How do you find the area inside of the circle #r = 3sin(theta)# and outside the cardioid #r = 1 + sin(theta)#?
1 Answer
Explanation:
Draw both curves on the same graph paper.
Notice that the cardioid intersects with the circle at
The area of interest has been shaded above.
To find the area of a polar curve, we use
#A = 1/2 int r^2 "d"theta#
We find the area of the cardioid and the circle separately on the interval
For the circle,
#A = 1/2 int_{pi/6}^{(5pi)/6} r^2 "d"theta#
#= 1/2 int_{pi/6}^{(5pi)/6} (3sin(theta))^2 "d"theta#
#= 9/4 int_{pi/6}^{(5pi)/6} 2sin^2(theta) "d"theta#
#= 9/4 int_{pi/6}^{(5pi)/6} (1-cos(2theta)) "d"theta#
#= 9/4 [theta - sin(2theta)/2]_{pi/6}^{(5pi)/6}#
#= 9/4 ([(5pi)/6 + sqrt3/4]-[pi/6 - sqrt3/4])#
#= (3pi)/2 + (9sqrt3)/8#
For the cardioid,
#A = 1/2 int_{pi/6}^{(5pi)/6} r^2 "d"theta#
#= 1/2 int_{pi/6}^{(5pi)/6} (1+sin(theta))^2 "d"theta#
#= 1/2 int_{pi/6}^{(5pi)/6} (1+2sin(theta)+sin^2(theta)) "d"theta#
#= 1/4 int_{pi/6}^{(5pi)/6} (2+4sin(theta)+1-cos(2theta)) "d"theta#
#= 1/4 [3theta - 4cos(theta) - sin(2theta)/2]_{pi/6}^{(5pi)/6}#
#= 1/4 ([(5pi)/2 + 2sqrt3 + sqrt3/4]-[pi/2 - 2sqrt3 - sqrt3/4])#
#= pi/2 + (9sqrt3)/8#
The difference in area is
#((3pi)/2 + (9sqrt3)/8) - (pi/2 + (9sqrt3)/8) = pi#