Evaluate the area of a surface of revolution the curve given by a graph of f(x) = x^3 , 1 ≤ x ≤ 2, rotate about the y-axis.

Question

NEED ASAP! Please provide step by step solutions and explanations. Thank you!

Answer 1

S = 2π∫₁²xsqrt(1 + (f'(x))²)dx. Change of variables tanθ = 3x², sec²θdθ = 6xdx, xdx = sec²θdθ/6 and we get

S = π/3∫sec²θsqrt(1 + tan²θ)dθ ∫=π/3∫sec³θdθ from θ = tan^-13 to θ = tan^-112. Using integral table we get S = π/6(secθtanθ + ln(secθ + tanθ)) fro tan^-13 to tan^-112.

tan(tan^-13) = 3, tan(tan^-112) = 12;

sec(tan^-13) = sqrt(1 + 3²) = sqrt(10)

sec(tan^-112) = sqrt(1 + 12²) = sqrt(145)

S = π/6(((12sqrt(145) + ln(12 + sqrt(145)) - (3sqrt(10)+ ln(3 + sqrt(10))) ≈ 71.41