1. ## Integration.. mistake?

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?

2. You have set up the integral correctly, but your evaluation of the integral is wrong. It looks like you've expanded the brackets, but kept the square there as well, and then tried to use the chain rule...

$\displaystyle \displaystyle \int{(6-\sqrt{x})^2\,dx} = \int{36 - 12\sqrt{x} + x\,dx}$

$\displaystyle \displaystyle = \int{36 - 12x^{\frac{1}{2}} + x\,dx}$.

Now just integrate term-by-term.

3. So using the chain rule here is wrong?

4. Yes