Find the area inside the lemniscate r^2=96cos(2*theta) and outside the circle r=

Question

Find the area inside the lemniscate r^2=96cos(2*theta) and outside the circle r=4(sqrt3)

Explanation / Answer

The limits of integration can be figured out with these steps below: r >= 2 sqrt(3) We're talking about an area so consider r>0 only, no need to consider r= 12 Therefore cos(2t) >= 1/2 2t is in [-pi/3, pi/3] t is in [-pi/6, pi/6] r goes from 2 sqrt(3) to +sqrt(24 cos(2t)) the area element in polar coordinates is r dr dt or r dt dr because I remember it :) Better do the r dr dt form since the r limits of integration depend on t The antiderivative of the inner integration is r^2 Then the integrand for the outer integral becomes (24 cos(2t)) - (2 sqrt(3))^2 = 24 cos(2t) - 12 Integrate from t = -pi/6 to pi/6 Antiderivative is 12 sin(2t) Evaluate 12 sin(pi/3) - 12 sin(-pi/3) = 24 sin(pi/3) = 24 sqrt(3)/2 = 12 sqrt(3). This gives an area of 20.78

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