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: 34

$\text{The volume is}$ \begin{align} & V = \frac{135\pi}{2} \end{align}

$\text{It is given that}$ \begin{align} y = \sqrt x; y = 0; x = 9 \end{align} $\text{The function revolves around y = 3. Also, when x = 9}$ \begin{align} y = \sqrt 9 = 3 \end{align} $\text{The volume is}$ \begin{align} & V = \pi \int_0^9 (9-(\sqrt x - 3)^2) \ dx =\pi \int_0^9 (6\sqrt x - x) \ dx= \pi \left( 108 -\frac{81}{2}\right) = \frac{135\pi}{2} \end{align}