%
\section{Surfaces and Area Integrals}
The next selection of topics focuses on \textbf{parametric surfaces}. In $\mathbb{R}^3$, these will be given by continuous,  vector-valued functions $\mathbf{r} : D \rightarrow \mathbb{R}^3$, where $D \subset \mathbb{R}^2$ is a subset of \textit{two-dimensional} space. Usually, we will assume that $D$ is the closure of an open set. In terms of components, we write
$$
\mathbf{r}(u,v) = x(u,v) \mathbf{i} + y(u,v) \mathbf{j} + z (u,v) \mathbf{k}, \quad (u,v) \in D.
$$
The \textbf{surface} $\mathcal{S} \subset \mathbb{R}^3$ is the \textit{range} of the vector-valued function $\mathbf{r}$. We say that $\bfr$ is a parametrization of $\mathcal{S}$.



\begin{example}
Suppose that $f: D \rightarrow \mathbb{R}$ is a scalar function on some set $D \subset \mathbb{R}^2$.  Letting $z = f(x,y)$,  the graph of $f$ can be written as a parametric surface, with the $(x,y)$ variables as the parameters:
\begin{equation}\label{eq-31paragraphsurface}
\mathbf{r}(x,y) = x \mathbf{i} + y \mathbf{j} + f(x,y) \mathbf{k}.
\end{equation}
\end{example}

\begin{example}\label{31examp-sphere1}
The sphere $x^2 + y^2 + z^2 = a^2$ (for $a > 0$)  can be parametrized using spherical coordinates $(\theta, \phi)$ as
$$
x = a \cos \theta \sin \phi, \quad y = a \sin \theta \sin \phi, \quad z = a \cos \phi,
$$
where $(\theta, \phi) \in D$ live in the domain
$$
D : = \{(\theta, \phi) : 0 \leq \theta \leq 2 \pi, \, 0 \leq \phi \leq \pi \}.
$$
We could also use combinations of spherical or cylindrical coordinates to parametrize cylinders, cones, etc.
\end{example}


\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .8\linewidth]{figures/31fig0.png}
\caption{An example of a surface in $\mathbb{R}^3$.}
\end{minipage}
\end{figure}





We will usually assume that the function $\mathbf{r}$ and all its derivatives are continuous as needed.  We will also need a further condition, as follows. Let $P = \mathbf{r}(u_0, v_0)$ be a point on $\mathcal{S}$.  Notice that the equation
$$
\mathbf{r}(u, v_0) = x(u,v_0) \mathbf{i} + y(u,v_0) \mathbf{j} + z (u,v_0) \mathbf{k},
$$
and $\mathbf{r}(u_0,v)$ (defined similarly), are parametric equations for two curves $\mathcal{C}_u$ and $\mathcal{C}_v$ (respectively) that are contained in $\mathcal{S}$ and intersect at $P$.  At the point $P$, the curves $\mathcal{C}_u$ and $\mathcal{C}_v$ have their associated tangent vectors 
$$\mathbf{r}_u=\frac{\partial \bfr}{\partial u} (u_0,v_0)\ \ \hbox{ and }\ \ \mathbf{r}_v=\frac{\partial \bfr}{\partial u}(u_0 ,v_0).
$$
 Our additional assumption is that $\mathbf{r}_u$ and $\mathbf{r}_v$ are not parallel, except possibly when $(u_0,v_0)$ belongs to the boundary of $D$. We write this as  
\begin{equation}\label{eq-31nondeg}
 \mathbf{r}_u \times \mathbf{r}_v \neq \mathbf{0}.
\end{equation}
To see what could go wrong if (\ref{eq-31nondeg}) fails, consider the vector-valued function $\bfr(u,v)=(u+2v) \ii +(u+2v)\jj +(u+2v)\kk$, with $\bfr_v=2\bfr_u$. Then $x(u,v)=y(u,v)=z(u,v)$, so that $\bfr$ parametrizes the line $x=y=z$ rather than a surface. On the other hand, if (\ref{eq-31nondeg}) does hold, then the curves on $\mathcal{S}$ corresponding to fixed values of $u$ and $v$ form a nice grid on $\mathcal{S}$ as shown in the illustrations in this section.


\begin{definition}\label{31-def-survace}
Assume that $D \subseteq \mathbb{R}^2$ is the closure of an open set $U$, and that its boundary consists of finitely many piecewise smooth curves.

\smallskip\noindent
(a) We say that $\mathbf{r} : D \rightarrow \mathbb{R}^3$ is a \textbf{smooth parametrization} of its image $\mathcal{S}$ if
$\bfr$ is $C^1$ on $D$,  one-to-one on $U$, and (\ref{eq-31nondeg}) holds on $U$. 


\smallskip\noindent
(b) A surface $\mathcal{S}$ is \textbf{smooth} if it admits a smooth parametrization. 

\smallskip\noindent
(c) A surface is \textbf{piecewise smooth} if it consists of a finite number of smooth surfaces joined together along their common boundaries.

\smallskip\noindent
(d) A surface is \textbf{closed} if it bounds a closed and bounded region in $\mathbb{R}^3$. 
\end{definition}

If $\bfr$ is defined only on $D$ (and not on some neighbourhood of it), then the partial derivatives of $\bfr$ may not be well defined on its boundary. In that case, we understand the $C^1$ condition to mean that the first order partials of $D$ (initially defined on the interior of $D$) extend to continuous functions on all of $D$ including the boundary.


We will see below that a sphere, and the graph of a $C^1$ function $z=g(x,y)$, are smooth surfaces. 
Additionally, a sphere in $\mathbb{R}^3$ is a closed surface since it bounds a ball in 3 dimensions.



\subsection{Normal Vector for Parametric Surfaces}

We start with the formulas to normal vectors and tangent planes for parametric surfaces.
Suppose that $\mathbf{r}: D \rightarrow \mathbb{R}^3$ is a smooth parametrization of $\mathcal{S}$. 
Let $P=(x_0,y_0,z_0) \in \mathcal{S}$, so that $P = \mathbf{r}(u_0, v_0)$ for some fixed values $(u_0, v_0) \in D$.  
Define the curves $\mathcal{C}_u$ and $\mathcal{C}_v$, and the tangent vectors $\mathbf{r}_u (u_0,v_0)$ and $\mathbf{r}_v (u_0 ,v_0)$, as above. The normal vector at $P$ should then be perpendicular to these two vectors. Assuming (\ref{eq-31nondeg}), 
we may use
\begin{equation}\label{eq-31normalvec}
\boxed{\mathbf{n} = \mathbf{r}_u \times \mathbf{r}_v}
\end{equation}
as the equation for our normal vector at $P$. Since $\mathbf{r}_u$ and $\mathbf{r}_v$ are functions of $(u,v) \in D$, the vector $\mathbf{n} = \mathbf{n}(u,v)$ is a vector-valued function on $D$. 




\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/31fig1.png}
\caption{A surface (in blue with guide-lines) parametrized by $\mathbf{r} (u,v)$ with two curves $\mathbf{r}(u, v_0)$ and $\mathbf{r}(u_0, v)$. given in green.  The normal vector $\mathbf{n}$ at the point of intersection is perpendicular to both of the tangent vectors to these green curves.}
\end{minipage}
\end{figure}

Once we have found the normal vector $\mathbf{n}$, we can proceed with the equation of the tangent plane. We just write the equation of the plane that passes through $P=(x_0,y_0,z_0) $ and is
perpendicular to $\mathbf{n}=:\langle n_1,n_2,n_3\rangle$:
$$
n_1(x-x_0) + n_2(y-y_0) + n_3(z-z_0)=0.
$$





\subsection{The Area Element}
We also want to calculate the surface area element of parametric surfaces---again, in terms of the parametrizing function $\mathbf{r}$. We will use it to evaluate the area of surfaces and to set up surface integrals.


Assume that $\mathcal{S}$ is parametrized by some function $\mathbf{r}(u,v)$ on $D \subset \mathbb{R}^2$. 
We start by subdividing $\mathcal{S}$ using the (not necessarily rectangular) grid which consists of the curves $u =\textrm{constant}$ and $v =\textrm{constant}$.  This has the effect of dividing $\mathcal{S}$ into approximate parallelograms, spanned by vectors $\mathbf{r}_u \Delta u$ and $\mathbf{r}_v \Delta v$.  (We used a very similar argument in deriving the change-of-variables formula in Math 226.) This situation is illustrated in Figure 3.3 below.


\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .5\linewidth]{figures/31fig2.pdf}
\caption{The area of the indicated parallelogram is $\vert \mathbf{n} \vert \,  \Delta u \Delta v$.}
\end{minipage}
\end{figure}


The area of such a parallelogram can be calculated as
\begin{eqnarray*}
d S & = & \big\vert \big(\mathbf{r}_u \Delta u \big) \times \big( \mathbf{r}_v \Delta v \big)  \big\vert = \big\vert \mathbf{r}_u \times \mathbf{r}_v \big\vert \Delta u \Delta v \\[1ex]
\quad & = & \vert \mathbf{n} \vert \,  \Delta u \Delta v,
\end{eqnarray*}
where $\mathbf{n}$ is the normal vector from the previous subsection. This delivers the following \textbf{integration formula}:
\begin{equation}\label{eq-31integrationformulapara}
\boxed{
\iint_{\mathcal{S}} f(x,y,z) d S = \iint_{D} f \big( \mathbf{r}(u,v) \big) \big\vert \mathbf{r}_u \times \mathbf{r}_v \big\vert du dv
}
\end{equation}
where $dS$ denotes integration with respect to surface area.


The integration formula \eqref{eq-31integrationformulapara} shows us how to calculate some common quantities, such as:
\begin{itemize}
\item The area of $\mathcal{S}$ is given by
$$
(\textrm{Area of } \mathcal{S}) = \iint_{\mathcal{S}} 1 d S = \iint_{\mathcal{S}} \big\vert \mathbf{r}_u \times \mathbf{r}_v \big\vert d u d v
$$

\medskip

\item If $\mathcal{S}$ is a surface made of a material of variable density $f(x,y,z)$, then the integral $\iint_{\mathcal{S}} f(x,y,z) d S$ is the \textit{total mass} of $\mathcal{S}$.

\medskip

\item The \textbf{average value} of a function $f$ over the surface $\mathcal{S}$ is given by
$$
\frac{1}{\textrm{(Area of } \mathcal{S})} \iint_{\mathcal{S}} f (x,y,z) d S = \frac{\iint_{\mathcal{S}} f (x,y,z) d S}{\iint_{\mathcal{S}} 1 \, d S}
$$

\medskip

\item The \textbf{centroid} of $\mathcal{S}$ has coordinates $(\overline{x}, \overline{y}, \overline{z})$ where
$$
\overline{x} = \frac{1}{(\textrm{Area of } \mathcal{S})} \iint_{\mathcal{S}} x \, dS, \textrm{ and similarly for } \overline{y}, \overline{z}.
$$
\end{itemize}

Finally, it is very important that we realize: if $\mathbf{n} = \mathbf{r}_u \times \mathbf{r}_v$ is the normal vector found above, then the vectors $\lambda \mathbf{n}$ are also normal vectors, for any $\lambda \in \mathbb{R} \setminus \{0\}$. But, for the purpose of \textit{integration}, we must use $\vert \mathbf{n} \vert = \vert \mathbf{r}_u \times \mathbf{r}_v \vert$ with \textit{no rescaling}.


We now give several more specialized formulas for important classes of surfaces.

\subsection{Area of Graphs of Functions}\label{31-sub-graph}
Suppose that $\mathcal{S}$ is the graph of a function $z = g(x,y)$, where $(x,y) \in D$.  We parametrize it by
$$\mathbf{r}(x,y) = x \mathbf{i} + y \mathbf{j} + g(x,y) \mathbf{k}.$$
The curves $\mathcal{C}_x$ and $\mathcal{C}_y$ have the tangent vectors
$$
\mathbf{r}_x = \mathbf{i} + g_x \mathbf{k}, \quad \mathbf{r}_y = \mathbf{j} + g_y \mathbf{k},
$$
which produce the normal vector
\begin{eqnarray*}
\mathbf{n} = \mathbf{r}_x \times \mathbf{r}_y & = &
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 0 & g_x \\
0 & 1 & g_y
\end{vmatrix}
= - g_x \mathbf{i} - g_y \mathbf{j} + \mathbf{k}.
\end{eqnarray*}
Hence
\begin{equation}\label{31-eq-angle}
\boxed{ dS=\vert \mathbf{n} \vert dxdy = \sqrt{1 + g_x^2 + g_y^2}\, dxdy}
\end{equation}
Therefore the area of the part of the graph of $z = g(x,y)$ with $(x,y) \in D$ is
\begin{equation}\label{eq-31areaformulaparagraph}
\iint_{\mathcal{S}} 1 d S = \iint_{D} \sqrt{1 + g_x^2 + g_y^2} \, dx dy.
\end{equation}


\subsection{Geometric Interpretation of the Area Element}
We continue to work with the same set-up as in Section \ref{31-sub-graph}, and we would like to understand the geometric meaning of 
the formula (\ref{31-eq-angle}).
Let $\gamma$ denote the angle between $\mathbf{n}$ (the normal vector defined above) and $\mathbf{k}$ (the unit vector in the vertical direction).  Then, using the definition of the dot product, we have
$$
\mathbf{n} \cdot \mathbf{k}=\vert \mathbf{n} \vert \vert \mathbf{k} \vert \cos\gamma.
$$
We have $|\kk|=1$ and, from the computation of ${\bf n}$ in Section \ref{31-sub-graph}, 
$$
\mathbf{n} : = - g_x \mathbf{i} - g_y \mathbf{j} + \mathbf{k} \Rightarrow \mathbf{n} \cdot \mathbf{k}  = 1.
$$
Therefore 
$\vert \mathbf{n} \vert = \frac{1}{\cos\gamma} = \sec \gamma$, and 
$$
\boxed{ d S = |{\bf n}|\, dxdy = \frac{dx dy }{\cos\gamma} = \frac{dA}{\cos  \gamma}\ ,}
$$
where $dS$ and $dA$ denotes the respective area differentials on the surface and on its projection to the $xy$-plane. This whole situation is illustrated in Figure 3.3. Intuitively, if ${\bf n}$ is close to vertical, then $\mathcal{S}$ is close to being horizontal and $dS$ is close to $dA$; if on the other hand the angle between ${\bf n}$ and $\kk$ is close to $\pi/2$ (so that $\cos\gamma$ is close to 0), then the surface is almost vertical, so that a large part of $\mathcal{S}$ can project to a very small area in the $xy$-plane.

\begin{figure}[h]
\begin{minipage}{\linewidth}
\centering
\includegraphics[width = .6\linewidth]{figures/31fig3.png}
\caption{The normal vector (in red) makes angle $\gamma$ with the $\mathbf{k}$ direction. The area element on the surface is represented by the green parallelogram.}
\end{minipage}
\end{figure}

We can carry this one step further by considering a surface $\mathcal{S}$ given by the equation $G(x,y,z) = 0$.  
In this scenario, $\nabla G$ (if non-zero) is normal to the surface. It does not necessarily have the ``right'' magnitude for the area element.  However, a clever calculation fixes this issue.  Notice that
\begin{eqnarray*}
\vert \cos \gamma \vert & = & \frac{\vert \nabla G \cdot \mathbf{k} \vert}{\vert \nabla G \vert \vert \mathbf{k} \vert } = \frac{\vert G_z \vert}{\vert \nabla G \vert} = \frac{\vert G_z \vert}{\sqrt{G_x^2 + G_y^2 + G_z^2}}.
\end{eqnarray*}
This, in turn, implies that
\begin{equation}\label{31eq-differentialformdS}
\boxed{
dS = \frac{\vert \nabla G \vert}{\vert G_z \vert} \, dx dy = \frac{\sqrt{G_x^2 + G_y^2 + G_z^2}}{\vert G_z \vert}\, dxdy, \textrm{ if } G_z \neq 0.
}
\end{equation}
If $G_z=0$ at some point of the domain, it may still be possible to set up and calculate the surface area as an improper integral. Other times, it may be preferable to use some other parametrization of the surface near the troublesome point. For example, if $G_z=0$ but $G_y\neq 0$ on some part of the surface, we could parametrize that part of it by the $x$ and $z$ variables instead of $x$ and $y$.

\subsection{Surface Area for Sphere and Cone}
We conclude this section with calculating the surface elements for two ``canonical surfaces''.

\begin{example}
Let $\mathcal{S}$ be the sphere of radius $a>0$ centered at the origin, parametrized by $(\theta, \phi)$ as in Example \ref{31examp-sphere1}. Then, a calculation (which you should do!) shows that
\begin{equation}\label{31eq-sphericalnorm}
\vert \mathbf{n} \vert = \vert \mathbf{r}_{\theta} \times \mathbf{r}_{\phi} \vert = a^2 \sin \phi.
\end{equation}
Note the similarity to the Jacobian $r^2 \sin \phi$ which appears when we change from Cartesian to Spherical Coordinates.
\end{example}



The identity \eqref{31eq-sphericalnorm} is useful when calculating integrals over the sphere.  For example,
\begin{align*}
\iint_{\{x^2 + y^2 + z^2 = 1\}} z^2 \, d S & = \int_0^{2 \pi} \int_0^{\pi} \underbrace{\cos^2 \phi}_{z = \cos \phi} \underbrace{\sin \phi \,d \phi d \theta}_{(*)} 
\\
&= \int_0^{2 \pi} \bigg[- \frac{\cos^3 \phi}{3} \bigg]_0^{\pi} d \theta = \int_0^{2 \pi} \frac{2}{3} d \theta = \frac{4 \pi}{3}.
\end{align*}
The expression $(*)$ is the area element with $a = 1$.


\begin{example}
Consider the positive half-cone symmetric about the $z$-axis,
$$
z = b\sqrt{ x^2 + y^2},  \quad b > 0.
$$
We use the parametrization based on cylindrical coordinates,
$$
x(\theta, t) = t \cos \theta, \quad y(\theta, t) = t \sin \theta, \quad z (\theta, t) = bt,
$$
for $0 \leq \theta \leq 2 \pi$ and $t\geq 0$. Then
$$
\mathbf{r}_{t} = \langle \cos \theta, \sin \theta, 1 \rangle, \quad \mathbf{r}_{\theta} = \langle - t \sin \theta, t \cos \theta, 0 \rangle,
$$
so that
$$
\mathbf{n} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\cos \theta & \sin \theta & 1 \\
- t \sin \theta & t \cos \theta & 0
\end{vmatrix}
=
-b t \cos \theta \, \mathbf{i} - bt \sin \theta \, \mathbf{j} + t \, \mathbf{k}.
$$
Hence,
$$
\vert \mathbf{n} \vert = \bigg(b^2 t^2 (\cos^2 \theta + \sin^2 \theta) + t^2 \bigg)^{1/2} = t \sqrt{b^2 + 1}.
$$


\end{example}


We use this to calculate integrals over the cone.  For instance, if $\mathcal{S}$ is the conical segment
$$
z = b\sqrt{x^2 + y^2}, \quad x^2 + y^2 \leq a^2, 
$$
for some $a,b > 0$, we have
\begin{eqnarray*}
\iint_{\mathcal{S}} z^3 d S & = & \int_0^a \int_0^{2 \pi} (\underbrace{b t}_{z})^3 \underbrace{(t \sqrt{b^2 + 1}) d \theta d t}_{d S} = 2 \pi b^3 \sqrt{b^2 + 1} \int_0^a t^4 dt \\
\quad & = & 2 \pi b^3 \sqrt{b^2 + 1} \, \frac{t^5}{5}\Big\vert_0^a \\
\quad & = & 2 \pi b^3 \sqrt{b^2 + 1} \, \frac{a^5}{5}.
\end{eqnarray*}


\bigskip



\section*{Worked Problems}
\begin{enumerate}[1.]
\item
{\it Find the area of the part of the plane $2x - y + z = 10$ that lies above the disc $(x - 3)^2 + y^2 \leq 1$.}

\bigskip

We consider our plane as a parametric surface, parametrized by
$$
\mathbf{r}(x,y) = x \mathbf{i} + y \mathbf{j} + (10 - 2x + y) \mathbf{k}.
$$
Then
$$
\mathbf{r}_x = \mathbf{i} - 2 \mathbf{k}, \quad \mathbf{r}_y = \mathbf{j} + \mathbf{k},
$$
so that
$$
\mathbf{n} = \mathbf{r}_x \times \mathbf{r}_y =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 0 & -2 \\
0 & 1 & 1
\end{vmatrix}
=
2 \mathbf{i} - \mathbf{j} + \mathbf{k} \Rightarrow \vert \mathbf{n} \vert = \sqrt{6}.
$$
Hence, we have
$$
\textrm{(Area of } \mathcal{S}) = \sqrt{6} \cdot \textrm{(Area of disc)} = \sqrt{6} \pi.
$$

\bigskip

(For a more general plane $Ax+By+Cz=D$ with $C\neq 0$, a similar calculation shows that
$$
dS=\frac{\sqrt{A^2+B^2+C^2}}{|C|}\, dxdy.
$$
The area element is a constant multiple of $dxdy$ because the normal vector to the plane makes a constant angle with the $z$-axis. A good exercise is to try to work out the details.) 


\bigskip

\item{\it Find the area of the part of the paraboloid $z = x^2 + y^2$ that lies below the plane $z = 9$.}

\bigskip

Call the surface described above $\mathcal{P}$.  In particular, if $(x,y,z) \in \mathcal{P}$, then we must have $z \leq 9$, and also $x^2 + y^2 \leq 9$.  Set $g(x,y) = x^2 + y^2$ with associated domain $D : = \{(x,y) : x^2 + y^2 \leq 9 \}$. Then $\mathcal{P}$ is the graph of $g$. Using that $g_x = 2x$ and $g_y = 2y$, formula \eqref{eq-31areaformulaparagraph} gives
$$
\textrm{(Area of } \mathcal{P}) = \iint_{D} \vert \mathbf{n} \vert dx dy = \iint_{\{x^2 + y^2 \leq 9 \}} \sqrt{1 + 4x^2 + 4 y^2} dx dy
$$
We evaluate this integral in polar coordinates
$$
\int_0^{2 \pi} \int_0^3 \sqrt{1 + 4 r^2}\, r \, dr d \theta = \int_0^{ 2 \pi} \frac{1}{12} \big(1 + 4 r^2 \big)^{3/2} \big\vert_0^3 d \theta = \underbrace{\frac{\pi}{6} \big( 37^{3/2} - 1 \big)}_{\textrm{Area of } \mathcal{P}}
$$

\bigskip

\item{\it Assume that $\mathcal{P}$ is the paraboloid defined in the previous problem.  Calculate $\iint_{\mathcal{P}} z d S$.}

\bigskip

This is a modification of the previous example.  We have
\begin{eqnarray*}
\iint_{\mathcal{P}} z \, dS & = & \iint_{D} \underbrace{(x^2 + y^2)}_{z(x,y)} \cdot \underbrace{\sqrt{1 + 4x^2 + 4 y^2} \, dx dy}_{dS} = \int_0^{2 \pi} \int_0^3 r^2 \cdot \sqrt{1 + 4 r^2}\, r \, dr d \theta \\
\quad & = & \frac{\pi}{8} \bigg[\frac{37^{5/2}}{5} - \frac{37^{3/2}}{3} - \frac{1}{5} + \frac{1}{3} \bigg].
\end{eqnarray*}
The last calculation can be done using the substitution $u = 1 + 4r^2$. (Exercise: complete the details!)


\bigskip

\item
{\it A parametric surface is given by the equations $x=u^2, y=uv, z=\frac{1}{2}v^2$, with $u,v\geq 0$.

\medskip

(a) Find the equation of the tangent plane to the surface at the point  
$P$ where $x=4$, $y=4$, $z=2$. }


\medskip

We have
$\bfr_u=\langle 2u, v, 0\rangle$ and $\bfr_v=\langle 0, u, v\rangle$, so that the normal vector is
$${\bf n}=\bfr_u\times \bfr_v=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
2u& v& 0\\
0 & u& v\\
            \end{matrix}\right|
 =v^2 \ii -2uv \jj + 2u^2 \kk \ , $$
The point $P$ corresponds to $u^2=4$, $uv=4$, $v^2=4$, so that the normal vector at $P$ is
${\bf n}= 4\ii -8\jj +8\kk$. (Optional: we can also solve these equations to get $u=v=2$.) 
Therefore the tangent plane is
$$
4(x-4)-8(y-4)+8(z-2)=0,
$$
which can be simplified to $(x-4)-2(y-4)+2(z-2)=0$, $x-2y+2z=0$.


\medskip

{\it (b) Find the area of the part of the surface where
$0\leq u\leq 1$ and $0\leq v\leq 2$. }

\medskip


Using ${\bf n}$ from part (a), we have
$$
|{\bf n}| = \sqrt{ v^4+4u^2v^2 + 4u^4 } = v^2+2u^2\ .
$$
The area is
\begin{align*}
\int_0^2 \int_0^1 (v^2+2u^2) du\,dv 
&= \int_0^2 \left[ uv^2+ \frac{2u^3}{3} \right]_{u=0}^1 dv
\\
&= \int_0^2 (v^2+ \frac{2}{3}) dv = \left[ \frac{v^3}{3}+ \frac{2v}{3} \right]_{v=0}^2
= \frac{8}{3}+ \frac{4}{3}=4.
\end{align*}





\end{enumerate}


\bigskip

\section*{Practice Problems}
\begin{enumerate}[1.]


\item
{Evaluate the surface integral
$\iint_{\mathcal S}\sqrt{x} \, dS$, if 
$\mathcal{S}$ is the parametrized surface ${\bf r}(u,v)=(u^2,v,u+v )$ with $0\leq u\leq 1$, $0\leq v\leq 1$.}

\bigskip

\item
{A parametric surface is given by the equations $x=uv, y=u^2-v^2, z=u^2+v^2$, with $u,v\geq 0$.
Find the equation of the plane tangent to the surface at the point  
$P$ where $(u,v)=(2,1)$. }

\bigskip

\item
Let $\mathcal{S}$ be the surface ${\bf r}(u,v)=\langle u \cos v, u\sin v, v\rangle$ with $0\leq u\leq 1$, $0\leq v\leq \pi$.

\medskip
\noindent
(a) Find the tangent plane to the surface at the point where $u=1/2,v=\pi/6$.

\medskip
\noindent
(b) Evaluate the surface integral $\iint_{\mathcal{S}} (x^2+y^2)^{1/2} dS$.

 
\bigskip
 
\item
{
Let $\mathcal{S}$ be the surface parametrized by ${\bf r}(u.v)=(u+\sin u,v+\sin v,1-\cos(u+v))$, $0<u<\pi$,
$0<v<\pi$ (the boundary of the rectangle is not included). 


\medskip
\noindent
(a) Find all $(u,v)$ such that the tangent plane to the surface at ${\bf r}(u,v)$ is parallel
to the $xy$-plane. }


\medskip
\noindent
{(b) 
Prove that the area of $\mathcal{S}$ is less than $5\pi^2$. }

\bigskip


\item {Evaluate the integral
$\iint_{\mathcal S} (x^2+4y^2+1) \, dS$, if 
$\mathcal{S}$ is the part of the paraboloid $z=\frac{x^2}{2}+y^2$ that lies above the region 
$D=\{(x,y):\ x^2+4y^2\leq 1$ and $y\geq 0\}$
in the $xy$-plane.}

\bigskip

\item {
Let $\mathcal{C}$ be a smooth simple curve in the plane $z=0$, parametrized by ${\bf r}(t)=x(t)\ii+y(t)\jj$ with $a\leq t\leq b$. Construct a surface $\mathcal{S}$ in three dimensions by taking the union of all line segments that connect points ${\bf r}(t)$ on the curve to the fixed point $P=(0,0,1)$. Prove that
$$
\frac{1}{2}(\hbox{length of }\mathcal{C}) \leq  (\hbox{area of }\mathcal{S}).
$$
Give an example of $\mathcal{C}$ such that equality holds in the above formula.

(Hint: How would you parametrize the surface? You may want to start by parametrizing each line segment separately.)}

\bigskip


\item Let $D$ be a planar domain such that 
$$
D\subset \{(x,y):\ x^2+y^2\leq 64\hbox{ and } \frac{x^2}{36}+\frac{y^2}{49}\geq 1\}.
$$
(This is just an inclusion and $D$ does not have to be equal to the set on the right.)

Let $\mathcal{S}$ be the part of the sphere $x^2+y^2+z^2=64$ that lies above $D$. Prove that
$$
(\hbox{area of }D)\leq\frac{3}{4} (\hbox{area of }\mathcal{S}).
$$



\item {
Let $\mathcal{S}$ be the part of the sphere $x^2+y^2+z^2=1$ that lies above the cone $z=\sqrt{x^2+y^2}$. Consider the integral
$$
\iint_{\mathcal{S}} \sqrt{x^2+y^2} dS.
$$
A student has evaluated this integral as follows.



\medskip\noindent
\fbox{
\begin{minipage}{0.8\textwidth}

The unit normal to $\mathcal{S}$ is the vector ${\bf N}=x\ii +y\jj +z\kk$. The angle $\gamma$ between ${\bf N}$ and $\kk$ is $\gamma=\phi$ in spherical coordinates, so that the area element is $\frac{1}{\cos\gamma}=\frac{1}{\cos\phi}$. Also in spherical coordinates, $\sqrt{x^2+y^2}=\sin\phi$ on the sphere. Therefore
\begin{align*}
\iint_{\mathcal{S}} \sqrt{x^2+y^2} dS
&= \int_0^{2\pi} \int_0^{\pi/4} \frac{\sqrt{x^2+y^2}}{\cos\gamma } d\phi\, d\theta\\
&= \int_0^{2\pi} \int_0^{\pi/4} \frac{\sin\phi}{\cos\phi} d\phi\, d\theta\\
&= 2\pi \Big[ -\ln(\cos\phi) \Big]_{0}^{\pi/4}\\
&= 2\pi \Big( -  \ln(\cos(\pi/4))  + \ln(\cos 0)\Big) \\
& =2\pi (- \ln(\sqrt{2}/2)- \ln(1) ) = - 2\pi \ln(\sqrt{2}/2).
\end{align*}
\end{minipage}
}

Is this solution correct? If not, explain where the student has made a mistake (if there is more than one error, indicate all of them), and provide the correct solution.}




\end{enumerate}

