I wish to find a curve $u(x)$ (symmetrical about the y-axis), such that the surface area of revolution about the x-axis: $$A[u]=2\pi \int_{x_1}^{x_2} u(x) \sqrt{1+u'(x)^2} dx$$ is minimised.
The constraint for this problem is that the volume of the solid of revolution must be: $$\pi\int_{x_1}^{x_2}u(x)^2 dx = V$$
This seems like a pretty common problem found in textbooks. However, the reason why I am asking this on stack exchange is because $x_1$ and $x_2$ in the equations above are not constrained(unlike classic problems where they let $u(x_1)=a$ and $u(x_2)=b$). They are instead the $x$ intercepts of $u(x)$. In other words, $x_1$ and $x_2$ have to satisfy the following relationships: $$u(x_1)=0, u(x_2)=0$$
How can I arrive at the desired result? Thank you very much.