Answer
$\dfrac{ 49 \pi }{30}$
Work Step by Step
The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
Now, $V=\pi \int_{0}^{1} [(\sqrt y+2)^2-(y^2+2)^2] \ dy \\ = \pi \int_0^1 [(y+4 \sqrt y+4) -(y^4+4y^2+4) ] \ dy \\=\pi \int_0^1 [-y^4-4y^2+y+4 \sqrt y ] \ dy \\= \pi[\dfrac{-y^5}{5}-\dfrac{4y^3}{3}+\dfrac{y^2}{2}+\dfrac{8y^{3/2}}{3}]_0^1 \\=\dfrac{ 49 \pi }{30}$
