s= π/27 [ 2305^(3/2) - 1]

= 2π ∫ x^3 √(1 + (3x^2)^2) dx = 2π ∫ x^3 √(1 + 9x^4) dx let u = 1 + 9x^4 .. . . du = 36x^3 dx .. . .. . S = 2π (1/36) ∫ u^(1/2) du = π/18 u^(3/2) / (3/2) = π/27 (1 + 9x^4)^(3/2) ... from 0 to 4

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