The area of the region that lies inside both curves r² = 2sin(2θ), r = 1 is -0.1287 square units.
Since the curves are
We find their point of intersection.
So, r² = r²
2sin(2θ) = 1²
2sin(2θ) = 1
sin(2θ) = 1/2
2θ = sin⁻¹(1/2)
2θ = π/6
θ = π/12
So, we integrate the area from θ = 0 to θ = π/12
Now the area A of the region between two curves between θ = α to θ = β is
[tex]A = \int\limits^{\beta }_\alpha ({r^{2} - r^{'2} }) \, d\theta[/tex]
So, the area betwwen the curves r² = 2sin(2θ), r = 1 between θ = 0 to θ = π/12 is
[tex]A = \int\limits^{\frac{\pi}{12} }_0 ({r^{2} - r^{'2} }) \, d\theta \\= \int\limits^{\frac{\pi}{12} }_0 ({2sin2\theta - 1^{2} }) \, d\theta\\= \int\limits^{\frac{\pi}{12} }_0 ({2sin2\theta - 1}) \, d\theta\\= \int\limits^{\frac{\pi}{12} }_0 {2sin2\theta}\, d\theta - \int\limits^{\frac{\pi}{12} }_0 {1} \, d\theta\\= [2(-cos2\theta)/2 - \theta]_{0}^{\frac{\pi}{12} } \\= [-cos2\theta - \theta]_{0}^{\frac{\pi}{12} } \\= -cos2\frac{\pi}{12} - \frac{\pi}{12} - ( -cos2(0) - 0)\\[/tex]
[tex]= -cos2(\frac{\pi}{12}) - \frac{\pi}{12} - ( -cos2(0) - 0)\\= -cos\frac{\pi}{6}- \frac{\pi}{12} - ( -cos0 - 0)\\= -cos\frac{\pi}{6}- \frac{\pi}{12} - ( -1)\\= -(\frac{\sqrt{3} }{2} ) - \frac{\pi}{12} + 1\\= -0.8660 - 0.2618 + 1\\= -1.1278 + 1\\= -0.1278[/tex]
So, the area of the region that lies inside both curves r² = 2sin(2θ), r = 1 is -0.1287 square units.
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