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The area of the closed region?

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2 Answers

area = ∫ (1/2)r^2 dθ from 0 to 2pi
= ∫ (1/2)(3+cosθ) dθ from 0 to 2pi
= ∫ 2(1/2)(3+cosθ) dθ from 0 to pi
= ∫ (3+cosθ) dθ from 0 to pi
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Attn: ∫ cosθ dθ from 0 to pi = 0

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   2π                                     2π
 ∫   ( 3 + cos θ )dθ  = 3θ - Sin θ l     = 6π - 0 = 6
   0                                     l  0 

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You misunderstood the question. What I meant was that how to find the answer I provided above?

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