Answer to Question #176652 in Calculus for Joshua

Question #176652

Find the area of the smaller loop r=2-4sin0, symmetrical with Oy.


1
Expert's answer
2021-04-14T14:57:50-0400

We know,


Area can be evaluated as rdθ\int rd \theta


We have given that r=24sinθr = 2-4sin\theta


Hence I=02π(24sinθ)dθI = \int_{0}^{2\pi} (2-4sin\theta) d\theta


I=[02π(2θ+4cosθ)I = [_{0}^{2\pi} (2\theta+4cos\theta)


We get I=4π+404I = 4\pi + 4 - 0 - 4


Hence, I=4πI = 4\pi



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