The area inside a polar curve is
∫1/2 r^2 dθ
The circles intersect where
15sinθ = 1
θ = 0.0667
So, from 0 to 0.0667, we have r = 15sinθ
From 0.0667 to π/2, we have to subtract the area inside r=1 from r=15sinθ, so the area is twice the area given below (due to symmetry)
∫[0,0.0667] 1/2 (15sinθ)^2 dθ
+ ∫[0.0667,π/2] 1/2 ((15sinθ)^2-1^2) dθ
or,
2(∫[0,π/2] 1/2 (15sinθ)^2 dθ - ∫[0.0667,π/2] 1/2 dθ)
Hint: the area inside 15sinθ is just the area of a circle of radius 15/2.
Find the length of the entire perimeter of the region inside r=15sin(theta)
but outside r=1.
I am not sure how to start the problem
2 answers
Oops. The part from 0 to 0.0667 is not included, because it's all inside r=1. So, the real area is just
2∫[0.0667,π/2] 1/2 ((15sinθ)^2-1^2) dθ
2∫[0.0667,π/2] 1/2 ((15sinθ)^2-1^2) dθ