Answer is: option3
\( \frac{3\pi}{2} \)Solution:
\( r = 1 + \sin\theta \) is a cardioid.
The area is
\[ \frac{1}{2}\int_{0}^{2\pi}(1+\sin\theta)^2\,d\theta = \frac{1}{2}\int_{0}^{2\pi}(1+2\sin\theta+\sin^2\theta)\,d\theta \]
Since \[ \sin^2\theta = \frac{1}{2}-\frac{1}{2}\cos(2\theta), \]
\[ = \frac{1}{2}\int_{0}^{2\pi}\left(1+2\sin\theta+\frac{1}{2}-\frac{1}{2}\cos(2\theta)\right)d\theta \]
\[ = \frac{1}{2}\int_{0}^{2\pi}\left(\frac{3}{2}+2\sin\theta-\frac{1}{2}\cos(2\theta)\right)d\theta \]
Because of the periodicity of \( \sin\theta \) and \( \cos(2\theta) \),
\[ \int_{0}^{2\pi}\sin\theta\,d\theta = 0 \qquad \text{and} \qquad \int_{0}^{2\pi}\cos(2\theta)\,d\theta = 0. \]
Hence the area is:
\[ \frac{1}{2}\int_{0}^{2\pi}\frac{3}{2}\,d\theta = \frac{1}{2}\cdot\frac{3}{2}\cdot 2\pi = \frac{3\pi}{2}. \]
