Use integration to find the area of the region outside the circle r=1/2 and inside the circle of r=cos(theta)?

They intercept where r=r , so

1/2= cos T

T=+- pi/3 ( It is +- because cos A= cos -A)

So

dA= rdrdT

A= INT INT rdrdT

1/2<r<cosT

-pi/3<T<pi/3

A= INY (1/2)r^2 dT

A= (1/2( (NT ( cos^2T -(1/2)^2) dT

A= (1/2) INT cos^2T dT - (1/8) INT dT

A= (1/2) INT ( 1+cos2T)/2 dT - (1/8) T

A= (1/4) T +(1/8) sin2T -(1/8) T

A= (1/8) ( T +sin 2T )

A= ( 1/8) ( pi/3+sin 2pi/3- (-pi/3 +sin -2pi/3))

A= (1/8) ( 2pi/3 +2 sin 2pi/3)

A= (1/4) ( pi/3+sin2pi/3)

sin2(pi/3) = 2sinpi/3 cos pi/3 = 2 (1/2) sqrt3 *(1/2)= (1/2) sqrt3

A= (1/4) ( pi/3 +(1/2) sqrt3)