In order to solve the isoperimetric problem, I am extremising the functional:
$$A[x,y] = \frac{1}{2}\int_0^{2\pi}(x \dot y - y \dot x) \;dt\tag{1}$$
where $x = x(t)$, $y = y(t)$, and the Lagrangian is:
$$f = \frac{1}{2}(x \dot y - y \dot x) - \lambda\sqrt{\dot x^2 + \dot y^2}. \tag{2} $$
I have already successfully solved the problem using the standard Euler-Lagrange equations, which give the parametric equations for a circle with radius $r = \lambda$. This time, I was trying to use the following first integral, which is applicable since $\frac{\partial f}{\partial t} = 0.$
$$f - \dot x \frac{\partial f}{\partial \dot x} - \dot y \frac{\partial f}{\partial \dot y} = K.\tag{3} $$
Evaluating this seems to give $K = 0$. However, plugging in the solution curves for perimeter $2\pi r$,
$$x(t) = r\cos(t)\tag{4}$$ $$y(t) = r\sin(t)$$
gives:
$$f = \frac{r^2}{2} - r\lambda\tag{5}$$
$$\implies \dot x \frac{\partial f}{\partial \dot x}, \;\;\dot y \frac{\partial f}{\partial \dot y} = 0 \tag{6}$$
$$\implies f \equiv 0 \implies \lambda = \frac{r}{2}\tag{7}$$
I have been staring at this for a long time, but I can't find the error. Could anyone help please?
FWIW, there are several issues with OP's method:
In eq. (2) the square root term should in principle be replaced by the full perimeter constraint, not just part of it.
In eq. (6) it does not make sense to partial differentiate the Lagrangian (2) wrt. $\dot{x}$ and $\dot{y}$ after it has been evaluated (5) on the solution (4).
Boths issues can be remedied relatively easy.