So here's the graph of the relevant portion of $f(x) = \sqrt x$:
!enter image description here
And here is a pic of the resulting solid with the same portion shaded for comparison:
!enter image description here
Imagine taking slices of this solid as you move along the x-axis from 0 to 3. At each point x, the cross-sectional area $A(x)$ is $(\sqrt x)^2$. Now integrate $A(x)$ over the interval $[0, 3]$, so $V = \int _0^3(\sqrt x)^2dx = \int_0^3 xdx = \frac 9 2$