Answer to Question #257425 in Calculus for putri

Question #257425
find the area outside circle r = 1/2 and inside circle r = cosθ



1
Expert's answer
2021-10-27T15:42:29-0400

The two curves intersect where 1/2=cosθ.1/2=\cos \theta. So θ=π/3\theta=-\pi/3 or π/3.\pi/3.

The area we want is then


12π/3π/3((cosθ)2(1/2)2)dθ\dfrac{1}{2}\displaystyle\int_{-\pi/3}^{\pi/3}((\cos \theta)^2-(1/2)^2)d\theta


=18π/3π/3(2(1+cos(2θ))1)dθ=\dfrac{1}{8}\displaystyle\int_{-\pi/3}^{\pi/3}(2(1+\cos(2\theta))-1)d\theta

=18π/3π/3(1+2cos(2θ))dθ=\dfrac{1}{8}\displaystyle\int_{-\pi/3}^{\pi/3}(1+2\cos(2\theta))d\theta

=18[θ+sin(2θ)]π/3π/3=\dfrac{1}{8}[\theta+\sin(2\theta)]\begin{matrix} \pi/3 \\ -\pi/3 \end{matrix}


=18(π3+sin(2π3)(π3)sin(2π3))=\dfrac{1}{8}(\dfrac{\pi}{3}+\sin(\dfrac{2\pi}{3})-(-\dfrac{\pi}{3})-\sin(\dfrac{-2\pi}{3}))

=(π12+38)(units2)=(\dfrac{\pi}{12}+\dfrac{\sqrt{3}}{8})({units}^2)

The area outside circle r=1/2r = 1/2 and inside circle r=cosθr = \cosθ is (π12+38)(\dfrac{\pi}{12}+\dfrac{\sqrt{3}}{8}) square units.



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