Question

Find the area of the region inside: $r = 5 \sin \theta$ but outside:

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Find the area of the region outside $r = 6 + 6 \sin \theta$ , but inside $r = 18 \sin \theta$ .

Answer 1

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r = 5sinθ
r = 2



A = (1/2) , where outer and inner radii aredetermined like (top - bottom) for regular area
outer radius = 5sinθ
inner radius = 2

finding limits of integration:
5sinθ = 2
sinθ = 2/5
θ = arcsin(2/5) = .4115, 2.73

A =
A =
power-reducing identity: sin²θ =
A =
A =
A = (1/2)[(25/2)[θ - sin(2θ)/2] - 4θ]
A = (25/4)[θ - sin(2θ)/2] - 2θ
A = (25/4)θ- (25/8)sin(2θ) - 2θ

Evaluate: A = 13.894 - (-.5423) = 14.4363

r = 6 + 6sinθ
r = 18sinθ

A = (1/2)
outer radius = 18sinθ
inner radius = 6 + 6sinθ

solving for limits of integration:
18sinθ = 6 + 6sinθ
18sinθ - 6sinθ = 6
12sinθ = 6
sinθ = 1/2
θ = π/6, 5π/6

A =
A =
A =
A =
A =
A =
A =
A =
A =
A = -72[sin(2θ)/2] + 36cosθ + 54θ = -36sin(2θ)+ 36cosθ + 54θ

Evaluating:
A = 45π - 9π = 36π