Find the area inside of the region inside r=6sin(t)

http://i41.tinypic.com/20ijf5f.jpg

Note that r = 6sin(t) is a circle of radius 3 centered at (0, 3) and r = 1 is a circle of radius 1 centered at the origin.

Both r = 6sin(t) and the top-half of r = 1 gets traced out once on the interval [0, π]. If you draw a line from the origin that intersects both circles, it will intersect r = 1 first and then r = 6sin(t); thus, the inner region is r = 1 and the outer region is r = 6sin(t).

Now, note that r = 6sin(t) and r = 1 intersect when:
6sin(t) = 1 ==> t = arcsin(1/6) and t = π - arcsin(1/6).

The region in question lies between these two t values.

Using the formula for the area of a polar curve, the required area is:
A = 1/2 ∫ {[6sin(t)]^2 - 1^2} dt (from t=arcsin(1/6) to π - arcsin(1/6))
= 1/2 ∫ [36sin^2(t) - 1] dt (from t=arcsin(1/6) to π - arcsin(1/6))
= 1/2 ∫ {(36)(1/2)[1 - cos(2t)] - 1} dt (from t=arcsin(1/6) to π - arcsin(1/6))
= 1/2 ∫ [18 - 18cos(2t) - 1] dt (from t=arcsin(1/6) to π - arcsin(1/6))
= 1/2 ∫ [17 - 18cos(2t)] dt (from t=arcsin(1/6) to π - arcsin(1/6))
= (1/2)[17t - 9sin(2t)] (evaluated from t=arcsin(1/6) to π - arcsin(1/6))
= (1/2){17[π - arcsin(1/6)] - 9sin{2[π - arcsin(1/6)]} - 17arcsin(1/6) + 9sin[2arcsin(1/6)]}
= (1/2){17π - 17arcsin(1/6) - 9sin[2π - 2arcsin(1/6)] - 17arcsin(1/6) + 9sin[2arcsin(1/6)]}
= (1/2){17π - 34arcsin(1/6) - 9sin[-2arcsin(1/6)] + 9sin[2arcsin(1/6)]}, due to periodicity
= (1/2){17π - 34arcsin(1/6) + 18arcsin[2arcsin(1/6)]}, since sine is odd
= (1/2)[17π - 34arcsin(1/6) + √35].

I hope this helps!

12