Answer to Question #125301 in Analytic Geometry for Samuel kassapa

Question #125301
Find the area of the region inside both the rose r =sin(2Φ) and the circle r=cosΦ.The appropriate graph and intervals to the best of your interest to solve this question 
1
Expert's answer
2020-07-06T19:02:43-0400
sin(2ϕ)=cos(ϕ)\sin(2\phi)=\cos(\phi)

2sin(ϕ)cos(ϕ)=cos(ϕ)2\sin(\phi)\cos(\phi)=\cos(\phi)

cos(ϕ)=0 or sin(ϕ)=12\cos(\phi)=0\text{ or } \sin(\phi)={1\over 2}




A1=120π/2(sin(2ϕ))2dϕ=A_1={1\over2}\displaystyle\int_{0}^{\pi/2}(\sin(2\phi))^2d\phi=

=140π/2((1cos(4ϕ))dϕ=={1\over4}\displaystyle\int_{0}^{\pi/2}((1-\cos(4\phi))d\phi=

=14[ϕ14sin(4ϕ)]π/20=π8(units2)={1\over4}\big[\phi-{1\over 4}\sin(4\phi)\big]\begin{matrix} \pi/2 \\ 0 \end{matrix}={\pi\over 8}(units^2)


A2=12π/6π/2((sin(2ϕ))2(cosϕ)2)dϕ=A_2={1\over2}\displaystyle\int_{\pi/6}^{\pi/2}\bigg((\sin(2\phi))^2-(\cos\phi)^2\bigg)d\phi=

=14π/6π/2(1cos(4ϕ)1cos(2ϕ))dϕ=={1\over4}\displaystyle\int_{\pi/6}^{\pi/2}\bigg(1-\cos(4\phi)-1-\cos(2\phi)\bigg)d\phi=

=14[14sin(4ϕ)+12sin(2ϕ)]π/2π/6==-{1\over4}\bigg[{1\over 4}\sin(4\phi)+{1\over 2}\sin(2\phi)\bigg]\begin{matrix} \pi/2 \\ \pi/6 \end{matrix}=

=14(038+034)=3332(units2)=-{1\over4}(0-{\sqrt{3}\over 8}+0-{\sqrt{3}\over 4})={3\sqrt{3}\over 32}(units^2)


Area=2(A1A2)=2(π83332)=4π3316(units2)Area=2(A_1-A_2)=2({\pi\over 8}-{3\sqrt{3}\over 32})={4\pi-3\sqrt{3}\over 16} (units^2)

Area=4π3316 square unitsArea=\dfrac{4\pi-3\sqrt{3}}{16} \text{ square units}



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