The leaf in the first quadrant touches r = 0 when 2 sin(2θ) = 0
==> 2θ = 0, π ==> θ = 0, π/2.
Hence, the area of this leaf equals ∫∫ 1 dA
= ∫(θ = 0 to π/2) ∫(r = 0 to 2 sin(2θ)) 1 * (r dr dθ).
I hope this helps!
==> 2θ = 0, π ==> θ = 0, π/2.
Hence, the area of this leaf equals ∫∫ 1 dA
= ∫(θ = 0 to π/2) ∫(r = 0 to 2 sin(2θ)) 1 * (r dr dθ).
I hope this helps!