Answer
$6 \pi$
Work Step by Step
Consider the shell model to compute the volume:
$V=\int_{m}^{n} (2 \pi) (\space Radius \space of \space shell) \times ( height \space \text{of} \space \text {shell}) dx \\=\int_0^{2} (2 \pi) \cdot x (1+\dfrac{x^2}{4}) dx \\= \int_0^{2} (2 \pi ) (x+\dfrac{x^3}{4}) dx\\= [\dfrac{(2\pi ) x^2}{2}+\dfrac{(2\pi ) x^4}{16}]_0^{2} dx \\=6 \pi$