Area Under Polar Curve
The area of the region bounded by a curve r = ƒ(θ) and the radii vectors θ = and θ = is
1/2 ∫ r^{2} dθ.
Proof: Let ∠XOA = , ∠XOB = .
Let P (r, θ) and Q (r + Δr, θ + Δ θ) be two neighbouring points on the given curve such that the function r = ƒ(θ) is increasing (or decreasing) in the interval [θ, θ + Δ θ]. With O as centre and OP, OQ as radii, draw arcs PR and QS as shown in the figure. Suppose that
S = Area of the region OAP,
S + ΔS = Area of the region OAQ,
∴ ΔS = Area of the region OPQ.
We have PR = r Δ θ, QS = (r + Δr) Δ θ. Clearly, 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 r^{2} ΔS/(Δ θ) < 1/2 (r + Δr)^{2}. (1)
Let QP so that Δ θ0.
Then from (1) we obtain
dS/(d θ) = 1/2 r^{2}.
∴ 1/2 ∫ r^{2} dθ = ∫ dS/(d θ) dθ = |S|
= [S]θ = - [S]θ =
= Area of AOB – 0.
Hence the required area = 1/2 ∫ r^{2} dθ
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