= Solution
For a regular <isoperimetric constraint>, introduce a constant <Lagrange multiplier> $\lambda$ and set $L=F-\lambda G$. With fixed endpoint values, the <Euler-Lagrange equation> is
$$
\boxed{L_y-\frac d{dx}L_{y'}=0.}
$$
If $F$ and $G$ have no explicit $x$ dependence, the <Beltrami identity> gives the <first integral>
$$
L-y'L_{y'}=\text{constant}.
$$
The regularity assumption excludes the exceptional case in which the constraint itself has zero <first variation> for every admissible perturbation.
Parametrize the <surface of revolution> by $r(z,\theta)=(f(z)\cos\theta,f(z)\sin\theta,z)$. The magnitude of $r_z\times r_\theta$ is $f\sqrt{1+f'^2}$. Integrating the resulting <surface area> element over $\theta$, and integrating the circular cross-sectional areas for the <volume>, gives
$$
A=2\pi\int_a^b f\sqrt{1+f'^2}\,dz,
\qquad V=\pi\int_a^b f^2\,dz.
$$
Here $f\geq0$ because it is a radius. For a regular connected closed surface, take $f>0$ in the interior; the closing endpoints can have unbounded $f'$ even when the surface itself is smooth.
After dividing $A-\lambda V$ by $\pi$, its integrand is $L=2f\sqrt{1+f'^2}-\lambda f^2$. The <Euler-Lagrange equation> and <Beltrami identity> become
$$
\sqrt{1+f'^2}-\lambda f-
\frac d{dz}\left(\frac{ff'}{\sqrt{1+f'^2}}\right)=0,
\qquad
\frac{2f}{\sqrt{1+f'^2}}-\lambda f^2=C.
$$
As $f$ tends to zero at an endpoint, $0\leq2f/\sqrt{1+f'^2}\leq2f$ also tends to zero, so $C=0$. On the positive interior this implies $\lambda>0$ and, writing $R=2/\lambda$,
$$
1+f'^2=\frac{R^2}{f^2}.
$$
The <Euler-Lagrange equation> also yields $f''=-R^2/f^3$. Therefore
$$
(f^2)''=2(f'^2+ff'')=-2,
$$
including at the maximum where $f'=0$. Integrating and using the <first integral> gives
$$
f(z)^2=R^2-(z-z_0)^2.
$$
Thus the regular extremal is a <sphere>. The two zeros fix its center and radius:
$$
\boxed{z_0=\frac{a+b}2,\qquad R=\frac{b-a}2,\qquad\lambda=\frac2R=\frac4{b-a}.}
$$
Equivalently, the prescribed <volume> fixes $R=(3V/(4\pi))^{1/3}$. If $a,b$ are fixed in advance, their separation and the prescribed <volume> must be compatible with these equations; there is no positive regular stationary profile of the stated type for an incompatible <volume>.
For minimality, the three-dimensional <isoperimetric inequality> states that every sufficiently regular closed surface enclosing <volume> $V$ has
$$
A^3\geq36\pi V^2,
$$
with equality precisely for a <sphere>. Our extremal has $A=4\pi R^2$ and $V=4\pi R^3/3$, so it attains equality and is a \b[global minimum of area at the fixed <volume>]. This also makes clear the connected-surface assumption behind uniqueness: allowing several disconnected components introduces stationary collections of spheres, rather than the single positive generating profile used here.
Back to article page
