Find the surface area of revolution about the x-axis of y=6x+6 over the interval 0 <= x <= 2

Answer 1

The surface area is found by combining a lot of areas: all will have the circumference of 2πy multiplied by the length of the segment. You can find the length of the segment by using the Pythagorean theorem on the change in x (dx/dx = 1) and change in y (dy/dx = 6); this makes length = √(1²+6²) = √37. This means we'll integrate these areas of 12π√37(x+1) over the bounds of x=0 to 2.∫

12π√37 ∫₀²x+1 dx = 12π√37 (x²/2 + x)|₀²

⁼ 12π√37 (4/2 + 2)

= 48π√37

= 917.259 sq. units

Answer 2

S=∫2πyds, ds=√1+(dy/dx)² dx

dy/dx = 6, ds=√1+6² dx, ds=√37dx,

S=∫2π(6x+6)√37dx(0 to 2)

S=2π√37∫(6x+6)dx(0 to 2)

S=2π√37(3x²+6x)Ι(0 to 2)

S=48√37 π≈917.259