$\int_0^{2\pi} \int_0^1 2r^2 \, dr \, d\theta$

$\int_0^{2\pi} \int_0^1 4\sqrt{r^2 + 1} \cdot r \, dr \, d\theta$

$\int_0^{2\pi} \int_0^1 \sqrt{4 + 16r^2} \, dr \, d\theta$

$\int_0^{2\pi} \int_0^1 4r \, dr \, d\theta$