Volume of revolution
= π ∫ y² dx
= π ∫x²√(a²-x²) dx

Use the substitution x = a sin u

dx/du = a cos u
dx = a cos u du

And,
x = a sin u → x/a = sin u → u = sin-¹(x/a)

The original upper bound is x = ½a and the lower bound is x = 0

When x = ½a, u = sin-¹ (½a/a) = sin-¹(½) = π/6

When x = 0, u = sin-¹(0/a) = sin-¹ 0 = 0

So our upper limit and lower limit has been changed.


So, π ∫x²√(a²-x²) dx

= π ∫ a² sin² u √(a² - a²sin² u) (a cos u) du

= π ∫ a³ sin² u cos u √(a² cos² u) du

= π ∫ a³ sin² u cos² u (a cos u) du

= π ∫ a⁴ sin² u cos² u du

= a⁴π ∫ (sin u cos u)² du

= a⁴π ∫ (½sin 2u)² du

= a⁴π ∫ ¼ sin² 2u du

= a⁴π ∫ ¼(1 - cos 4u)/2 du

= ⅛ a⁴π [u - ¼ sin 4u]

= ⅛ a⁴ π ([π/6 - ¼ sin 4π/6] - [0 - ¼ sin 4(0)] )

= ⅛ a⁴ π [π/6 - ¼ (√3 / 2)]

= ⅛ a⁴ π [π/6 - √3 / 8]

= 1/16 a⁴ π [π/3 - √3 / 4]


So, 1/16 a⁴ π [π/3 - √3 / 4] = 567

a⁴ = 567 / (1/16 π [π/3 - √3 / 4])

a⁴ = 9072 / (π [π/3 - √3 / 4])

a = ⁴√(9072 / (π [π/3 - √3 / 4])

(Since a is positive, we don't consider the negative solution of a)


a ≈ 8.28063262688

a = 8.28 (2.dp)