= Solution
For a normal constrained extremal, introduce a constant <Lagrange multiplier> and use $L=F-\lambda G$. The <Euler-Lagrange equation> is
$$
\boxed{L_y-\frac d{dx}L_{y'}=0.}
$$
When both integrands have no explicit $x$ dependence, the <Beltrami identity> gives $L-y'L_{y'}=\text{constant}$, as follows by differentiating this expression and using the <Euler-Lagrange equation>.
Parametrize the surface by $(f(z)\cos\theta,f(z)\sin\theta,z)$. The <cross product> of its two coordinate tangents has magnitude $f\sqrt{1+(f')^2}$, so integration in $\theta$ gives $A=2\pi\int_a^bf\sqrt{1+(f')^2}\,dz$. Integrating the cross-sectional disk areas gives $V=\pi\int_a^bf^2\,dz$.
Dividing the augmented functional $A-\lambda V$ by $\pi$, use
$$
L=2f\sqrt{1+(f')^2}-\lambda f^2.
$$
The <Euler-Lagrange equation> is
$$
\boxed{\frac d{dz}\left(\frac{2ff'}{\sqrt{1+(f')^2}}\right)
-2\sqrt{1+(f')^2}+2\lambda f=0.}
$$
The <Beltrami identity> becomes $2f/\sqrt{1+(f')^2}-\lambda f^2=C$. At an endpoint $f\to0$, the absolute value of the first term is at most $2f$, so $C=0$, even if the profile has an infinite endpoint slope. For a nondegenerate regular closed surface take $f>0$ on $(a,b)$. Then $\lambda>0$, and, with $R=2/\lambda$,
$$
f^2(1+(f')^2)=R^2.
$$
Combining this identity with the <Euler-Lagrange equation> yields $f''=-R^2/f^3$. Consequently $(f^2)''=2((f')^2+ff'')=-2$. The two zero endpoint values uniquely determine
$$
\boxed{f(z)^2=(z-a)(b-z)
=\left(\frac{b-a}{2}\right)^2-\left(z-\frac{a+b}{2}\right)^2.}
$$
The positive radius profile is therefore a semicircle and generates a <sphere>, centred at height $(a+b)/2$, with radius $R=(b-a)/2$. In particular
$$
\boxed{\lambda=\frac2R=\frac4{b-a}.}
$$
The <graph> is smooth in the interior; its vertical slopes at the two poles are compatible with a smooth spherical surface. Its <volume> is $4\pi R^3/3$, the <volume> associated with this stationary candidate.
For minimality, the three-dimensional <isoperimetric inequality> says that every closed surface enclosing <volume> $V$ has $A^3\geq36\pi V^2$, with equality for a <sphere>. The displayed extremal attains equality, proving that it minimizes area at its fixed enclosed <volume>. The uniqueness above concerns the regular connected positive profile with the two stated endpoint heights.
Back to article page
