Find the volume of the solid that results when the region enclosed by $\displaystyle y=\sqrt{x}$, y=0 and x=9 is revolved about the line y=3.

Here is the image:

Now the problem is that I can't even imagine the revolution. What figure would come up when the revolution is done?

Should the volume be $\displaystyle \int_{0}^{9} \pi (3-\sqrt{x})^2dx$

**That's the volume of the internal conical part, because the radius of the discs is the difference between y=3 and the function.**
This will calculate the "conical" volume first, which can be subtracted from a cylinder.

An indirect way to calculate the volume you want.

or maybe $\displaystyle \int_{0}^{9} \pi (3^2 - (3-\sqrt{x})^2)$

This is so strange.

It seems like the graphic of $\displaystyle \sqrt{x}$ is below the y=9 axis, which is not actually consistent with the definition.

It is like revolution of negative function around the x-axis. Very strange.