# 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

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).`