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