Solution
= Solution
Parametrize the surface by $r(x,\theta)=(x,y(x)\cos\theta,y(x)\sin\theta)$, with $0\le\theta<2\pi$. Its tangent vectors are <orthogonal>, of lengths $\sqrt{1+y'^2}$ and $y$. Thus the area element is $y\sqrt{1+y'^2}\,dx\,d\theta$. Integrating over the angle gives
$$
\boxed{J[y]=2\pi\int_{x_1}^{x_2}y\sqrt{1+y'^2}\,dx}.
$$