Find the area of the region which is inside the polar curve

Solution

Find the area of the region which is inside the polar curve

r = 6cos

and outside the curve

r = 4 - 2cos

First find the points of intersection between the two curves so we know what to integrate over.

r = 6cos = 4 - 2cos
8cos = 4
cos = 1/2
= ± /3

Integrate over -/3 <= <= /3.

Area A = int((1/2)(R^2 - r^2) d)

= (1/2)int[(6cos)^2 - (4 - 2cos)^2] d

= (1/2)int[36cos^2 - (16 - 16cos + 4cos^2)] d

= (1/2)int [36cos^2 - 16 + 16cos - 4cos^2 ] d

= (1/2)int[32cos^2 + 16cos - 16] d

= int(16cos^2 + 8cos - 8) d

= 8int(2cos^2 + cos - 1) d

= 8int(cos 2 + cos) d

= 8[(1/2)[(sin 2) + sin] | [Evaluated from -/3 to /3]

= 8[sqrt(3)/4 + sqrt(3)/2] - 8[-sqrt(3)/4 - sqrt(3)/2] = 12sqrt(3)