Answer
$$\dfrac{7\pi}{15}$$
Work Step by Step
Consider the shell model to compute the volume:
$$Volume=\int_{m}^{n} (2 \pi) (\space Radius \space of \space shell) \times ( height \space \text{of} \space \text {shell}) dx \\= \int_0^{1} (2 \pi) \cdot (x)[\sqrt x-(2x-1)] dx \\ =2 \pi [(\dfrac{2}{5})x^{5/2}-(\dfrac{2}{3})x^3+\dfrac{x^2}{2}]_0^1\\ = 2 \pi \times (\dfrac{2}{5}-\dfrac{2}{3}+\dfrac{1}{2}) \\=\dfrac{7\pi}{15}$$
