Find the area of the smaller loop r=2-4sin0, symmetrical with Oy.
We know,
Area can be evaluated as ∫rdθ\int rd \theta∫rdθ
We have given that r=2−4sinθr = 2-4sin\thetar=2−4sinθ
Hence I=∫02π(2−4sinθ)dθI = \int_{0}^{2\pi} (2-4sin\theta) d\thetaI=∫02π(2−4sinθ)dθ
I=[02π(2θ+4cosθ)I = [_{0}^{2\pi} (2\theta+4cos\theta)I=[02π(2θ+4cosθ)
We get I=4π+4−0−4I = 4\pi + 4 - 0 - 4I=4π+4−0−4
Hence, I=4πI = 4\piI=4π
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