The area of the closed region bounded by the polar graph of r=sqrt(3+cos theta) is given by what integral?

Answer: ∫ (3+cos theta) dθ from 0 to pi

How can you find this?

The area of the closed region bounded by the polar graph of r=sqrt(3+cos theta) is given by what integral?

Answer: ∫ (3+cos theta) dθ from 0 to pi

How can you find this?

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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

2π 2π

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

0 l 0

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

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