I spent a few more hours on the problem, and eventually found the solution. No major breakthroughs, but a lot of algebra.
First, I reversed the problem, instead maximizing the volume subject to a surface area constraint. Not strictly necessary, but it simplifies the computation. I also gave the surface area a magnitude $k$, per user7530's comment. The new action is:
$$S=\int_{-\lambda/2}^{\lambda/2}\pi y^2 + 2\lambda\pi y\sqrt{1+\dot{y}^2}-\lambda k~dx$$
The integration boundaries were also changed for convenience.
Then, because the Lagrangian does not depend explicitly on the variable $x$, I was able to apply the Beltrami identity instead of the traditional EL equation.
$$\frac{d}{dx}\left(\pi y(x)^2 + 2\lambda\pi y(x) \sqrt{1+y'(x)^2}-\lambda k - \frac{2\pi\lambda y'(x)^2y(x)}{\sqrt{1+y'(x)^2}}\right) = 0$$
Integrating both sides, we find that the functional inside the derivative is equal to a constant. Because $y$ is equal to $0$ at $-\lambda/2$ and $\lambda/2$, and each term in the derivative depends on $y(x)$ except for the $\lambda k$ term, the arbitrary additive constant must be $-\lambda k$, which cancels with the $-\lambda k$ on the left side.
The DE becomes:
$$\pi y(x)^2 + 2\lambda\pi y(x) \sqrt{1+y'(x)^2} - \frac{2\pi\lambda y'(x)^2y(x)}{\sqrt{1+y'(x)^2}} = 0$$
Multiplying the second term by $\frac{\sqrt{1+y'(x)^2}}{\sqrt{1+y'(x)^2}}$, that in turn simplifies to:
$$y(x)^2 + \frac{2\lambda y(x)}{\sqrt{1+y'(x)^2}}=0$$
This can be solved for $y(x)$, and with the constant chosen to equal zero, the solution is the equation of a circle of radius $2\lambda$.
$$y(x)=\pm\sqrt{4 \lambda ^2-x^2}$$
Your first equation seems to be wrong
The Area is given by
$$J[y]=\int_0^\ell y dx$$
The length of the differential element:
$$(ds)^2=(dx)^2+(dy)^2$$
$$dx=\sqrt{1-(y'(s))^2}ds$$
You get
$$J[y]=\int_0^\ell y(s) \sqrt{1-(y'(s))^2}ds$$
Also see Isoperimetric problem in the calculus of variations
Best Answer
Actually after checking out "Mathematical Methods for Students of Physics and Related Fields" by Sadri Hassani and a website I concluded that the answer is a semicircle whose radius is given by $ \lambda^2 $
I am mirroring the treatment in the book by Hassani The treatment is as follows,
The function to be minimized is $$ \mathbf{L}[y] = \int_{-a}^{a}ydx $$ Under the following conditions and constraints $$ y(-a) = 0 = y(a) $$ $$ \mathbf{K}[y]=\int_{-a}^{a}\sqrt{1+y'^2}dx=L $$ Using Euler-Lagrange equation is more convenient in this case and it is given by: $$ \frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} + \lambda \left(\frac{\partial G}{\partial x} - \frac{d}{dt}\frac{\partial G}{\partial \dot{x}}\right)=0 $$ In our case $L=y$ and $G=\sqrt{1+y'^2}$, substituting thse into above equation gives $$ 1 + \lambda \frac{d}{dx} \frac{y'}{\sqrt{1+y'^2}}=0 $$ Integrating wrt x yields $$ x + \lambda \frac{y'}{\sqrt{1+y'^2}} = C_1 $$ Solve for $y'$ $$ y'=\pm \frac{C_1-x}{\sqrt{\lambda ^2-(C_1-x)^2}}=\frac{dy}{dx} $$
Multiply each side by $dx$ and integrate $$ y = \pm \sqrt{\lambda ^2-(C_1-x)^2}+C_2$$ Or $$ (x-C_1)^2+(y-C_2)^2=\lambda^2 $$ The unknowns are to be determined from the boundary conditions and the length constraint.