Can anyone help me please, thank you so much.

Find the area of the region that lies inside the circle r =1 and outside the cardioid r = 2-2cos(teta).

I like to sketch the region if possible. If you sketch the cardioid, you'll find that it is a sideways heart-shaped region; it passes through the points (0,0), (0,2), (-4,0), (0,-2). Since we want the area of the region inside the circle r=1 and outside of the cardioid, we have the find the angles theta for which the curves intersect. They can be found by solving 1=2-2cos(theta) for theta; the solutions in [-pi,pi] are -pi/3 and pi/3. For area of curves given in polar coordinates, recall the formula
Area=1/2 int from a to b f(theta)^2 d(theta).
Since we want the area inside of one curve and outside another, we have to use a modification of this formula. The area we want is
1/2 int from -pi/3 to pi/3 [1^2-(2-2cos(theta))^2] d(theta)=
=int from 0 to pi/3 (1+2-2cos(theta))(1-2+2cos(theta)) d(theta)=
=int from 0 to pi/3 (3-2cos(theta))(-1+2cos(theta)) d(theta)=
=int from 0 to pi/3 (-3+8cos(theta)-4cos^2(theta)) d(theta)=
=int from 0 to pi/3 (-3+8cos(theta)-4(1+cos(2theta))/2) d(theta)=
=int from 0 to pi/3 (-5+8cos(theta)-2cos(2theta)) d(theta)=
I believe that the final answer is -5pi/3+7sqrt(3)/2