For each x in the interval 0 ≤ x ≤ 5, the shell at that point has
• radius = 5 - x, which is the distance from x to x = 5
• height = x ² + 2
• thickness = dx
and hence contributes a volume of 2π (5 - x) (x ² + 2) dx.
Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:
[tex]\displaystyle 2\pi \int_0^5 (5-x)(x^2+2)\,\mathrm dx=2\pi\int_0^5 (10-2x+5x^2-x^3)\,\mathrm dx=\boxed{\frac{925\pi}6}[/tex]