Integral check to find the area generated by revolving y = cosx, 0<=x<=pi/4 about the x-axis?

Question

I belive if I take 2pi integral from 0 to pi/4 cosx sqrt[1 + sin^2x] dx I will get the correct area is my thinking accurate?

Answer 1

your disc will have a radius of y = cosx
Area of the disc is πcos^2x, so you have
π/4
∫cos^2x dx =
0

π/4
∫(1/2(1 + cos2x) dx =
0
π/8 - 0 + (1/4)sin(π/4) - 0 =
π/8 + (1/8)√2 =
V = 0.56948

but you wanted area:
dS = 2π√(1 + sin^2x) dx
S = π√2

You are correct.

Answer 2

i believe if you take 2pi internally, that you will have a 2x belly ache in the morning