Answer
$\dfrac{2 \pi m^5}{15}$
Work Step by Step
It has been noticed that the points of intersection are at $(0,0)$ and $(m^2, m^2)$.
Apply the shell method to compute the volume of the region.
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy \\=2\pi \int_0^{m^2} y \times (\sqrt y-\dfrac{y}{m}) \ dy \\=2 \pi [\dfrac{2y^{5/2}}{5} -\dfrac{y^3}{3m}]_0^{m^2} \\=2 \pi [\dfrac{2m^5}{5}-\dfrac{m^5}{5}] \\=\dfrac{2 \pi m^5}{15}$
