Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 363: 33

$\text{The volume is}$ \begin{align} V = \frac{648\pi}{5} \end{align}

$\text{It is given that}$ \begin{align} y = \sqrt x; y = 0; x = 9 \end{align} $\text{The function revolves around x = 9. Thus,}$ \begin{align} y = \sqrt x \Rrightarrow f(y) = x = y^2 \end{align} $\text{The volume is}$ \begin{align} & V = \int_0^3 \pi (f(y) - 9)^2 \ dy = \int_0^3 \pi (y^2 - 9)^2 \ dy = \int_0^3 \pi (y^4-18y^2+81) \ dy = \\ & = \pi \left(\frac{y^5}{5} -6y^3+81 \right)_0^3 = \frac{648\pi}{5} \end{align}