how do I find the integral of sqrt(16x - x^2)?
Not sure I could do it by hand anymore, but in a book of integrals, we find INT(SQRT(2ax-x^2)dx. That's your integral, with a=8. The answer is quite lengthy; too long for me to type. Worst case, get it from integrals.com or wolframalpha.com.
Like LBickford the answer is kind of complicated but here are the steps. (1) Complete the square inside the square root--SQRT(64- (x-8)^2) (2) Use u substitution-- u=x-8 (3) Use z substitution-- z = \[\sin \theta\]