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Let A ( θ ) = 1 2 ∫ 0 θ [ 2 ( 1 + cos φ ) ] 2 d φ A(\theta)=\dfrac{1}{2}\displaystyle\int_{0}^{\theta}\bigl[2\bigl(1+\cos\varphi\bigr)\bigr]^2\,d\varphi A ( θ ) = 2 1 ∫ 0 θ [ 2 ( 1 + cos φ ) ] 2 d φ …
Mathematics
Problem Let A ( θ ) = 1 2 ∫ 0 θ [ 2 ( 1 + cos φ ) ] 2 d φ A(\theta)=\dfrac{1}{2}\displaystyle\int_{0}^{\theta}\bigl[2\bigl(1+\cos\varphi\bigr)\bigr]^2\,d\varphi A ( θ ) = 2 1 ∫ 0 θ [ 2 ( 1 + cos φ ) ] 2 d φ denote the area swept out by the curve r = 2 ( 1 + cos θ ) r=2\bigl(1+\cos\theta\bigr) r = 2 ( 1 + cos θ ) as φ \varphi φ increases from 0 0 0 to θ \theta θ , where 0 ≤ θ ≤ π 0\le\theta\le\pi 0 ≤ θ ≤ π . For which value of θ \theta θ is A ( θ ) A(\theta) A ( θ ) maximized?
a) 3 π 4 \dfrac{3\pi}{4} 4 3 π
b) π 2 \dfrac{\pi}{2} 2 π
c) π \pi π
d) 2 π 3 \dfrac{2\pi}{3} 3 2 π
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