I use $\begin{cases} x=R\cos \theta\\ y=R\sin \theta\\ z=z \end{cases}$ , then $dS=Rd\theta dz$, and I find $$\int_{0}^{2\pi}\int_0^RR(R^2\cos^2\theta+z^2)d\theta dz=\frac{5\pi R^4}{3}.$$
But my classmate got $\frac{4\pi R^4}{3},$ which is right? Can anyone help me, thank you!
The correct result is yours: $$\int_{0}^{2\pi}\left(\int_0^RR(R^2\cos^2\theta+z^2)dz\right)d\theta = \int_{0}^{2\pi}(R^4\cos^2\theta+\frac{R^4}{3})d\theta=\pi R^4+\frac{2\pi R^4}{3}=\frac{5\pi R^4}{3}.$$