交点:(0,0),(1,1)
A = π∫(0~1) [(√y)² - (y²)²] dy
= π∫(0~1) (y - y⁴) dy
= π[y²/2 - y⁵/5] |(0~1)
= π[1/2 - 1/5]
= 3π/10
或
A = 2π∫(0~1) x(√x - x²) dx
= 2π∫(0~1) (x√x - x³) dx
= 2π[(2/5)x^(5/2) - x⁴/4] |(0~1)
= 2π[2/5 - 1/4]
= 3π/10
交点:(0,0),(1,1)
A = π∫(0~1) [(√y)² - (y²)²] dy
= π∫(0~1) (y - y⁴) dy
= π[y²/2 - y⁵/5] |(0~1)
= π[1/2 - 1/5]
= 3π/10
或
A = 2π∫(0~1) x(√x - x²) dx
= 2π∫(0~1) (x√x - x³) dx
= 2π[(2/5)x^(5/2) - x⁴/4] |(0~1)
= 2π[2/5 - 1/4]
= 3π/10