A calculus of variation problem with an obligation: minimizing $I(x,y,z)=\int_0^1 \sqrt{\dot x^2 +\dot y^2 +\dot z^2} dt$ I have to minimize this :
$$I(x,y,z)=\int_{0}^{1} \sqrt{\dot x^2 +\dot y^2 +\dot z^2} dt$$
with this obligation :
$$z=x^2+y^2$$
And :
$$x(0)=y(0)=z(0)=0$$
$$x(1)=y(1)=z(1)/2=1$$
I tried some strategies , but they weren't useful.
 A: We use polar coordinate $x = r\cos \theta$, $y=r\sin \theta$ here. So 
$$\dot x = \dot r\cos\theta - r\dot \theta\sin\theta ,\ \ \dot y = \dot r \sin\theta + r\dot\theta \cos\theta,\ \ \ \dot z = 2r \dot r.$$
Thus 
$$\begin{split}
I(r, \theta) &= \int_0^1 \sqrt{\dot r^2 + r^2 \dot\theta^2 + 4r^2 \dot r^2} dt \\
&= \int_0^1 \sqrt{(1+4r^2)\dot r^2 + r^2 \dot\theta^2 }dt \\
&\ge \int_0^1 \sqrt{(1+4r^2)\dot r^2} dt \\
&= \int_0^1 \sqrt{1+4r^2)} |\dot r| dt \\
&\ge \int_0^1 \sqrt{1+4r^2} \dot r dt \\
&= \int_0^{\sqrt 2} \sqrt{1+4r^2} dr
\end{split}$$
Note that in the last step we used substitution and had taken into account the initial conditions. The last integral equals $\frac 14(6\sqrt 2 + \sinh^{-1}(2\sqrt 2))$ and from the above, equality holds if and only if both $\dot r \ge 0$ and $\dot \theta  =0$. That is, all minimizers are of the form $(r(t), \pi/4)$, where $r(t)$ satisfies $\dot r \ge 0$. This, in particular, tells you why it is hard to tackle the problem using Euler-Lagrange equation - the solution space is infinite! 
Remark If you think of $I$ as the length functional defined on curves $(x(t), y(t), z(t))$ on the surface $\{z = x^2 + y^2\}$, then geometrically it is obvious that the minimizer must have constant angle. 
If you are interested, you can also try the minimization of the Energy functional
$$E(x, y, z) = \frac 12\int_0^1 (\dot x^2 + \dot y^2 + \dot z^2) dt$$
with the same contraints. You will find that geometrically you got the same curve, but now that the parametrization $r(t)$ is not arbitrary. In this case the solution is called a geodesic in the surface $\{z = x^2 + y^2\}$. 
