The figure shows the curve represented by the polar equation r=cos 3θ ,0≤θ≤π. Find the area of the region enclosed by the curve?

If you read literature

the area inside a curve

r = f(θ)

is given by

Area =∫ (1/2) (f(θ)^2 dθ

If you look at the figure and data

you can see one half of one of the three identical leaves

is found for θ = 0 to pi/6 (from 0 to pi/6 draws

the top half of 1st leaf )

so (1/2) of a leaf = this section

and you have 3 leave *2 *size of half of leaf = 6 size of half a leaf

total Area = 6 * From 0 to pi/6 ∫ (1/2) (cos(3θ ))^2 dθ

total Area = 6*(1/2) From 0 to pi/6 ∫ (1/2) cos^2(3θ ) dθ

total Area = 3* From 0 to pi/6 ∫ cos^2(3θ )dθ

since cos(2x) = 2*cos^2(x) - 1

cos^2(x) = (1/2)cos(2x) + 1/2

so cos^2(3θ) = (1/2) cos(6θ) + 1/2

total Area = 3* From 0 to pi/6 ∫ ((1/2)cos(6θ) +1/2 )dθ

total Area = 3* (1/2) *From 0 to pi/6 ∫ (cos(6θ) + 1) dθ

total Area = (3/2) *( (1/6)sin(6θ) + θ ) | From 0 to π/6

total Area = (3/2) * ( (1/6)*sin(π) + π/6) - (0 + 0)

total Area = (3/2) *( 0 + π/6) = (3/2)*(π/6) = (1/2)*(π/2) = π/4

total Area = π/4