Answer to Question #125319 in Calculus for gyamfi francis

Question

Find the area of the region inside both the rose curve r=sin⁡〖(2θ)〗 and the circle r=cos⁡θ. Choose the most appropriate graph and intervals to the best of your interest to solve this question

Answer 1

rose curve r=sin(2$\theta$)

circle r=cos$\theta$

$\implies$ sin(2$\theta$)=cos$\theta$

2sin$\theta$ cos$\theta$ =cos$\theta$

2sin$\theta$cos$\theta$ -cos$\theta$ =0

cos$\theta$ (2sin$\theta$ -1)=0

cos$\theta$ =0 , $\theta$ = $\pi$/2

2sin$\theta$ -1=0

2sin$\theta$ =1 , sin$\theta$ =1/2 ,$\theta$ =($\pi$/6)

now we got the best interval for integrated function is (0,$\pi$/6),($\pi$/6,$\pi$/2)

Area of bounded reason is

A= $\int^{\pi/6}_0$ {sin (2$\theta$}²d$\theta$ + $\int^{\pi/2}_{\pi/6}$ (cos$\theta$)²d$\theta$

A= $\int^{\pi/6}_0$ 1/2 {1-cos(4$\theta$) d$\theta$ + $\int^{\pi/2}_{\pi/6}$ 1/2 (1+cos(2$\theta$)d$\theta$

A=1/2[$\theta$-(sin(4$\theta$)/4)]$^{\pi/6}_0$ + 1/2[$\theta$+(sin2$\theta$)/2]$^{\pi/2}_{\pi/6}$

A=1/2[{($\pi$/6)-($\sqrt3$ )/2/4}-(0-0) + 1/2[{($\pi$/2)+1/2} - {($\pi$/6)+($\sqrt3$ )/2/2}

A=[($\pi$/12)-($\sqrt3$ )/16] + [($\pi$/4)+(1/4)-($\pi$/12)-($\sqrt3$)/8]

A=($\pi$/4)+(1/4)-($\sqrt3$ )/16 - ($\sqrt3$)/8

A=$\pi$/4 + 1/4 -($\sqrt3$)/16 - ($\sqrt3$)*2/8*2

A=$\pi$/4 + 1/4 -($\sqrt3$ )/16 - 2($\sqrt3$ )/16

A=$\pi$/4 + 1/4 + 3($\sqrt3$ )/16