Example Video Example ) Find the area of the region that lies inside the circle r = 6 sin(0) and outside the cardioid r = 2 + 2 sin(0). Solution 1 The cardioid (in blue) and the circle (in red) are...

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Extracted text: Example Video Example ) Find the area of the region that lies inside the circle r = 6 sin(0) and outside the cardioid r = 2 + 2 sin(0). Solution 1 The cardioid (in blue) and the circle (in red) are sketched in the figure above. The value of a and b in A = de are determined by finding the points of intersection of the two curves. They intersect when 6 sin(0) = which gives la 5л , so 0 = 6 sin(0) The desired area can be found by subtracting the area inside the cardioid between 0 = 6 from the area inside the circle from 6 Thus we have the following. to -5л /6 1 5n / 6 1 A = 2 (6 sin(®))? do - (2 + 2 sin(0))2 de Since the region is symmetric about the vertical axis 0 = we can write the following. I / 2 1 4 36 sin?(0) de (1 + 2 sin(0) + sin?(0)) de A = 2 n / 2 n/ 6 - L *n / 2 r/ 6 16 cos(20) · sin(0) ) de because sin (0) = (1 – cos(20))| - л /2 Jr / 6 - " [32 sin°(0) – 4 - de Jr / 6