I read that the surface of minimum area, generated by considering curves joining two points and then rotating them about the $x$-axis, which can be expressed as $$ 2\pi\int_{x_0}^{x_1} y \sqrt{1 + (y')^2}dx $$ is given by a catenary unless the two points are very far apart compared to their distance to the $x$-axis. In that extreme case the area of the surface consisting of the circles on the endpoints plus the segment of the axis between them will be less than any surface of revolution generated by a smooth curve.
But I don't see how this is possible, as at the endpoints the circles are defined by revolving about a single point. In essense, the $dx$ in the integral above is only 'used' at $x_0$ and at $x_1$ and hence area of two circles at the endpoints should be zero?