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Find the area of the polar region enclosed by f(θ)=sin(θ) for 0≤θ≤π/6.?
Update:
I don't understand why the answer isn't 0.0226. I would really appreciate some help with this problem!
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- kbLv 76 years agoFavorite Answer
The area equals
∫ (1/2) r^2 dθ
= ∫(θ = 0 to π/6) (1/2)(sin θ)^2 dθ
= ∫(θ = 0 to π/6) (1/2) * (1/2)(1 - cos(2θ)) dθ, via half angle identity
= (1/4)(θ - sin(2θ)/2) {for θ = 0 to π/6}
= (1/4)(π/6 - √3/4)
= (1/4) * (1/12)(2π - 3√3)
= (1/48)(2π - 3√3).
I hope this helps!
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