Find the area of the region that lies inside both curves r = 3 sin θ, r = 3 cos θ.

Answer 1

Polar angle of intersection: 3sinθ = 3cosθ; tanθ = 1; θ = π/4

A =1/2∫₀^π/49sin²θdθ + 1/2∫_π/4^π/29cos²θdθ = 9/4∫₀^π/4(1 - cos2θ)dθ + 9/4 ∫_π/4^π/2(1 + cos2θ)dθ =

9/4(θ - 1/2sin2θ)₀^π/4 + 9/4(θ + 1/2sin2θ)_π/4^π/2 = 9π/16 - 1/2 + 9π/8 - 9π/16 - 1/2 = 9π/8 - 1

Answer 2

First we need to set these equations equal to one another and solve for theta and we will find that these two polar curves intersect at pi/4

Now we need to use integration to find the area and this can be set up by doing 1/2 times the integral from 0 to pi/4 of (3sinθ)^2 + 1/2 times the integral from pi/4 to pi/2 of (3cosθ)^2 which will give you 1.284. This can also be evaluated by hand using power reducing formulas if that's how they want the answer.