How do you find the area between the loops of #r=2(1+2sintheta)#?

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Answer 1 · 2018-07-24T02:36:53

#r = sqrt( x^2 + y^2 )>= 0. So, there is only one loop.

See graph.

#0 <= r = 2 ( 1 + 2 sin theta ) rArr sin theta >= - 1/2#

#rArr theta notin (7/6pi, 11/6pi )# .

graph{x^2+y^2-2sqrt(x^2+y^2)-4y=0[-14 14 -7 7]}

Of course, just as an exercise, r-negative loop can be inserted, and

the area in between can be evaluated..