Find the exact length of the polar curve. r = θ^2, 0 ≤ θ ≤ 3π/4
Good work. A small correction, in the integration step:
= integral(0 to 3π/4) θ sqrt(θ^2 + 4) dθ
= 1/2 * integral(0 to 3π/4) 2θ sqrt(θ^2 + 4) dθ
= 1/2 * (2/3) (θ^2 + 4)^(3/2) (for θ from 0 to 3π/4)
= (1/3) (θ^2 + 4)^(3/2) (for θ from 0 to 3π/4)
= (1/3) [((9π^2 /16) + 4)^(3/2) – 4^(3/2)]
You can’t simplify more than that, you have to use a calculator and find a decimal approximation.
dr/dθ = 2θ
and the arc length from θ = 0 to 3π /4 is:
∫√(1 + (2θ)²) dθ
let 2θ = tan(α)
so 2 dθ = sec²β dβ
and the integral is:
∫√(1 + tan²β) * (1/2)sec²β dβ
=
(1/2) ∫√(1 + tan²β) * sec²β dβ
using the identity:
1 + tan²β = sec²β
We have:
(1/2) ∫sec³β dβ
=
(1/2) ∫secβ (1 + tan²β) dβ
=
(1/2) ∫secβ dβ + (1/2) ∫secβ tan²β dβ
=
(1/2) ∫ secβ dβ + (1/2) ∫ sin²β / cos³β dβ
the first integral is:
(1/2) ln|secβ + tanβ|
I can come back later, it’s a difficult question.