Q.439. Area of the surface of revolution

Question 439.
Find area of the surface of revolution generated by revolving around x axis
one loop of (8y^2) = (x^2)(1-x^2).

Answer 439.
Plugging y = 0
=> x^2 (1 - x^2) = 0
=> x = -1, 0 or 1
=> one loop is from x = 0 to x = 1

8y^2 = x^2 (1 - x^2) = x^2 - x^4
=> 16ydy/dx = 2x - 4x^3
=> ydy/dx = (1/8) (x - 2x^3)

Surface area, S
= 2π ∫ (x = 0 to 1) y √[1 + (dy/dx)^2] dx
= 2π ∫ (x = 0 to 1) √[y^2 + (ydy/dx)^2] dx
= 2π ∫ (x = 0 to 1) √[ (x^2/8)(1 - x^2) + (1/64)(x - 2x^3)^2] dx
= (π/4) ∫ (x = 0 to 1) √(8x^2 - 8x^4 + x^2 - 4x^4 + 4x^6) dx
= (π/4) ∫ (x = 0 to 1) √(9x^2 - 12x^4 + 4x^6) dx
= (π/4) ∫ (x = 0 to 1) (3x - 2x^3) dx
= (π/4) (3x^2/2 - x^4/2) ... (x = 0 to 1)
= π/4.

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