Isoperimetric inequality on a sphere Say we draw a smooth curve $\gamma\subset S^2$ so as to maximise the area (on $S^2$) enclosed by $\gamma$.
Will $\gamma$ be a circle? If so, does anyone know where I can find a proof?
If $\gamma$ has fixed length $L<2\pi$, is there a result about the maximum area $\gamma$ can enclose in terms of $L$?
 A: Let first consider a spherical cap. Once undestood the spherical cap case we can find a general solution for a generic curve on a sphere. What we want to determine is the relationship between the area of the cap and the length of the boundary circle. The area can be easily be computed by noting that the area of a zone cut out by a couple of parallel planes is proportional to the distance h between the planes. We can then write that $A_{cap}= const \cdot h$. In order to determine the value of the constant we can consider the case when the two planes are tangent to the sphere. In this case their distance is 2R, where R is the sphere radius. When $h= 2R$ we get the full sphere area, i.e. $A=4 \pi R^2$. Hence the constant is $2 \pi R$, and the area of the cap is $A_{cap}= 2 \pi R h$. Now if only one plane is tangent to the sphere, we get a circular cap with a radious equal to the geometrical mean between $h$ and $2R - h$. Then the circle length is $L_{cap}=2 \pi \sqrt{h (2 R -h)}$. We can then write
\begin{equation}
L^2 - 4 \pi A = - (\frac{A}{R})^2
\end{equation}
The left hand side of the above relation is negative or zero. Now if a circle has the smallest length among all curves bounding a fixed area, then the previous relation for a generic curve should take the form
\begin{equation}
L^2 - 4 \pi A \geq - (\frac{A}{R})^2
\end{equation}
It is possible to show that:
1.A smooth closed curve on a surface bounds a domain of maximum area if the curve as a constant geodesic curvature. To prove this, we want to find that a closed curve of fixed arclength which maximizes the surface area of the region of the sphere which it encloses has constant geodesic curvature. The constrained problem of maximizing surface area while keeping arclength fixed gives:
\begin{equation}
\int_{\gamma} (1-\lambda k_g)(N \times T) \cdot V ds = 0
\end{equation}
where $\gamma$ is the curve on the sphere, $k_g$ the geodesic curvature, $\lambda$ a constant, $N$ the unit normal, $T$ the tangent vector, and V an appropriate vector field. This equation has solution $1-\lambda k_g = 0$, i.e. the geodesic curvature is constant.


*A curve of constant geodesic curvature on a sphere is a circle of radius $\rho$ given by the equation:
\begin{equation}
k_g \rho = \sqrt{1-(\frac{\rho}{R})^2}
\end{equation}


Proof
Consider a circle of radius $\rho$ on a sphere and project it onto the tangent plane at some point of the circle. The curvature of the resulting curve at the point of contact between the tangent plane and the sphere is the geodesic curvature $k_g$. In other words, the geodesic curvature $k_g$ of a curve on a surface, is the component of the curvature vector occurring in the tangent plane.

We can then write:
\begin{equation}
\frac{1}{k_g}  = \frac{\rho}{\cos{\theta}}
\end{equation}
where $\rho = R \sin{\theta}$. We can then eliminate $\theta$ by using the fact that $\cos(\theta)^2+\sin(\theta)^2=1$ to obtain the relation between the radius $\rho$, the radius of the sphere $R$, and the geodesic curvature $k_g$.
An alternative way to investigate the problem can be done by using the Gauss-Bonnet theorem. he Gauss–Bonnet Theorem consists of a formula for the integ
ral of the Gaussian curvature over all or part of an abstract surface. 
Let $\gamma$ be a smooth simple closed curve on a patch $r(U)$ of a sphere surface, with anticlockwise parametrisation, enclosing a region D. Then the Gauss Bonnet theorem can be written as:
\begin{equation}
\iint_D K dA = 2 \pi- \int_{\gamma} k_g ds
\end{equation}
where $\gamma(s) = \partial D$ i.e. the boundary of D, and $\gamma:[0,l]\rightarrow S$. Now for a sphere the Gaussian curvature is constant ($K=\frac{1}{R^2}$) so we can write:
\begin{equation}
\iint_D  dA = 2 \pi R^2- R^2 \int_{\gamma} k_g ds=2 \pi R^2- R \int_{\gamma} \cot{\theta} ds = 2 \pi R^2- R \int_{\gamma} \cot{\theta} \sin{\theta}dt = 2 \pi R^2-2 \pi R \cos{\theta}
\end{equation} 
