Find the area of the region enclosed by the inner loop of the curve r = 4 + 8sin(θ) ?

Answer 1

The loop is generated for θ in [7π/6, 11π/6].
(this is from setting r = 0).

So, A = ∫(7π/6 to 11π/6) (1/2)(4 + 8 sin θ)^2 dθ
= ∫(7π/6 to 11π/6) (8 + 32 sin θ + 32 sin^2(θ)) dθ
= ∫(7π/6 to 11π/6) (8 + 32 sin θ + 16(1 – cos(2θ))) dθ
= ∫(7π/6 to 11π/6) (24 + 32 sin θ – 16 cos(2θ)) dθ
= (24θ – 32 cos θ – 8 sin(2θ)) {for θ = 7π/6 to 11π/6}
= (44π – 16√3 + 4√3) – (28π + 16√3 – 4√3)
= 16π – 24√3.

I hope this helps!

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