Question: The region R is bounded by the x-axis and the line $\displaystyle x=16$ and the curve with the equation $\displaystyle y=6-\sqrt{x}$ where $\displaystyle 0\leq{x}\leq36$. Find in terms of $\displaystyle \pi$ the volume of the solid generated when R is rotated through one revolution about the x axis.

Here's what I've done:

$\displaystyle \int (6-\sqrt{x})^2 \pi dx$ (Integrating between the bounds 16 and 0 didn't know how to display that with latex)

After integrating I have

$\displaystyle [\frac {1}{2} (36-12\sqrt{x}+x]^2 \pi$ again for x= 16 and 0

solving that gives me the wrong answer

any ideas?