This was posed as an extra credit problem. I tried a hodge podge
explanation with a picture, some shading, and some integral signs, but it
didn't fly. Think Riemann sums he said. So I have. but I'm not sure
what to do. I used definition of the indefinite integral using Riemann
sums. I cannot really see what happens to the n. This is tough
expressing without proper symbols but Ill do my best.
heres what I did:
limit Summation i=1 n=infinity f(x)*(b-a)/n
f(x)= +/-sq rt(r^2-x^2) and (b-a)/n = 2r/n
then I said x=r*sin(@[EMAIL PROTECTED]
) so then f(x)= r*cos(@[EMAIL PROTECTED]
)...I used a triangle and went
back into x, but am I at least on the right track? Isnt the goal hear to
get 1/n to go to zero? But not the whole thing, because I know that I
want to arrive at 2pi*r is the trick to multiply by the conjugate or
something?
