I asked the same question but get no answer in other place. Here is the following.


For a compact Riemannian surface $\Sigma$. For an initial embedded closed curve $\gamma\_0$ in $\Sigma$, a family $\gamma\_t$ $(0\leq t<T)$ is parametrized by
\begin{equation}
F : S^{1} \times[0, T) \rightarrow \Sigma, 
\end{equation}
it is called a curve shortening flow, if
\begin{equation}
\frac{\partial}{\partial t} F(\theta, t)=-\kappa\_{t}(F(\theta, t)) v\_{t}(F(\theta, t))
\end{equation}
P. Topping states on page 51 in [1] that
\begin{equation}
\frac{d A\_{t}}{d t}=-\int\_{\gamma\_{t}} \kappa\_{t}. ~~~(1)
\end{equation}
where $A\_t$ is the area of the set bounded by the curve $\gamma\_t$.


I know how to derive this for $\Sigma=\mathbb{R}^2$. How to prove (1) for $\Sigma$ being a surface? Thank you very much.


**References**


[1] P. Topping, [Mean curvature flow and geometric inequalities](https://www.degruyter.com/view/j/crll.1998.1998.issue-503/crll.1998.099/crll.1998.099.xml), J. Reine Angew. Math. 503, 47-61, 1998



Time derivative of area under curve shortening flow
A late answer if the OP is still interested in a solution. In $\mathbb{R}^2$, the area $A$ enclosed by the closed embedded curve $\gamma \colon I \subset \mathbb{R} \to \mathbb{R}^2$ is given by (using the Green's identity) $$A = \frac 12\oint x\,dy - y\,dx = \frac 12\int\_I \left(x\frac{dy}{dz} - y\frac{dy}{dz}\right)dz = \frac 12\int\_I RX\cdot \frac{dX}{dz},$$ where $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},$ $X = (x,y)$. Thus, we deduce that
\begin{align\*}
\frac{d}{dt}A &= \frac 12\int\_I RX\cdot \frac{dX}{dz} \\
&= \frac 12\int\_I \left(R\frac{dX}{dt}\cdot\frac{dX}{dz} + RX\frac{d^2 X}{dzdt}\right) dz \\
&= \frac 12\int\_I \left(R\frac{dX}{dt}\cdot\frac{dX}{dz} - R\frac{dX}{dz}\frac{dX}{dt}\right) dz, 
\end{align\*}
where the last identity follows from integration by parts. Now recall that under the MCF, we have $\frac{dX}{dt} = \kappa N$ for $N$ being a unit normal vector, and we also have $\frac{dX}{dz} = T\,\frac{ds}{dz}$ for $T$ being the unit tangent vector (here $\frac{ds}{dz} = \|\frac{dX}{dz}\|$ measures the speed of the curve). Note that $RN = -T$ and $RT =N$. Finally, putting all these together, you will arrive at
$$ \frac{d}{dt}A = -\int\_{\gamma} \kappa \, ds,$$ which is the advertised conclusion .  

Remark: The last display, i.e., $\int\_{\gamma} \kappa$, will equal to $2\pi$ thanks to Gauss-Bonnet formula (if the enclosed region is compact and convex).


