Answer
$\dfrac{9 \pi }{8}$
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, $V=\pi \int_{1/2}^{2} [(\dfrac{5}{2}-y)^2-(\dfrac{1}{y})^2] \ dy \\ = \pi \int_{1/2}^{2} [(\dfrac{25}{4}-5y+y^2-y^{-2}] \ dy \\=\pi [\dfrac{25y}{4}-\dfrac{5y^2}{2}+\dfrac{y^3}{3}-\dfrac{y^{-1}}{(-1)}]_{1/2}^2\\=\dfrac{9 \pi }{8}$
