Answer
$$\dfrac{\pi}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{2\pi}_0 \int^1_0 \int^{1/2}_{-1/2}(r^2 \times \sin^2\theta+z^2) \space dz \space r \space dr \space d\theta =\int^{2\pi}_0 \int^1_0 (r^3 \times \sin^2\theta+\dfrac{r}{12}) \space dr \space d\theta \\=\int^{2\pi}_0 (\dfrac{\sin^2\theta}{4}+\dfrac{1}{24}) \space d\theta \\=\dfrac{\pi}{3}$$
