Answer
$$\dfrac{\pi}{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^1_0 \int^{\sqrt{z}}_0 \int^{2\pi}_0 (r^2cos^2\theta+z^2)r \space d\theta \space dr \space dz =\int^1_0 \int^{\sqrt{z}}_0 [\dfrac{r^2\theta}{2}+\dfrac{r^2 }{4} (\sin2\theta)+z^2 \theta]^{2\pi}_0 \space r \space dr \space dz \\=\int^1_0 \int^{\sqrt{z}}_0(\pi \times r^3+(2\pi r) \times z^2) \space dr \space dz \\=\int^1_0 [\pi \times \dfrac{ r^4}{4}+\pi \times r^2 \times z^2]^{\sqrt{z}}_0 \space dz \\=
\int^1_0 (\dfrac{\pi z^2}{4}+\pi z^3) \space dz \\=\dfrac{\pi}{3}$$
