Calculus volume question The region enclosed by the circle defined by the equation $(x-2)^2+y^2 = 4$ is rotated about the y-axis to generate a solid of revolution. Find the volume of the solid.

The suggest answer is
$\int_0^4$ $2{\pi}x2{\sqrt{(4-(x-2)^2)}}$
and I want to know why it is $2{\sqrt{(4-(x-2)^2)}}$ instead of ${\sqrt{(4-(x-2)^2)}}$, where the 2 came from? Thanks.
The area will be given by

$$A=\int_0^42\pi rh\,dx$$

where $r=x$ and $h=2y=2\sqrt{4-(x-2)^2}$



thus

$$A=\int_0^44\pi x\sqrt{4-(x-2)^2}\,dx$$
Because the circle is symmetric. When you use shell method to calculate the volume, you need to consider the area below x axis,