Answer
$$\dfrac{5 \pi}{6}$$
Work Step by Step
Consider the shell model to compute the volume:
$$V=\int_{m}^{n} (2 \pi) (\space Radius \space of \space shell) \times ( height \space \text{of} \space \text {shell}) dx \\= \int_0^{2} (2 \pi) \cdot (x)[2-x-x^2] \space dx \\ =2 \pi \times [x^2-\dfrac{x^3}{3}-\dfrac{x^4}{4}]_0^1 \\= (2 \pi) (1-\dfrac{1}{3}-\dfrac{1}{4}) \\=\dfrac{5 \pi}{6}$$