Answer is: option2
\( \int_{0}^{\pi} (1 + \cos \theta)^2 \, d\theta \)Solution:
The polar curve \( r = 1 + \cos \theta \) is a cardioid. The part produced as \( \theta \) goes from \( 0 \) to \( \pi \) encloses the same area as that generated as \( \theta \) goes from \( \pi \) to \( 2\pi \).
Although we would usually compute the area as \( A = \frac{1}{2} \int_{0}^{2\pi} (f(\theta))^2 \, d\theta \), in this case we can use the interval \( [0, \pi] \) instead of \( [0, 2\pi] \) and eliminate the factor \( \frac{1}{2} \).
Thus, \( A = \int_{0}^{\pi} (1 + \cos \theta)^2 \, d\theta \).
