Find the area of the surface generated by revolving the given curve about the x-axis . Y=√x,0 ≤x≤2

Question

I am having trouble using the u substitution to solve the problem. First, I used the s=integral 2pi√1+(dx/dy)^2 dx formula but got stuck on the u substitution section.

Answer 1

Area of the surface = 2π Int (yds) = 2π Int from 0 to 2 of (y sqrt(1 + (dy/dx)²)dx) You are rotating the s(x) arc length about the x axis, so it looks like you are solving the wrong integral. Check the derivation of the differential of the arclength formula and also the surface area of revolution formula.

2π Int from 0 to 2 of (sqrt(1+1/(4x))sqrt(x)dx) 1/4x = (dy/dx)²

the integrand can be simplified to sqrt(x)*(sqrt(x + 1/4)/sqrt(x) or just sqrt(x+ 1/4) which is easily integrable.