Final answer:
The area of the region enclosed by both curves r = 6sinθ and r = 6 - 6sinθ is 12√3π square units.
Step-by-step explanation:
To find the area of the region enclosed by both curves, we need to determine the points of intersection between the two curves. We can set the equations r = 6sinθ and r = 6 - 6sinθ equal to each other and solve for θ.
6sinθ = 6 - 6sinθ
12sinθ = 6
sinθ = 0.5
θ = π/6 or 5π/6
The area enclosed by the curves can be found by integrating the difference between the two curves with respect to θ, from π/6 to 5π/6.
A = ∫(6sinθ - (6 - 6sinθ)) dθ
A = ∫(12sinθ - 6) dθ
A = [-12cosθ - 6θ] from π/6 to 5π/6
A = [-12cos(5π/6) - 6(5π/6)] - [-12cos(π/6) - 6(π/6)]
A = 6√3π - 6π - (-6√3π + 6π)
A = 12√3π square units