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...that lies inside the first curve and outside the second curve

r= 1-sin
r=1

2007-11-04 19:33:51 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

The curve r = 1 is the circle of radius 1 centered at the origin. A point (r, t) lies outside this circle and inside the curve r = 1 - sin t if, and only 1 < r < 1 - sin(t), which implies sin(t) < 0 and t is in (pi, 2pi).

So, our area A is given by

A = Int (pi to 2pi) (1 - sin(t))^2 dt - Int (pi to 2pi) dt =

Int (pi to 2pi) (1 - 2 sin(t) + sin^2(t)) dt) - pi =

[t + 2 cos(t) + (1/2) (t - (sin(2t)/2)] (pi to 2 pi) - pi =

Now, all you have to do is plug in the integration limits.f

2007-11-04 23:09:49 · answer #1 · answered by Steiner 7 · 0 0

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