### Instruction: The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$  What is the area of the region it sweeps over?### Response: To find the area of the region swept by the polar curve, I need to use the formula for the area of a polar curve rotated around a point.
The general formula is $A = \frac{1}{2} \int_a^b \left( r(\theta)^2 - r_0^2 \right) \, d\theta,$ where $r(\theta)$ is the polar curve, $r_0$ is the distance from the pole to the point of rotation, and $a$ and $b$ are the angles that bound the region.
In this case, $r(\theta) = 1 + \cos \theta,$ $r_0 = 2,$ and the region is bounded by $0 \leq \theta \leq 2\pi,$ since the curve is rotated once.
Plugging these values into the formula, I get $A = \frac{1}{2} \int_0^{2\pi} \left( (1 + \cos \theta)^2 - 2^2 \right) \, d\theta.$
To evaluate this integral, I need to expand the square and simplify the integrand.
I get $A = \frac{1}{2} \int_0^{2\pi} \left( 1 + 2\cos \theta + \cos^2 \theta - 4 \right) \, d\theta = \frac{1}{2} \int_0^{2\pi} \left( -3 + 2\cos \theta + \cos^2 \theta \right) \, d\theta.$
Now I can use some basic trigonometric identities and properties of integrals to further simplify the expression.
I recall that $\cos^2 \theta = \frac{1}{2} + \frac{1}{2} \cos 2\theta,$ so I can rewrite the integrand as $-3 + 2\cos \theta + \frac{1}{2} + \frac{1}{2} \cos 2\theta.$
Then I can split the integral into four parts and use the fact that $\int_0^{2\pi} \cos k\theta \, d\theta = 0$ for any nonzero integer $k,$ since the cosine function is periodic.
I get $A = \frac{1}{2} \left( \int_0^{2\pi} -3 \, d\theta + \int_0^{2\pi} 2\cos \theta \, d\theta + \int_0^{2\pi} \frac{1}{2} \, d\theta + \int_0^{2\pi} \frac{1}{2} \cos 2\theta \, d\theta \right).$
The first and third integrals are easy to evaluate, since they are constants.
The second and fourth integrals are zero, as explained before.
I get $A = \frac{1}{2} \left( -6\pi + 0 + \pi + 0 \right) = -\frac{5}{2} \pi.$