Area Under Polar Curve:

The area of the section bounded by a curve r = ƒ(θ) and the radii vectors θ = α and θ = β is

1/2 ∫β  r² dθ.

Proof: Let ∠XOA = α , ∠XOB = β.

Let P (r, θ) and Q (r + Δr, θ + Δ θ) be two neighboring points on the provided curve such that the function r = ƒ(θ) is decreasing or increasing in the interval [θ, θ + Δ θ]. With O as centre and OP, OQ as radii, draw arcs QS and PR as given in the figure. Consider that

S = Area of the section OAP,

S + ΔS = Area of the section OAQ,

∴ ΔS = Area of the section OPQ.

We have PR = r Δ θ, QS = (r + Δr) Δ θ. obviously area of the circular sector OPR < ΔS < area of the circular sector OQS.

i.e. 1/2 OP.PR < ΔS < 1/2 OQ.QS

i.e. 1/2 r.r Δ θ < ΔS < 1/2 (r + Δr) (r + Δr) Δ θ

i.e. 1/2 r² ΔS/(Δ θ) < 1/2 (r + Δr)².                                   (1)

Let Q -> P so that Δ θ -> 0.

Then from (1) we get

dS/(d θ) = 1/2 r².

∴ 1/2 ∫β  r² dθ = ∫βα  dS/(d θ) dθ = |S|βα

= [S]θ =β- [S]θ = α

= Area of AOB - 0.

Hence the essential area = 1/2 ∫βr²dθ