AREA BOUNDED BY A POLAR CURVE

The area of the region bounded by a polar curve, r = f (θ ) and the lines θ = α and θ = β is given by: β α A = 1 2 r 2 dθ θ = β θ = α θ = 0 O

Example 1: Find the area bounded by the Polar curve shown, with equation r = 3 sin 2θ ; 0 ≤ θ ≤. π2π     – π2π2 0 θ – sin 4θ 4   – ( 0 – 0 ) 9π 8 = π2π2 –   = 0 A = 1 2 (3sin 2θ ) 2 dθ π2π2 – 0 A = 9 2 sin 2 2θ dθ π2π2 – = A = – π2π2 1 – cos 4θ dθ cos 2A = 1 – 2sin 2 A Using: β α A = 1 2 r 2 dθ

Example 2: The diagram shows a sketch of the curves C 1 and C 2 with polar equations C 1 : r = 1 + cos θ ; 0 ≤ θ ≤ π and C 2 : r = 3 (1 – cos θ ) ; 0 ≤ θ ≤ π. The curves intersect at the point A as shown. The region inside C 1 and outside C 2 is shown shaded. Find the polar coordinates of A and the area of the shaded region. A For A: 1 + cos θ = 3 (1 – cos θ ) Hence, A is the point 3232 π3π3, 3232 cos θ = 1 2 θ = π3π3 r = 1 + cos θ = 3 – 3 cos θ 4 cos θ = 2

A = – π3π3 (1 + cos θ ) 2 dθ – – π3π3 9(1 – cos θ ) 2 dθ A = 4 π – 9 – 4π 2 = 36 π – cos 2A = 2cos 2 A – 1 Expand the brackets and use: θ = π3π3 r = 1 + cos θ r = 3 (1 – cos θ )

Summary of key points: This PowerPoint produced by R.Collins ; Updated Sep The area of the region bounded by a polar curve, r = f (θ ) and the lines θ = α and θ = β is given by: β α A = 1 2 r 2 dθ cos 2A = 1 – 2sin 2 A It is often necessary to use cos 2A = 2cos 2 A – 1 or to integrate sin 2 A or cos 2 A.