# Calculus HELP plz

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Find the area of the surface obtained by rotating the curve y=sqrt(4x) from x=0 to x=1 about the -axis.

surface area=
⌠ 2pi * f(x)*sqrt[1 + (f'(x))^2] dx from a to b

it's been 45 years since I used that formula.

I ended up with the integral of (2pi(4x + 4)^(1/2) dx from 0 to 1

looks pretty straighforward after this, let me know whether it worked out.

BTW, what level is this from, surely not highschool.

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