# r=6-6sin(theta), r=6, find the area of the region that lies inside the first curve and outside the second. as you solve please explain

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You should solve the following equation to check where the curves intersect each other such that:

`6 = 6 - 6sin theta => -6sin theta = 0 => sin theta = 0 => {(theta = 0),(theta = pi),(theta = 2pi):}`

Using the symmetry yields:

`A = 2(1/2) int_(-pi/2)^0 ((6 - 6sin theta)^2 - 6^2)d theta`

`A = int_(-pi/2)^0 (36 - 72sin theta + 36 sin^2 theta - 36) d theta`

`A = int_(-pi/2)^0 (36 sin^2 theta - 72 sin theta)`

Using the property of linearity of integrals yields:

`A = int_(-pi/2)^0 36 sin^2 theta d theta- int_(-pi/2)^0 72 sin theta d theta`

You need to use half angle identity to replace `sin^2 theta` such that:

`sin^2 theta = (1 - cos 2 theta)/2`

`A = 36int_(-pi/2)^0 (1 - cos 2 theta)/2 d theta+ 72 cos theta|_(-pi/2)^0`

`A = 18 theta|_(-pi/2)^0 - 9 sin2 theta|_(-pi/2)^0 + 72 cos theta |_(-pi/2)^0`

`A = 18(0 + pi/2) - 9(sin 0 + sin (pi/2)) + 72 (cos 0 - cos(pi/2))`

`A = 9 pi - 9 + 72 => A = 63 + 9pi => A = 9(7 + pi)`

**Hence, evaluating the area under the given conditions yields `A = 9(7 + pi).` **