You need to evaluate the double integral, hence, you need to convert the double integral into an iterated integral such that:

`int_0^(2pi) int_(-2(sin^2(theta/2)))^(2(sin^2(theta/2))) r dr d theta`

Since the cardioid function is even yields:

`int_0^(2pi) (2*int_0^(2(sin^2(theta/2))) r dr d theta)`

Evaluating the inner integral, using the fundamental theorem of calculus, yields:

`int_0^(2pi) 2*r^2/2 |_(0)^(2(sin^2(theta/2))) d theta`

Substituting `1 - cos theta`  for `2 sin^2(theta/2)`  yields:

`int_0^(2pi) ((1 - cos theta)^2 - 0^2) d theta`

`int_0^(2pi) (1 - 2cos theta + cos^2 theta) d theta`

Using the property of linearity of integral yields:

`int_0^(2pi) (1 - 2cos theta + cos^2 theta) d theta = int_0^(2pi) d theta - 2 int_0^(2pi) cos theta d theta + int_0^(2pi) cos^2 theta d theta`

You need to use the following identity such that:

`cos^2 theta = (1 + cos 2 theta)/2`

`int_0^(2pi) (1 - 2cos theta + cos^2 theta) d theta = theta|_0^(2pi) - 2 sin theta|_0^(2pi) + int_0^(2pi) (1 + cos 2 theta)/2 d theta`

`int_0^(2pi) (1 - 2cos theta + cos^2 theta) d theta = theta|_0^(2pi) - 2 sin theta|_0^(2pi) + theta/2|_0^(2pi) + (sin 2 theta)/4|_0^(2pi) `

`int_0^(2pi) (1 - 2cos theta + cos^2 theta) d theta = 2pi - 0 - 2 sin (2pi) + 2 sin 0 + (2pi)/2 - 0/2 + (sin (4pi))/4 - (sin 0)/4`

`int_0^(2pi) (1 - 2cos theta + cos^2 theta) d theta = 3pi`

Hence, evaluating the region enclosed by the given cardioid yields `int_0^(2pi) int_(-2(sin^2(theta/2)))^(2(sin^2(theta/2))) r dr d theta = 3pi`