfrac{8\pi a^5}{15}$$
