$\newcommand{\R}{\mathbb R }$ $\newcommand{\N}{\mathbb N }$ $\newcommand{\Z}{\mathbb Z }$ $\newcommand{\bfa}{\mathbf a}$ $\newcommand{\bfb}{\mathbf b}$ $\newcommand{\bfc}{\mathbf c}$ $\newcommand{\bfe}{\mathbf e}$ $\newcommand{\bft}{\mathbf t}$ $\newcommand{\bfn}{\mathbf n}$ $\newcommand{\bff}{\mathbf f}$ $\newcommand{\bfF}{\mathbf F}$ $\newcommand{\bfk}{\mathbf k}$ $\newcommand{\bfg}{\mathbf g}$ $\newcommand{\bfG}{\mathbf G}$ $\newcommand{\bfH}{\mathbf H}$ $\newcommand{\bfh}{\mathbf h}$ $\newcommand{\bfu}{\mathbf u}$ $\newcommand{\bfv}{\mathbf v}$ $\newcommand{\bfw}{\mathbf w}$ $\newcommand{\bfx}{\mathbf x}$ $\newcommand{\bfp}{\mathbf p}$ $\newcommand{\bfy}{\mathbf y}$ $\newcommand{\ep}{\varepsilon}$
Suppose that $S$ is a surface in $\R^3$ parametrized by a function $$ \bfx = \bfG(\bfu), \quad \bfu \in R $$ where $R$ is a measurable subset of $\R^2$. We will usually assume that $\bfG$ is of class $C^1$ and one-to-one in $R$, and that $\{ \frac{\partial \bfG}{\partial u}, \frac{\partial \bfG}{\partial v}\}$ are linearly independent at every point in $R$, except possibly on a set of zero content.
Also assume that $f$ is a continuous (scalar) function on $\R^3$
and that $\bfF$ is a continuous vector field on $\R^3$.
If $|\frac{\partial \bfG}{\partial u}\times \frac {\partial \bfG}{\partial v}|$ is integrable on $R$, then
we define
$$
\mbox{area of }S \ :=
\ \iint_R |\frac{\partial \bfG}{\partial u}\times \frac {\partial \bfG}{\partial v}|\, du\,dv\, ,
$$
(Here and below, we write $du\, dv$ to indicate an integral with respect to the $(u,v)$ variables. We do not mean that it has to be to be written as an iterated integral in that order.)
$$
\mbox{surface integral of }f\mbox{ over }S := \iint_S f\, dA \ := \ \iint_R f(\bfG(\bfu))
\left|\frac{\partial \bfG}{\partial u}\times \frac {\partial \bfG}{\partial v}\right|\, du\,dv\, ,
$$
and
$$
\mbox{surface integral of }\bfF\mbox{ over }S := \iint_S \bfF\cdot \bfn dA \ := \ \iint_R
\bfF(\bfG(\bfu)) \cdot
\left(\frac{\partial \bfG}{\partial u}\times \frac {\partial \bfG}{\partial v}\right)
du\,dv\, .
$$
We also sometimes say or write integral
instead of surface integral
when referring to the above integrals, if it is clear that we are integrating over a surface.
Note that the area of $S$ equals $\iint_S 1\, dA$.
Although we have used a parametrization in the above definitions, we will see below that, exactly as with line integrals,
surface integrals of functions are independent of the choice of parametrization, and
the choice of a parametrization can change the sign of the surface integral of a vector field, so we will need to pay attention to orientation when carrying out such integrals. (In our definition of $\iint_S \bfF\cdot \bfn \,dA$, there is an implicit orientation determined by the parametrization.) These issues wil be illustrated by examples and discussed in more detail below.
The integral of a function $f$ over a surface $S$ is interpreted as: $$ \mbox{surface integral of }f \mbox{ over } S \ = \ ( \mbox{area of }S )\times( \mbox{average of } f \mbox{ over }S). $$ The rigorous meaning of the above formula is that it is the definition of the average of $f$ over $S$ (i.e., as integral/area). But since we have an intuitive sense of what both terms on the right-hand side should mean, its non-rigorous meaning is that it tells us how to think about the surface integral of a scalar function.
Example 1.
Let $S$ be the set $\{ (x,y,z)\in \R^3 : x^2+y^2+z^2 = 1, \ z\ge 0\}$
Compute the area of $S$, as well as the integrals
$$
\iint_S z\, dA, \quad\mbox{ and }\quad \iint_S (0,0,z)\cdot \bfn \, dA,
$$
with the $\bfn$ oriented upwards. (We will later discuss in detail what $\bfn$ oriented upwards
means. This example will illustrate its meaning in this spcificc ase.)
Solution.
First we have to parametrize $S$. We can do this by
$$
\bfx = ( \cos\theta \sin\phi , \sin\theta\sin\phi ,\cos \phi) =: \bfG(\theta,\phi), \qquad 0\le\theta\le 2\pi, \ \ 0\le \phi\le \pi/2 .
$$
Now we work out the values of the various integrals in this specific case via routine computations.
First,
\begin{align}
\partial_\theta \bfG &= ( -\sin\theta \sin\phi , \cos\theta\sin\phi ,0)
\nonumber \\
\partial_\phi\bfG &=( \cos\theta \cos\phi , \sin\theta\cos\phi ,-\sin \phi),\quad\mbox{ and thus }
\nonumber \\
\partial_\theta \bfG\times \partial_\phi \bfG &=(-\cos\theta\sin^2\phi, -\sin\theta\sin^2\phi, -\cos\phi\sin \phi) \nonumber
\end{align}
Then a short computation shows that
$$
|\partial_\theta \bfG\times \partial_\phi \bfG| = |\sin \phi| = \sin \phi
$$
(since $\sin \phi\ge 0$ for $\phi\in [0,\pi/2]$.)
It follows that
\begin{align}
&\mbox{area}(S) = \int_0^{2\pi}\int_0^{\pi/2} \sin \phi \, d\phi\, d\theta = \cdots = 2\pi \nonumber \\
&\iint_S z \, dA =\int_0^{2\pi}\int_0^{\pi/2} \cos\phi \sin\phi\, d\phi \, d\theta = \cdots = \pi.
\end{align}
(The substitution $u = \sin \phi$ is helpful in
evaluating the second integral.)
Before evaluating the integral of the vector field, we have to pay attention to
orientation. The problem says that $\bfn$ should be oriented
upwards. What this means is that $\frac{\partial \bfG}{\partial u}\times \frac {\partial \bfG}{\partial v}$ should point upwards. Point upwards
means that the $z$-component of the vector should be positive. We can see
that for the parametrization we have chosen, the $z$-component
of $\frac{\partial \bfG}{\partial u}\times \frac {\partial \bfG}{\partial v}$is $-\cos\phi \sin\phi$, which is negative for $0\le \phi < \pi/2$, hence points
downwards. So to correct for the wrong orientation, we should multiply by $-1$.
Thus
$$
\iint_S (0,0,z)\cdot \bfn \, dA =
-\int_0^{2\pi}\int_0^{\pi/2} -\cos^2\phi\sin \phi \, d\phi\,d\theta = \cdots = +\frac 2 3 \pi,
$$
(using the substitution $u = \cos\phi$.)
When $\frac{\partial \bfG}{\partial u}$ and $\frac{\partial \bfG}{\partial v}$ are linearly independent (as we typically assume to be the case, except possibly for a set of zero content), then
normal, which means
orthogonal, is often used in this context.) As a result of these considerations, $$ \frac{ \frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v} } {|\frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}|} \qquad\mbox{ and }\qquad \frac{ -\frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v} } {|\frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}|} = \frac{ \frac{\partial \bfG}{\partial v}\times \frac{\partial \bfG}{\partial u} } {|\frac{\partial \bfG}{\partial v}\times \frac{\partial \bfG}{\partial u}|} %\quad \mbox{ (order of } u,v\mbox{ reversed on right)} $$ are unit normal vectors to $S$ --- unit vectors that are normal to $S$ at the point $\bfG(u,v)$.
We therefore sometimes write $ \frac{ \frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v} }
{|\frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}|}
=:
\bfn(\bfG(\bfu))$, where $\bfn$
stands for normal
.
With this notation,
for any vector field $\bfF$,
\begin{align}
\iint_S \bfF\cdot \bfn \,dA
&\ = \ \iint_R \bfF(\bfG(\bfu))\cdot
( \frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v} )\, du\, dv\nonumber \\
&\ = \ \iint_R \bfF(\bfG(\bfu))\cdot \bfn(\bfG(\bfu))
\left| \frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v} \right|
\, du\, dv\nonumber \\
&=
\mbox{surface integral of the scalar function }(\bfF\cdot \bfn)\mbox{ over }S.\nonumber
\end{align}
For this reason, the notation $\iint_S \bfF\cdot \bfn\, dA$ for a surface integral
of a vector field is entirely consistent with the notation for the surface integral of a scalar function.
So far we have discussed the direction of $ \frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}$.
What about its magnitude? Properties of the cross product imply that
its magnitude is exactly the area of the parallelogram in $\R^3$ with
sides
$\frac{\partial \bfG}{\partial u}$ and $\frac{\partial \bfG}{\partial v}$.
This leads to a non-rigorous but suggestive interpretation, which is that
\begin{align}
| \frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}|\,du\,dv
& =
| (\frac{\partial \bfG}{\partial u} du) \times (\frac{\partial \bfG}{\partial v} dv)|\nonumber \\
&\sim \mbox{ area of the infinitessimal parallelogram with sides }
\frac{\partial \bfG}{\partial u}du\mbox{ and }
\frac{\partial \bfG}{\partial v}dv.
\end{align}
Thus, it is traditional to imagine that when we integrate
$\iint_R | \frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}|\,du\,dv$, we are adding up the areas of the infinitely many
infinitessimal parallelogram that make up the surface $S$
.
This can be viewed as a sort of justification for our defintion of
area. One can also make similar arguments with rectangles of
very small but but finite size. Other justifications will be presented below.
Example 2. Let $S$ be the boundary of the set $\{ (x,y,z)\in \R^3 : x^2+y^2\le 1, |z|\le 1\}$, and compute $\iint_S \bfF\cdot \bfn\, dA$ for $\bfF = (x^2z,y,0)$, and with $\bfn$ oriented outwards.
Solution. A feature of this problem is that $\partial S$ is not a smooth surface everywhere. However, it is a a piecewise smooth surface, that is, a union of smooth surfaces: $$ \begin{aligned} S_{top} &:= \{(x,y,z)\in \R^3 : \ x^2+y^2\le 1, \ z = 1\} \\ S_{side} &:= \{(x,y,z)\in \R^3 : \ x^2+y^2 = 1, \ |z|\le 1\} \\ S_{bottom} &:= \{(x,y,z)\in \R^3 : \ x^2+y^2\le 1, \ z = - 1\} \end{aligned} $$ that intersect only along lower-dimensional sets. For example, $S_{top}\cap S_{side} = \{(x,y,z) : x^2+y^2 = 1, z = 1\}$ which is a smooth (1-dimensional) curve in $\R^3$), hence has zero (2-dimensional) area. In cases like this, we simply integrate along each of the surfaces and add up the results.
integral over $S_{top}$.. With some practice you can see right away that this integral must equal $0$, because we can see geometrically that $\bfn = (0,0,1)$ on $S_{top}$, and hence that $\bfF\cdot \bfn = 0$ there, due to the form of $\bfF$. But for practice, let's do it:
integral over $S_{bottom}$.. For exactly the same reason, $$ \iint_{S_{bottom}} \bfF\cdot \bfn \, dA = 0. $$
integral over $S_{side}$..
The above formulas simplify if $S$ is the graph of some function $\phi$, i.e. $S = \{ (x,y, \phi(x,y) ): (x,y)\in R\}$. Then we can use the parametrization $$ \bfx = (u , v , \phi(u,v)) =: \bfG(u,v), \qquad (u,v)\in R. $$ With this choice, $$ \partial_u\bfG = (1, 0, \partial_u \phi), \qquad \partial_v\bfG = ( 0,1, \partial_v \phi), \qquad \partial_u\bfG\times \partial_v\bfG=(-\partial_u \phi, -\partial_v\phi, 1). $$ and thus $|\partial_u\bfG\times \partial_v\bfG| = \sqrt {1+|\nabla\phi|^2}$. So we get the simpler formulas $$ \mbox{area}(S) \ = \ \iint_R \sqrt {1+|\nabla\phi|^2}\, dA $$ $$ \iint_S f\, dA \ = \ \iint_R f(u,v,\phi(u,v)) \sqrt {1+|\nabla\phi|^2} \, dA $$ and $$ \iint_S \bfF\cdot \bfn \, dA \ = \ \iint_R \bfF(u,v,\phi(u,v)) \cdot (-\partial_u\phi, -\partial_v\phi, 1) \, dA $$ if $S$ is oriented with the unit normal pointing upwards; for the downward unit normal, one would have to multiply by $-1$.
(Since $u=x$ and $v=y$, these formulas are often written in terms of variables $(x,y)$ rather than $(u,v)$.)
It follows from a short argument that the integral of a scalar function $f$ over a surface $S$ is independent of the parametrization A more precise version of this statement appears in Theore 1 below.
The special case $f=1$ reduces to the area of a surface $S$ is independent of the parametrization
This is nice to know, because if our definition of area
depended on the
parametrization, then it would probably not be a good definition.
For integrals of vector fields, as already discussed, we have to pay attention to the orientation, which is the choice of the direction in which the unit normal $\bfn$ points, as we have already seen in examples. That is, the integral of a vector field $\bfF$ over a surface $S$ depends on the orientation of $S$ but is otherwise independent of the parametrization. In fact, changing the orientation of a surface (which amounts to multiplying the unit normal $\bfn$ by $-1$, changes the sign of the surface integral of a vector field. See Theorem 1 below for a precise statement.
Because of the above facts, if we want to integrate a scalar function over a surface, or (in particular) to find the area of a surface, we do not need to know how the surface is oriented. but to compute the surface integral of a vector field, we need to specify the orientation of the surface. So if someone asks you to compute $\iint_S \bfF\cdot \bfn\, dA$, it is their job to specify not only the vector field $\bfF$ and the surface $S$, but also the orientation of the unit normal $\bfn$. One that is done, we can speak unambiguously about $\iint_S \bfF\cdot \bfn \, dA$, without having to say precisely which parametrization we have in mind. Of course, we have already done this above.
Assume that $S$ is a surface parametrized by $$ (x,y,z) = \bfG(u,v) \ \ \ \mbox{ for }(u,v) \in R, $$ and that $\varphi:W\to R$ is a function that is one-to-one, onto, of class $C^1$, and with $C^1$ inverse. Here both $W$ and $R$ are assumed to be measurable subsets of $\R^2$. Let us say that the variables in $W$ are called $(s,t)$, so that the $(s,t)$ and $(u,v)$ variables are related by $(u,v) = \varphi(s,t)$.
Now define $\bfH(s,t) = \bfG\circ\varphi(s,t)$. Then $S$ is also parametrized by $\bfx = \bfH(s,t), (s,t)\in W$.
Our assumptions about $\varphi$ imply that $\det D\varphi(s,t)\ne 0$ for all $(s,t)\in W$. Also, by a short computation that uses the chain rule, one can check that \begin{equation} \frac{\partial \bfH}{\partial s}\times \frac{\partial \bfH}{\partial t} = \left(\frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}\right) \cdot \det D\varphi \label{cvsurf}\end{equation} where derivatives of $\bfH, \varphi$ are evalated at $s,t$ and derivatives of $\bfG$ are evaluated at $(u,v) = \varphi(s,t)$. It follows that
$$ \begin{aligned} \det D\varphi >0 & \mbox{ everywhere in }W \\ &\ \iff \ \frac{\partial \bfH}{\partial s}\times \frac{\partial \bfH}{\partial t} \mbox{ and } \frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}\mbox{ point in the same direction} \\ &\ \iff \ \mbox{ parametrizations }\bfG\mbox{ and }\bfH\mbox{ induce the same orientation}, \end{aligned} $$ where the orientation induced by a parametrization $\bfG$ is the orientation of $\bfn := \partial_u\bfG\times \partial_v \bfG/|\partial_u\bfG\times \partial_v \bfG|$.
Theorem 1.
Under the above circumstances,
$$
\iint_R f(\bfG(u,v))\left|\frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}\right| du\,dv =
\iint_W f(\bfH(s,t))\left|\frac{\partial \bfH}{\partial s}\times \frac{\partial \bfH}{\partial t}\right|
ds\,dt
$$
and
$$
\iint_R \bfF(\bfG(u,v))\cdot\left(\frac{\partial \bfG}{\partial u}\times \frac{\partial \bfG}{\partial v}\right) du\,dv = \pm
\iint_W \bfF(\bfH(s,t))\cdot \left(\frac{\partial \bfH}{\partial s}\times \frac{\partial \bfH}{\partial t}\right)
ds\,dt
$$
with $+$
if $\bfG,\bfH$ induce the same orientation and $-$
if they induce the opposite orientation.
Proof. This can be checked using \eqref{cvsurf} and the change of variables theorem for multiple integrals.
This material was not covered in any lecture and you should read it only if you are interested.
Surface area is a subtler concept than volume. If you are asked to measure the volume of a physical object, it is (in principle) easy to do: dunk the object in a bucket of water, and measure how much water it displaces.
Measuring surface area is much more difficult. How might you do it?
One possibility is to paint the object on one side with a thin coat of paint. Then the area of the surface should be proportional to the volume of paint used, which we can measure. For example, if we are able to apply paint with a uniform thickness of exactly $h$ cm, (a realistic number might be $h = 1 \,mm = 0.1\, cm$) and if we use $V$ $cm^3$ of paint, then the area should be about $$ \frac{ V \, cm^3}{h \, cm} = \frac V h \, cm^2. $$
Example 3. Let $S$ denote the unit sphere in $\R^3$, that is $$ S :=\{\bfx \in \R^3 : |\bfx| = 1\}. $$ If we coat $S$ with a layer of paint of thickness $h$, the painted sphere occupies the region $$ S_h := \{ \bfx\in \R^3 : 1 \le |\bfx| \le 1+h\}. $$ The volume is easily computed and equals $$ \mbox{Vol}(S_h) = \frac 43 \pi [(1+h)^3 - 1^3] = \frac 43\pi(3h +3h^2+h^3). $$ So the area of the sphere should be approximately $$ \mbox{area}(S) \approx \frac {\frac 43\pi(3h^2 +3h+h^3)}h = 4\pi + 4\pi h + \frac 43\pi h^2. $$ If $h$ is small, this is very close to the number you have probably computed in first-year calculus or elsewhere, $4\pi$.
If we had (somehow) painted $S$ on the inside, we would have gotten a different approximation, area$(S)\approx 4\pi - 4\pi h + \frac 43 \pi h^2$. But these two different approximations would yield the same number for the area if we take the limit $h\to 0$.
These considerations suggest a way of computing area of a surface $S$ that is a mathematical idealization of the procedure of measuring the volume of a thin coat of paint.
choose an orientation $\bfn$ for the surface. (It will not matter which you choose.)
For $h>0$, define \begin{equation}\label{Sh.def} S_h := \{ \bfx + s \bfn : \bfx \in S, 0 \le s \le h \}. \end{equation} This is the set of all points on the side of the surface in the $\bfn$ direction, that are a distance as most $h$ from the surface, or if you prefer, the very thin region occupied by a coat of paint of thickness $h$.
Then we expect that $$ \mbox{area}(S) = \lim_{h\to 0}\frac{\mbox{Vol}(S_h)}{h}. $$ As the example of the sphere suggests, this procedure should yield a result that does not depend on the orientation of $\bfn$.
In fact this is true:
Theorem 2. Let $S$ be a surface parametrized by $\bfG:R\to \R^3$, where $R$ is compact and measurable, and $\bfG$ is $C^2$ on an open set containing $R$. Also assume that $\{ \partial_u \bfG, \partial_v\bfG\}$ are linearly independent everywhere in $W$, and hence induce an orientation $\bfn$ (as discussed above). Then $$ \mbox{area}(S) = \lim_{h\to 0}\frac{\mbox{Vol}(S_h)}{h} $$ for $S_h$ as defined in \eqref{Sh.def} (using the orientation induced by $\bfG$).
We will never use the theorem; its only role for us is as a motivation/justification for the definition of area. You may find it to be a more convincing justification than the hand-waving arguments about adding up areas on infinitessimal parallelograms alluded to earlier.
The proof is a little too involved for this class, but we can at least describe the outline; see below if interested.
Sketch of the proof.
For $s>0$ define $$ \widetilde \bfG(u,v,s) := \bfG(u,v) + s \bfn(\bfG(u,v)), $$ where $$ \bfn(\bfG(u,v)) = \frac{\partial_u\bfG\times \partial_v \bfG}{|\partial_u\bfG\times \partial_v \bfG|}(u,v). $$ Also, let $R_h := \{(u,v,s) : (u,v)\in R, 0\le s \le h\}$. Then
check that $\widetilde \bfG$ is a transformation of $R_h$ onto $S_h$ (that is, one-to-one and onto, with $C^1$ inverse.)
Then in principle we can compute the volume of $S_h$ using the Change of Variables Theorem, leading to $$ \mbox{Vol}(S_h) = \iint_{R_h} |\det \widetilde \bfG(u,v,s)|\, du\,dv\,ds. $$
Next, one can prove that $$ \lim_{h\to 0}|\det \widetilde \bfG(u,v,s)| = |\partial_u\bfG\times \partial_v\bfG|(u,v). $$
Once this is known, then completing the proof of the theorem is relatively straightforward. $\quad\Box$
The details we have omitted are not at all easy, but they are in principle within reach for MAT237 students who have gotten this far. That is, you know all the mathematical techniques/definitions/theorems etc needed for the proofs; it's just that the arguments are long and complicated, and we have to cover other things, so we choose not to spend the (long) amounts of time that would be required for a proof.
You must know the definitions of the area of a surface $S$, and the surface integral of a scalar function or a vector field and you must be able to use them.
In all these questions, the most important part is to set up the integrals correctly, since this is the new part. We already know (or don't know, as the case may be) how to evaluate them.
Compute the area of the following surfaces. Do enough examples (and no more) to fix in your memory the definition of area of a surface.
$S = \{ (x,y,z)\in \R^3 : x^2+y^2 = z^2, \ 1\le z \le 2\}$.
$S = \{(x,y,z) \in \R^3 : x^2+y^2+z^2 = 4, (x-1)^2+ y^2 \le 1\}$.
$S$ is the boundary of the set of points in the first octant of $\R^3$ where $x+y+z\le 1$. In principle you can do all of this by elementry geometry, but using calculus might make things easier.
$S = \{(x,y,z) \in \R^3 : \sqrt{y^2+z^2} = 1-x^2\}$.
Derive a formula for the area of a surface $S$ of the form $S = \{(x,y,z)\in \R^3 : \sqrt{x^2+y^2} = f(z), a\le z\le b\}$.
Evaluate the following integrals. (Do enough examples to fix in your memory the definition of $\iint_S f\, dA$. Since it's similar to the definiton of area, not many examples should be needed.)
Find the centroid of the set $\{(x,y,z)\in \R^3 : x^2+y^2 + 4z^2 = 4, z\ge 0\}$. (The centroid of a set is defined in Section 4.4. When computing the centroid of a surface, you should replace $dV$ by $dA$.)
Compute $\iint_S xy \, dA$ when $S$ is the portion of the unit sphere in $\R^3$ that sits in the first octant (thatis, all coordinates are nonnegative.)
Compute the following integrals. (Again, do enough examples to make sure that you remember the definition of $\iint_S \bfF\cdot \bfn \,dA$.)
Let be a surface of the form $S = \{ (x,y,z) : (x,y)\in W, z= \phi(x,y)\}$, with the unit normal oriented upwards. where $W$ is a measurable subset of $\R^2$ and $\phi$ is $C^1$. Write down a general formula for $\iint_S \bfF\cdot \bfn \, dA$, where $\bfF(x,y,z) = (0,0,z)$.
For $S$ as in the previous question, write down a general formula for $\iint_S \bfF \cdot \bfn \, dA$, where $\bfF(x,y,z) = (0,0,1)$.
Let $S$ be the unit sphere in $\R^3$, with $\bfn$ oriented outwards, and compute $\iint_S \bfF\cdot \bfn\, dA$ where $\bfF = (x,y,z)$.
Let $S$ be the ellipse $(\frac x a)^2 + (\frac yb)^2 + (\frac zc)^2 = 1$, with $\bfn$ oriented inwards, and compute $\iint_S \bfF\cdot \bfn\, dA$ for $\bfF = (-a^2y,b^2x,z^2)$.
Let $S := \{(x,y,z)\in \R^3 : (x,y)\in [0,1]^2, z = xy \}$, with $\bfn$ oriented downwards, and compute $\iint_S \bfF\cdot \bfn\, dA$ for $\bfF = (zx, z^2, xy )$.
Prove \eqref{cvsurf}.
If you are interested, use \eqref{cvsurf} to prove Theorem 1.