Calculus help: Find the area between the inner and outer loops of the limacon r=2 - 4sinϴ?

I have gotten 3/4 of the way but one step is alluding me.


These equations should help if you know what they mean:


A=½(r²)ϴ


sin²ϴ = (1- cosϴ)/2

Calculus help: Find the area between the inner and outer loops of the limacon r=2 - 4sinϴ?
inner loop is


0.5 int [ ( 2 - 4sinA )^2 ]dA from pi / 6 to 5pi / 6





outer loop is the same, but 5pi/6 to 13pi/6 for limits ...





[ 2 - 4sinA ]^2 = 4 - 16sin(A) + 16[sin(A)]^2





and [ sin(A)]^2 = 0.5 [ 1 - cos(2A) ], by trig ident...





simplify, etc... integrate ...





I got, after integrating .... 6A + 8 cos(A) - 2sin(2A) , then use the limits given ...upper.. (5pi/6)...and lower..(pi/6) .....





my answer is..... 4pi - 6sqrt(3) = 2.174 for the inner loop... confirmed by TI graphic....





In case you haven%26#039;t used your TI Calculator in Polar Mode yet to integrate [ fnInt( ...] .....





TI Graphic... ..mode, set Pol.. ...Y=...enter.. r1 = 2 -4sin(theta), [[ for theta push x,T,theta,n button ]] ....quit....MATH,9...gives ..fnInt( ....enter0.5VARS,Y-VARS,3,selectr1, enter...push..x^2...comma...theta... comma ... theta...enter pi/6... comma, ... enter 5pi/6 , ...enter..... ans is 2.174065...








Your Answer is the outer loop area minus the inner loop area ...





PS.... limits found by 2 - 4sinA = 0, sinA = 1 / 2, so A = pi /6, 5pi /6, plus 2pi multiples ... thats how I got 13pi/6 .. = pi/6 + 2pi ... of course it helps to trace the graph on the TI calculator, use the TRACE and CALC..VALUE choices...



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