Answer: The area enclosed by the curve is \(6\pi\).

For a polar curve, the enclosed area is

\(A=\frac{1}{2}\int r^2\,d\theta\).

Here \(r=2(1+\cos\theta)\), so

\(A=\frac{1}{2}\int_0^{2\pi} \left(2(1+\cos\theta)\right)^2\,d\theta\)

\(=\frac{1}{2}\int_0^{2\pi} 4(1+2\cos\theta+\cos^2\theta)\,d\theta\).

Using \(\cos^2\theta=\frac{1+\cos 2\theta}{2}\),

\(A=\int_0^{2\pi}(3+4\cos\theta+\cos 2\theta)\,d\theta\).

Integrating,

\(A=\left[3\theta+4\sin\theta+\frac{1}{2}\sin 2\theta\right]_0^{2\pi}\).

Therefore,

\(A=6\pi\).