Answer to Question #118319 in Calculus for Lizwi

Question

Find the area enclosed by the curve r(θ)=1+2sinθ and the rays θ=0 and θ=π/3

Answer 1

$S=\frac{1}{2}\int_{0}^{\pi/3}(1+2\sin\theta)^2d\theta=\frac{1}{2}\int_{0}^{\pi/3}(1+4\sin\theta+4\sin^2\theta)d\theta$ then

$S=\frac{1}{2}(\theta-4\cos\theta+4\sdot\frac{1}{2}(\theta-\frac{1}{2}\sin2\theta))|_{0}^{\pi/3}=\frac{1}{2}(\pi/3+2+2(\pi/3-\frac{\sqrt3}{4})$ $)$ then

$S=\frac{1}{2}(\pi+\frac{4-\sqrt3}{2})$