How do you find the area inside of the circle #r = 3sin(theta)# and outside the cardioid #r = 1 + sin(theta)#?

1 Answer
Bio
Apr 24, 2016

#pi#

Explanation:

Draw both curves on the same graph paper.

enter image source here

Notice that the cardioid intersects with the circle at #(3/2,pi/6)#, #(3/2,(5pi)/6)# and the pole.

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 #pi/6 < theta < (5pi)/6# and take the difference.

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#

Related questions
See all questions in Integration: the Area Problem
Impact of this question
72803 views around the world
You can reuse this answer
Creative Commons License