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% 07/02/2022 10AM

\subsection{Surface Integrals}

We define surfaces in $\RR^3$ by
\begin{itemize}
    \item A function $F(x, y, z) = 0$
    \item A \emph{paramterised surface} is a map
        \[ \bf{x} : \RR^2 \mapsto \RR^3 \]
\end{itemize}
At each point on the surface, the \emph{normal
vector} $\bf{n}$ points away in a perpendicular
direction.

\begin{claim*}
    For the surface $F(\bf{x}) = 0$, $\bf{n}
    \parallel \nabla F$.
\end{claim*}

\begin{proof}
    $\bf{m} \cdot \nabla F$ is the rate of change
    of $F$ in the direction $\bf{m}$. There are
    two linearly independent vectors $\bf{m}_1$
    and $\bf{m}_2$ that lie tangent to the surface
    and obey $\bf{m}_i \cdot \nabla F = 0$, $i =
    1, 2$. The normal vector $\bf{n}$ is
    perpendicular to $\bf{m}_1$ and $\bf{m}_2$ and
    so $\bf{n} \parallel \nabla F$.
\end{proof}

\myskip
We usually define
\[ \bf{n} = \pm \frac{1}{|\nabla F|} \nabla F \]
For a parametrised surface $\bf{x}(u, v)$ the
tangent vectors are
\[ \pfrac{\bf{x}}{u} \qquad \text{and} \qquad
\pfrac{\bf{x}}{v} \]
The normal vector is $\bf{n} \parallel
\pfrac{\bf{x}}{u} \times \pfrac{\bf{x}}{v}$.

\begin{definition*}
    If $\bf{n} \neq 0$ at all points, the surface
    is \emph{regular}.
\end{definition*}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $F(\bf{x}) = x^2 + y^2 + x^2 - R^2 = 0$
        is a sphere of radius $R$. The normal
        vector is $\parallel$ to $\nabla F$ and is
        \[ 2
        \begin{pmatrix}
            x \\
            y \\
            z
        \end{pmatrix}
        \]
    \item A hyperboloid is defined by
        \[ F(\bf{x}) = x^2 + y^2 - z^2 - R^2 = 0 \]
        \[ \nabla F = 2
        \begin{pmatrix}
            x \\
            y \\
            -z
        \end{pmatrix}
        \]
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/3118642a8a2811ec.png}
        \end{center}
        \begin{note*}
            When $R = 0$ the surface is no longer
            regular:
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/5bfe4ccc8a2811ec.png}
            \end{center}
        \end{note*}
\end{enumerate}

\noindent
A surface $S$ can have a boundary. The boundary is
a closed curve $C$, denoted as $C = \partial S$.
\myskip
\textbf{Deep Fact}: The boundary curve $C$ is
closed, i.e. it has no end points. \myskip
\textbf{Another Deep Fact}: We denote the boundary
of something by $\partial$. The fact that the
boundary of a boundary vanishes is written
\[ \partial C = \partial^2 S = 0 \]

\begin{definition*}
    A surface is \emph{bounded} / \emph{unbounded}
    if it doesn't / does stretch to infinity. A
    bounded surface with no boundary is
    \emph{closed}.
\end{definition*}

\begin{note*}
    There is no canonical way to fix the $\pm$
    sign of $\bf{n}$. If there is a consistent
    choice over the surface $S$, then $S$ is
    \emph{orientable}. For example the sphere
    $S^2$ is orientable but the M\"obius $M$ with
    $\partial M = S^1$ (circle) is non-orientable.
    We will only work with orientable surfaces.
\end{note*}

\subsubsection*{Integrating Scalar Fields}

Consider a parametrised surface
\[ \bf{x}(u, v) \]
sit at some point $(u, v)$ and move a small amount
$\delta u$ or $\delta v$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/f5f96e008a2911ec.png}
\end{center}
The parallelogram defined by $\pfrac{\bf{x}}{u}$
and $\pfrac{\bf{x}}{v}$ has scalar area
\[ \delta S = \left| \pfrac{\bf{x}}{u} \times
\pfrac{\bf{x}}{v} \right| \delta u \delta v \]
The integral of a scalar field $\phi(\bf{x})$ over
a parametrised surface is
\[ \int_S \phi(\bf{x}) \dd S = \int_D \dd u \dd v
\left| \pfrac{\bf{x}}{u} \times \pfrac{\bf{x}}{v}
\right| \phi(\bf{x}(u, v)) \]

\begin{note*}
    This does not depend on the orientation of
    $S$. Also $\int_S \dd S$ is the area of the
    surface. Also, the integral does not depend on
    the choice of parametrisation.
\end{note*}