Question

find the area of the resulting surface.

x=sq.rt.(a^2-y^2), 0<=y<=a/2

I have it so the integral is set up at:
2pi(integrate) sq.rt.(a^2-y^2)sq.rt(1+(-y/sq.rt.(a^2-y^2))^2)
I do not know where to go from here.

Answer 1

\[2\pi \int\limits_{0}^{a/2}\sqrt{a^2-y^2}\sqrt{1+(\frac{ -y }{ \sqrt{a^2-y^2} })^2}\]

Answer 2

simplify the terms in the second square root :)

Answer 3

thats what I thought to do, but after you said that I looked at it again and I didn't get rid of the -y once i squared it, I kept it as -y^2, that was my problem, thank you.

Answer 4

:)