Answer
Area between $y= \sqrt{x}$, $y=0$ , $x=0$ and $x=4$ revolves about $y=3 $
Work Step by Step
Given
$$V= \pi \int_1^4\left[3^2-(3-\sqrt{x})^2\right] d x$$
Compare with
$$V= \pi \int_a^b [R^2(x )-r^2(x)]dx $$
where $R(x)$ is the outer radius and $r(x)$ is the inner, here
\begin{aligned}
R(x)&= 3\\
r(x)&= 3-\sqrt{x}
\end{aligned}
This means that the bounded area between $y= \sqrt{x}$, $y=0$ , $x=0$ and $x=4$ revolves about $y=3$.