% vim: tw=50
% 09/02/2022 10AM

\noindent
To see that this integral is parametrisation
invariant, suppose that $\bf{x}(\tilde{u},
\tilde{v})$
describes the same surface. Then
\[ \pfrac{\bf{x}}{u} = \pfrac{\bf{x}}{\tilde{u}}
\pfrac{\tilde{u}}{u} + \pfrac{\bf{x}}{\tilde{v}}
\pfrac{\tilde{v}}{u} \]
\[ \pfrac{\bf{x}}{v} = \pfrac{\bf{x}}{\tilde{u}}
\pfrac{\tilde{u}}{v} + \pfrac{\bf{x}}{\tilde{v}}
\pfrac{\tilde{v}}{v} \]
\[ \implies \pfrac{\bf{x}}{u} \times
\pfrac{\bf{x}}{v} = \pfrac{(\tilde{u},
\tilde{v})}{(u, v)} \pfrac{\bf{x}}{\tilde{u}}
\times \pfrac{\bf{x}}{\tilde{v}} \]
But from earlier,
\[ \dd \tilde{u} \dd \tilde{v} =
\pfrac{(\tilde{u}, \tilde{v})}{(u, v)} \dd u \dd v \]
\[ \implies \dd S = \left|
\pfrac{\bf{x}}{\tilde{u}} \times
\pfrac{\bf{x}}{\tilde{v}} \right| \dd \tilde{u}
\dd \tilde{v} \]
and the integral takes the same form for $(u, v)$
and $(\tilde{u}, \tilde{v})$.

\subsubsection*{An Example}

Let $s$ be the surface of a sphere of radius $R$
subtended by angle $\alpha$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/603627608b7c11ec.png}
\end{center}
In spherical polars,
\begin{align*}
    \bf{x}(\theta, \phi)
    &= R(\sin\theta \cos\phi, \sin\theta \sin\phi,
    \cos\theta) \\
    &:= R \bf{e}_r
\end{align*}
(with $\phi \in [0, 2\pi)$ and $\theta \in [0,
\alpha]$). We also write $\bf{e}_r =
\hat{\bf{r}}$. We have
\begin{align*}
    \pfrac{\bf{x}}{\theta}
    &= R(\cos\theta \cos\phi, \cos\theta \sin\phi,
    -\sin\theta) \\
    &:= R \bf{e}_\theta \\
    \pfrac{\bf{x}}{\phi}
    &= R(-\sin\theta \sin\phi, \sin\theta
    \cos\phi, 0) \\
    &:= R\sin\theta \bf{e}_\phi \\
    \implies \pfrac{\bf{x}}{\theta} \times
    \pfrac{\bf{x}}{\phi}
    &= R^2 \sin\theta \bf{e}_r \\
    \implies \dd S
    &= R^2 \sin\theta \dd \theta \dd \phi
\end{align*}
The area is now
\begin{align*}
    A
    &= \int_0^{2\pi} \dd \phi \int_0^\alpha \dd
    \theta R^2 \sin\theta \\
    &= 2\pi R^2 (1 - \cos\alpha)
\end{align*}

\subsubsection*{Integrating Vector Fields}

It is often useful to integrate a vector field
over a surface to yield a number. We do this by
\[ \int_S \bf{F}(\bf{x}) \cdot \bf{n} \dd S =
\int_D \dd u \dd v \left( \pfrac{\bf{x}}{u} \times
\pfrac{\bf{x}}{v} \right) \cdot \bf{F}(\bf{x}(u,
v)) \]
($\bf{n}$ is unit normal to the surface). This is
the \emph{flux} of $\bf{F}$ through $S$. Again, it
is reparametrisation invariant. \myskip
We define the \emph{vector area element}
\[ \bf{\dd S} = \bf{n} \dd S = \pfrac{\bf{x}}{u}
\times \pfrac{\bf{x}}{v} \dd u \dd v \]
Clearly $|\bf{\dd S}| = \dd S$. Then the flux can
be written as
\[ \int_S \bf{F} \cdot \bf{\dd S} \]
The flux depends on the orientation of $S$, i.e.
on the sign of $\bf{n}$. \myskip
\textbf{An Application}: Consider a fluid with a
velocity field $\bf{F}(\bf{x})$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9be9aa748b7d11ec.png}
\end{center}
In a small $\delta t$, the amount of fluid that
flows through $S$ is
\begin{align*}
    \text{Fluid flow}
    &= \bf{F} \delta t \cdot \bf{n} \delta S \\
    \text{Flow}
    &= \int \bf{F} \cdot \bf{\dd S}
    = \text{fluid crossing $S$ per unit time}
\end{align*}

\begin{example*}
    Let $\bf{F} = (-x, 0, z)$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/f75adc528b7d11ec.png}
    \end{center}
    We'll integrate this over the spherical cap $r
    = R$, $0 \le \theta \le \alpha$ and $0 \le
    \phi < 2\pi$. We know that
    \[ \bf{\dd S} = \RR^2 \sin\theta \dd \theta
    \dd \phi \bf{e}_r \]
    \[ \bf{e}_r \equiv \hat{\bf{r}} = (\sin\theta
    \cos\phi, \sin\theta \sin\phi, \cos\theta) \]
    \begin{align*}
        \bf{F} \cdot \bf{e}_r
        &= -x \sin\theta \cos\phi + z \cos\theta
        \\
        &= R(-\sin^2\theta \cos^2 \phi + \cos^2
        \theta)
    \end{align*}
    using $x, z$ polar coordinates.
    \begin{align*}
        \int \bf{F} \cdot \bf{\dd S}
        &= \int_0^\alpha \dd \theta \int_0^{2\pi}
        \dd \phi R^3 \sin\theta (-\sin^2
        \cos^2\phi + \cos^2\theta) \\
        &= \pi R^3 \cos\alpha \sin^2 \alpha
    \end{align*}
\end{example*}

\subsubsection*{The Gauss-Bonnet Theorem (non-examinable)}

Consider a surface $S$ with a \emph{normal}
$\bf{n}$ at some point.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/b24831548b7e11ec.png}
\end{center}
Draw a plane containing $\bf{n}$. The intersection
of the plane and $S$ gives a curve $C$, with
curvature $\kappa$ at the point. \myskip
Now we rotate the plane about $\bf{n} \implies$
the curve and $\kappa$ change. The \emph{Gaussian
curvature} of $S$ at the point is
\[ K = \kappa_{\min} \kappa_{\max} \]

\begin{theorem*}[Gauss-Bonnet v1]
    For a closed surface $S$,
    \[ \int_S \kappa \dd S = 4\pi(1 - g) \]
    where $g = \text{genus} = \text{number of holes}$.
    For example, for a sphere $g = 0$, for a torus $g
    = 1$ and a double torus has $g = 2$.
\end{theorem*}
