\noindent
\begin{center}
\begin{figure}[htbp]
\fbox{\resizebox{2.6in}{!}{\includegraphics{../images/instructors_guide_illustrations/uniform_contour_spacing}}} \hspace{.5cm}
\fbox{\resizebox{2.6in}{!}{\includegraphics{../images/instructors_guide_illustrations/varying_contour_spacing}}}
\caption[$d\vec{A}$ and level curves]{Surface area depends on the concentration of level curves, or, equivalently, the value of $\vec{\nabla} f$. Volume depends on the value of $f$, not the spacing of the level curves.}
\label{fig:surface_area}
\end{figure}
\end{center}
As indicated in Figure \ref{fig:surface_area}, the amount of surface area is associated with the spacing of the function's level curves, not their value. Not surprisingly, surface area is thus associated with partial derivatives or the gradient $\vec{\nabla} f$ of the function $f$. Indeed, this is the case:
\[ d\vec{A} = \left(\hat{n} - \vec{\nabla} f \right) \; dx \; dy\]
where $\hat{n}$ is normal to the domain with units compatible with $\vec{\nabla} f$ describes a small piece of area on the surface. This leads to the scalar description:
\[ dA = \left| \hat{n} - \vec{\nabla} f \right| \; dx \; dy = \sqrt{1^2 + \left|\vec{\nabla} f \right| } \;dx \;dy\]
Alternatively, as shown in Figure \ref{fig:surface_area}, the vector quantity $d\vec{A}$ can also be described as the cross product of two vectors $d\vec{r}_1$ and $d\vec{r}_2$ which are tangent to the surface. In this case,
\[ dA = \left| d\vec{r}_1 \times d\vec{r}_2 \right| \]
gives the scalar surface area element on the surface.
\noindent
\begin{center}
\begin{figure}[tbp]
\fbox{\resizebox{2.6in}{!}{\includegraphics{../images/instructors_guide_illustrations/surface_integral_da}}} \hspace{.5cm}
\fbox{\resizebox{2.6in}{!}{\includegraphics{../images/instructors_guide_illustrations/surface_integral_connect_to_gradient2}}}
\caption[Constructing $d\vec{A}$ for surface area]{The surface area element can be found from tangent vectors $d\vec{r}$ to the surface, or in terms of the gradient $\vec{\nabla} f$}
\label{fig:surface_area}
\end{figure}
\end{center}
From the above descriptions, it appears that the surface area element can be described by either the gradient vector or by the cross product of (infinitesimal) tangent vectors. Indeed, this is the case. To see this relationship, let $d\vec{r}_1$ and $d\vec{r}_2$ be differential tangent vectors to the surface. In components in rectangular coordinates, they are:
\[ d\vec{r}_1 = dx \hat{i} + dz \hat{k} = \left(1 \hat{i} + \frac{\partial z}{\partial x} \hat{k}\right) dx\]
\[ d\vec{r}_2 = dy \hat{j} + dz \hat{k} = \left(1 \hat{j} + \frac{\partial z}{\partial y} \hat{k}\right) dy\]
\noindent
The cross product of the differential quantities gives:
\begin{eqnarray*}
d\vec{A} & = & d\vec{r}_1 \times d\vec{r}_2 \\
& = & dx dy \hat{k} - dx df|_x \hat{j} - dy df|_y \hat{i} \\
% & = & \left(- dy df|_y \hat{i} - dx df|_x \hat{j}\right) + dx dy \hat{k} \\
& = & \left(- \frac{\partial f}{\partial x} \hat{i} - \frac{\partial f}{\partial y} \hat{j} + \hat{k}\right) dx dy \\
& = & \left( - \vec{\nabla} f + \hat{k} \right) dx dy \\
\end{eqnarray*}
\noindent
Hence,
\begin{eqnarray*}
dA &= &\left| - \vec{\nabla} f + \hat{k} \right| dx dy \\
&= &\sqrt{\left(-\frac{\partial f}{\partial x}\right)^2 + \left(-\frac{\partial f}{\partial y}\right)^2 + 1^2} \; dx dy\\
\end{eqnarray*}
In polar coordinates, the expression for $d\vec{r}_1$ and $d\vec{r}_2$ become
\[ d\vec{r}_1 = dr \; \hat{r} + dz \; \hat{z} = \left(1 \; \hat{r} + \frac{\partial z}{\partial r} \; \hat{z}\right) dr\]
\[ d\vec{r}_2 = rd\varphi \; \hat{\varphi} + dz \; \hat{z} = \left(r \; \hat{\varphi} + \frac{\partial z}{\partial \varphi} \; \hat{z}\right) d\varphi\]
\noindent
The cross product then leads to:
\begin{eqnarray*}
d\vec{A} & = & d\vec{r}_1 \times d\vec{r}_2 \\
% & = & r dr d\varphi \; \hat{z} - dr df|_r \; \hat{\varphi} - r d\varphi df|_\varphi \; \hat{r} \\
% & = & \left(- r d\varphi df|_\varphi \; \hat{r} - dr df|_r \; \hat{\varphi}\right) + r dr d\varphi \; \hat{z} \\
& = & \left(- \frac{\partial f}{\partial r} \; \hat{r} - \frac{\partial f}{r \partial \varphi} \; \hat{\varphi} + \; \hat{z}\right) r dr d\varphi \\
& = & \left( - \vec{\nabla} f + \; \hat{z} \right) r dr d\varphi \\
\end{eqnarray*}
\noindent
Thus, in polar coordinates, the resulting scalar quantity is:
\begin{eqnarray*}
dA &= &\left| - \vec{\nabla} f + \; \hat{k} \right| r dr d\varphi \\
&= &\sqrt{\left(-\frac{\partial f}{\partial r}\right)^2 + \left(-\frac{1}{r}\frac{\partial f}{\partial \varphi}\right)^2 + 1^2} \; r dr d\varphi\\
\end{eqnarray*}
%The surface area vector $d\vec{A}$ element can be constructed as a cross product of two vectors $d\vec{r}_1$ and $d\vec{r}_2$ which are tangent to the surface. The scalar quantity
%\[dA = \left| d\vec{r}_1 \times d\vec{r}_2 \right|\]
%gives the scalar surface area element on the surface. Alternatively, as this quantity clearly depends on the concentration of level curves, it can also be expressed in terms of the gradient vector, $\vec{\nabla} f$, and the unit vector $\hat{n}$ normal to the domain, as
%\[ d\vec{A} = \hat{n} - \vec{\nabla} f\]
%Hence,
%\[ dA = \left| \hat{n} - \vec{\nabla} f \right| \]
%
%
%
%In polar coordinates, the expression for $d\vec{r}_1$ and $d\vec{r}_2$ become
%
%\[ d\vec{r}_1 = dr \; \hat{r} + dz \; \hat{z} = \left(1 \; \hat{r} + \frac{\partial z}{\partial r} \; \hat{z}\right) dr\]
%\[ d\vec{r}_2 = rd\varphi \; \hat{\varphi} + dz \; \hat{z} = \left(r \; \hat{\varphi} + \frac{\partial z}{\partial \varphi} \; \hat{z}\right) d\varphi\]
%
%The cross product of the differential quantities gives:
%
%\begin{eqnarray}
%d\vec{A} & = & d\vec{r}_1 \times d\vec{r}_2 \\
% & = & r dr d\varphi \; \hat{z} - dr df|_r \; \hat{\varphi} - r d\varphi df|_\varphi \; \hat{r} \\
% & = & \left(- r d\varphi df|_\varphi \; \hat{r} - dr df|_r \; \hat{\varphi}\right) + r dr d\varphi \; \hat{z} \\
% & = & \left(- \frac{\partial f}{\partial r} \; \hat{r} - \frac{\partial f}{r \partial \varphi} \; \hat{\varphi} + \; \hat{z}\right) r dr d\varphi \\
% & = & \left( - \vec{\nabla} f + \; \hat{z} \right) r dr d\varphi \\
%\end{eqnarray}
%
%Hence,
%
%\begin{eqnarray}
%dA &= &\left| - \vec{\nabla} f + \; \hat{k} \right| r dr d\varphi \\
% &= &\sqrt{\left(-\frac{\partial f}{\partial r}\right)^2 + \left(-\frac{1}{r}\frac{\partial f}{\partial \varphi}\right)^2 + 1^2} \; r dr d\varphi\\
%\end{eqnarray}
%