Answer
$6 \pi$
Work Step by Step
Consider the shell model to compute the volume:
$V=\int_{m}^{n} (2 \pi) (\space Radius \space of \space shell) \times ( height \space \text{of} \space \text {shell}) dx \\V= \int_0^{2} (2 \pi) (x) \times (2-\dfrac{x^2}{4}) dx \\= \int_0^{2} 2 \pi (2x-\dfrac{x^3}{4}) dx \\=6 \pi$