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Section 6.6 Surface Integrals

Revised 2025-04-26, correcting the sentence below Equation (6.6.10).

References.

Topics.

Subsection 6.6.1 Parametric Surfaces

Just as not all plane curves can be conveniently described as the graph of a function \(y=f(x)\) and are instead best described parametrically as \(\vec{r}(t) = x(t) \veci + y(t) \vecj\text{,}\) so surfaces are often decribed parametrically, but now with two parameters:
\begin{equation} \vec{r}(u,v) = x(u,v) \veci + y(u,v) \vecj + z(u,v) \veck\tag{6.6.1} \end{equation}
for \((u,v)\) in some domain \(D\) in \(\reals^2\text{.}\)
The set of all points \((x,y,z)\) in \(\reals^3\) given by this form is the parametric surface \(S\text{,}\) with the parametric equations
\begin{equation} x = x(u,v), \; y=y(u,v), \; z = z(u,v)\tag{6.6.2} \end{equation}

Grid Curves: Curves on a Surface with Constant \(u\) or \(v\).

In visualizing and drawing surfaces it can be useful to consider
  1. lines with constant \(v\) value, \(v=v_0\text{,}\) which are the parameterized curves \(\vec{r}(u)=\vec{r}(u,v_0)\text{,}\) and
  2. lines with constant \(u\) value, \(u=u_0\text{:}\) \(\vec{r}(v)=\vec{r}(u_0,v)\text{.}\)
These are grid curves.

Surfaces of Revolution.

A familiar example is surfaces of revolution, given by rotating a curve around some axis. The surface given by rotating the curve \(y=f(x)\text{,}\) \(a \leq x \leq b\) about the \(x\)-axis can be described using the two parameters \(x\) and the angle \(\theta\) measuring rotation around that axis, starting at the \(y\) axis and going towards the \(z\)-axis:
\begin{equation*} x = x, y = f(x)\cos \theta, z = f(x)\sin \theta, \mbox{ domain } D = [a,b] \times [0, 2\pi]. \end{equation*}

Tangent Planes.

To find the tangent plane to \(S\) at the point \(P(x_0,y_0,z_0)\) given by parameter values \((u_0,v_0)\text{,}\) one can first find two tangent directions,so that their cross product is a normal to the surface and to the tangent plane.
Natural choices are the tangent lines given by using the linearizations of \(x(u,v)\text{,}\) \(y(u,v)\) and \(z(u,v)\)at this point and varying one parameter while holding the other constant. Note that these are tangent lines to the grid curves.
Varying \(u\) while fixing \(v=v_0\) gives linearizations
\begin{align*} x \amp= x(u_0,v_0) + \left[\frac{\partial x}{\partial u}(u_0,v_0)\right](u-u_0),\\ y \amp= y(u_0,v_0) + \left[\frac{\partial y}{\partial u}(u_0,v_0)\right](u-u_0),\\ z \amp= z(u_0,v_0) + \left[\frac{\partial z}{\partial u}(u_0,v_0)\right](u-u_0). \end{align*}
This gives the tangent vector in the \(u\) direction
\begin{equation*} \vec{r}_u = \frac{\partial x}{\partial u}(u_0,v_0) \veci + \frac{\partial y}{\partial u}(u_0,v_0) \vecj + \frac{\partial z}{\partial u}(u_0,v_0) \veck \end{equation*}
Similarly, varying \(v\) with fixed \(u=u_0\) gives tangent vector
\begin{equation*} \vec{r}_v = \frac{\partial x}{\partial v}(u_0,v_0) \veci + \frac{\partial y}{\partial v}(u_0,v_0) \vecj + \frac{\partial z}{\partial v}(u_0,v_0) \veck \end{equation*}
If the cross product \(\vec{r}_u(u_0,v_0) \times \vec{r}_v(u_0,v_0)\) is non-zero then it is a normal vector to \(S\) at point \(P\) and the tangent plane there is
\begin{equation*} (\vector{x,y,z} - \vector{x_0,y_0,z_0}) \cdot (\vec{r}_u(u_0,v_0) \times \vec{r}_v(u_0,v_0)) = \vec{0} \end{equation*}
If the cross product \(\vec{r}_u \times \vec{r}_v\) is non-zero for all \((u,v)\) in \(D\text{,}\) this is caled a regular parameterization and the surface \(S\) is called smooth: loosely, it has no corners or creases, and there is a well-defined tangent plane everywhere on the surface.
This is analagous to the condition \(\vecr'(t) \neq \vec{0}\) in Section 3.2 for a curve to be smooth and with well-defined tangent vector.

Subsection 6.6.2 Surface Area of a Parametric Surface

The area of the surface \(S\) can now be defined, and the intuitive approach is to look at the area of the infinitesimal part of the surface near a point with parameters \((u,v)\) given by varying \(u\) by an infinitesimal amount \(du\) and \(v\) by \(dv\text{.}\) This surface is an infinitesimal parallelogram with edges given by the above linearization as \(\vec{r}_u du\) and \(\vec{r}_v dv\text{,}\) and so of infinitesimal area \(dS\text{,}\)
\begin{equation} dS = \| \vec{r}_u du \times \vec{r}_v dv \| = \| \vec{r}_u \times \vec{r}_v \| du\ dv\tag{6.6.3} \end{equation}
Note that if the surface is a flat region in the \(x\)-\(y\) plane, so \(z(u,v)=0\text{,}\) then
\begin{equation*} \|\vec{r}_u \times \vec{r}_v\| = \left\| \left( \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u} \right) \veck \right\| = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \end{equation*}
and so this becomes the infinitesimal planar area \(dA\) of Equation (5.7.11) in Section 5.7.
Combining these infinitesimal areas by integration motivates the following definition.

Definition 6.6.1.

For a smooth parametric surface \(S\) given by
\begin{equation*} x = x(u,v), \, y = y(u,v), \, z = z(u,v) \end{equation*}
with \((u,v)\) in \(D\text{,}\) the surface area of \(S\) is
\begin{equation*} A(S) = \iint\limits_D dS = \iint\limits_D \| \vec{r}_u \times \vec{r}_v \| \ du\ dv = \iint\limits_D \| \vec{r}_u \times \vec{r}_v \| \ dA \end{equation*}
where \(dA\) refers to area in the \((u,v)\) plane.

Exercise 6.6.2.

Compute the surface area of a sphere, parameterizing it with spherical coordinates.
Alternatively, one can start from the beginning: approximate the areas of small parts of the surface, sum them, take a limit, and recognizing the limit as the above integral.

The Surface Area of the Graph of a Function.

In the case where the surface is the graph of a function \(z=g(x,y)\text{,}\) the parameters can be simply \(x\) and \(y\text{,}\) and straightforward calculations give
\begin{equation*} \| \vec{r}_x \times \vec{r}_y \| = \left\| \vector{-{\partial g}/{\partial x}, -{\partial g}/{\partial y}, 1} \right\| = \sqrt{1 + \left({\partial g}/{\partial x}\right)^2 + \left({\partial g}/{\partial y}\right)^2} \end{equation*}
so
\begin{equation} dS = \sqrt{1 + \left({\partial g}/{\partial x}\right)^2 + \left({\partial g}/{\partial y}\right)^2}\ dA\tag{6.6.4} \end{equation}
Thus the surface area is
\begin{equation} A(S) = \iint\limits_D dS = \iint\limits_D \sqrt{1 + \left({\partial g}/{\partial x}\right)^2 + \left({\partial g}/{\partial y}\right)^2}\ dx\ dy\tag{6.6.5} \end{equation}
(Compare this to the formula (1.2.4) for the arc length of the graph of a function in Section 1.2.)
Exercise 6.6.3.
Compute the surface area of a sphere again, this time by considering a hemisphere as the graph of a function.

Subsection 6.6.3 Surface Integral of a Scalar-Valued Function

The formula for surface area seen in the previous section was based on the formula
\begin{equation*} dS = \| \vec{r}_u \times \vec{r}_v \|\ dA \end{equation*}
for the area \(dS\) of an infinitesimal fragment of a parametric surface \(\vec{r}(u,v)\) in terms of the corresponding infinitesimal area \(dA = du\ dv\) in the \((u,v)\) plane of the parameters. To integrate a function over a surface, such as summing a density \(f=f(x,y,z)\) over the surface to compute a total mass, start with the idea that the mass on a small fragment of the surface is density times area:
\begin{equation*} f\ dS = f \| \vec{r}_u \times \vec{r}_v \|\ dA \end{equation*}
Thus the natural form for the integral of function \(f\) over surface \(S\) is
\begin{equation} \iint\limits_S f\ dS = \iint\limits_D f(\vec{r}(u,v)) \| \vec{r}_u \times \vec{r}_v \|\ dA\tag{6.6.6} \end{equation}
This can be taken as the definition of the integral over a surface, or one can define the integral in terms of limits of sums and derive this result, as already seen in various cases in these notes.
The formula for surface area in the previous section is just the case of integrating the constant function \(f=1\text{.}\)

Surface Integrals for Graphs of Functions.

When the surface is the graph of \(z=f(x,y)\) over domain \(D\text{,}\) using the form for the infinitesimal area \(dS\) in Equation (6.6.4) and \(f(x,y,z)=f(x,y,g(x,y))\) gives
\begin{equation*} \iint\limits_S f \, dS = \iint\limits_D f(x,y,g(x,y)) \sqrt{1 + ( {\partial g}/{\partial x} )^2 + ( {\partial g}/{\partial y} )^2}\ dA \end{equation*}

Subsection 6.6.4 Oriented Surfaces

To integrate the normal component of a vector field over a surface, we need something like the outward unit normal vector \(\N\) constructed for a positively oriented curve in Subsection 6.4.4.
There are two choices of unit normal at each point of a smooth surface, and we need to make a consistent choice over the whole surface which varies continuously, so that the normal can be considered as pointing "outward", or "upward" for example.

Definition 6.6.4.

A surface is oriented if there is a continuous function \(\N\) defined everywhere on the surface whose value at each point is a unit vector normal to the tangent plane at that point.
Note that for any oriented surface, there are two choices of orientation for the surface, since negating one suitable \(\N\) gives another. The two-way choice here is akin to the choice of positive (anti-clockwise) orientation for a curve.

Non-orientable Surfaces: The Möbius Strip etc..

Not all surfaces are oriented: the Möbius strip is a famous example of a surfaces where one can move from one side to the other by traveling along the surface, so starting on the side indicated by a continuous unit normal, as you move around without jumping sides, the normal should continue to point to your side, but you can move around returning to the same point but on the opposite side, so the normal at the same point must now point in the opposite direction.
This makes it impossible to choose between the two possible unit normals at each point in a way that is continuous over the whole surface.
Fortunately all the most familiar common types of surfaces are oriented: graphs of (differentiable) functions, level sets of (differentiable) functions, and ones with a single overall parameterization.

Oriented Surfaces: Globally Parameterized Surfaces.

Smooth surfaces that can be described by a single "global" parametrization \(\vec{r}(u,v)\) with \(\vec{r}_u \times \vec{r}_v \neq \vec{0}\) describing the whole surface are oriented. This is because that cross product is a continuous normal, giving unit normal
\begin{equation} \N = \frac{\vec{r}_u \times \vec{r}_v}{\| \vec{r}_u \times \vec{r}_v \|}\tag{6.6.7} \end{equation}
Note how the non-zero conditionfor smoothness is necessary and sufficient for this to make sense everywhere on the surface.
Exercise 6.6.5.
Compute this unit normal for the sphere or radius \(R\) parametrized with the spherical angular coordinates \(\phi\text{,}\) and \(\theta\text{,}\) \(\ds \vec{r}(\psi,\theta) = R \sin \phi \cos \theta \veci + R \sin \phi \sin \theta \vecj + R \cos \phi \veck\text{.}\)
What happens if we call this \(\vec{r}(\theta,\psi)\) instead?

Oriented Surfaces: Graphs of Functions.

The surface given by the graph of a function \(z=g(x,y)\) is globally parametrized by \(x\) and \(y\text{,}\) and so is oriented. Formula (6.6.7) applied to \(\vec{r}(x,y) = \vector{x,y,g(x,y)}\) gives
\begin{equation} \N = \frac{\vec{r}_x \times \vec{r}_y}{\|\vec{r}_x \times \vec{r}_y\|} = \frac{\vector{-{\partial g}/{\partial x}, -{\partial g}/{\partial y}, 1}} {\sqrt{1 + \left( {\partial g}/{\partial x} \right)^2 + \left( {\partial g}/{\partial y} \right)^2}}\tag{6.6.8} \end{equation}
Since the third component is always positive, this is the "upward" unit normal.
Exercise 6.6.6.
Compute this unit normal for the hemisphere given as the graph of \(z=\sqrt{R^2-x^2-y^2}\text{.}\)

Oriented Surfaces: Level Sets of Functions.

More generally surfaces defined as a level set by an equation \(f(x,y,z) = C\text{,}\) for \(C\) a constant, and with \(\del f \neq \vec{0}\) everywhere on the surface are oriented. For then \(\del f\) is a normal to the surface, and
\begin{equation} \N = \frac{\del f}{\| \del f \|}\tag{6.6.9} \end{equation}
is a suitable unit normal, pointing in the direction of increasing values of \(f\text{.}\)
It can be shown, using The Implicit Function Theorem in 3D, that such a surface is the union of a collection of parametric surfaces; indeed, as mentioned in that theorem, the parameters for each piece can be two of the cartesian coordinates: each piece can be the graph of a function \(z = g(x,y)\text{,}\) \(x = g(y,z)\) or \(y = g(x,z)\text{.}\)
Exercise 6.6.7.
Compute this unit normal for the sphere given as the level set \(x^2+y^2+z^2=R^2\text{.}\)

Choice of Orientation.

Note that there are two choices of unit normal vector function for any oriented surface, as we can always negate \(\N\text{.}\) Thus a particular choice of orientation has to be specified, such as outwards from or into a domain \(D\) when \(S\) is the boundary of \(D\text{.}\)
It will be then convenient to combine the infinitesimal area \(dS\) at a point on the surface with the chosen orientation \(\N\) of the surface there into the vector differential
\begin{equation} d\vec{S} = \N dS\tag{6.6.10} \end{equation}
For a parametrized surface \(\vec{r}(u,v)\) we then have \(dS = \| \vec{r}_u \times \vec{r}_v \| dA\) and choosing \(\ds \N = \frac{\vec{r}_u \times \vec{r}_v}{\|\vec{r}_u \times \vec{r}_v\|}\) as above,
\begin{equation} d\vec{S} = \N dS = (\vec{r}_u \times \vec{r}_v)\ dA\tag{6.6.11} \end{equation}
For the case of the graph of a function \(z = g(x,y)\text{,}\) combining Equation (6.6.8) for \(\N\) with Equation (6.6.4) for \(dS\) gives
\begin{equation} d\vec{S} = \N dS = \vector{-{\partial g}/{\partial x}, -{\partial g}/{\partial y},1}\ dA\tag{6.6.12} \end{equation}

Closed Surfaces and Their Outward Unit Normal.

A surface \(S\) that is the boundary of solid region \(E\) is called a closed surface, and then a natural choice of normal is outwards from the surface, which is called the positive orientation.
This is easily specified when the region is given by an inequality \(f(x,y,z) \leq k\) [or several such inequalities] because then the boundary is the surface is given by \(f(x,y,z) = k\) [or several such equations], and the outward normal is \(\N = \frac{\del f}{\|\del f\|}\) as seen above.
For example, the unit sphere centered at the origin considered as the boundary of the ball \(f(x,y,z)=x^2+y^2+z^2 \leq 1\) has outward unit normal \(\N=\vector{x,y,z}\text{.}\)
Exercise 6.6.8.
Evaluate \(d\vec{S}\) for the surface of the ball \(x^2+y^2+z^2 \leq R^2\text{.}\)

Subsection 6.6.5 Surface Integral of a Vector-Valued Function

The flux integral \(\oint_C \vec{F} \cdot \N\ ds\) around an oriented closed curve seen in the flux form of Green’s Theorem, Theorem 6.4.3, has a three dimensional counterpart for oriented surfaces:

Definition 6.6.9.

For \(\vec{F}\) a continuous vector field defined on an oriented surface \(S\) and \(\N\) a continuous unit normal vector for \(S\text{,}\) the surface integral of \(\vec{F}\) over \(S\) is
\begin{equation} \iint\limits_S \vec{F} \cdot \N \ dS, = \iint\limits_S \vec{F} \cdot d\vec{S}\text{.}\tag{6.6.13} \end{equation}
This is also called the flux of \(\vec{F}\) over \(S\).

The Integral of a Vector Field over a Parametrized Surface.

When surface \(S\) is parametrized as \(\vec{r}(u,v)\) for \((u,v)\) in domain \(D\text{,}\) we get a nice simplification:
\begin{equation*} d\vec{S} = \N \ dS = \frac{\vec{r}_u \times \vec{r}_v}{\| \vec{r}_u \times \vec{r}_v \|} \| \vec{r}_u \times \vec{r}_v \| \ dA = (\vec{r}_u \times \vec{r}_v) \ dA \end{equation*}
and so the flux is
\begin{equation} \iint\limits_S \vec{F} \cdot \N \ dS = \iint\limits_S \vec{F} \cdot d\vec{S} = \iint\limits_D \vec{F} \cdot (\vec{r}_u \times \vec{r}_v)\ dA\tag{6.6.14} \end{equation}
Avoiding the division and the square root in the formula for \(\| \vec{r}_u \times \vec{r}_v \|\) can help when it comes to evaluating the integral.
Remark 6.6.10.
Thus surface integral of vector functions are nicer than those of scalar functions in much the same way that path integrals of vector functions are nicer than those of scalar functions.
When the surface is the graph of a function \(z=g(x,y)\text{,}\) Equation (6.6.12) gives the form
\begin{align} \iint\limits_S \vec{F} \cdot d\vec{S} \amp= \iint\limits_D \vec{F} \cdot \vector{-{\partial g}/{\partial x}, -{\partial g}/{\partial y},1}\ dA\notag\\ \amp= \iint\limits_D -P\frac{\partial g}{\partial x} -Q\frac{\partial g}{\partial y} +R\ dA\tag{6.6.15} \end{align}

Study Guide.

Study Section 6.6 of OSC3
 9 
openstax.org/books/calculus-volume-3/pages/6-6-surface-integrals
; in particular
  • All the Definitions.
  • Examples 58–64 and 66–71, and the Checkpoints following each.
  • The T/F Exercises 269–271 and one or several exercises from each of the following ranges: 273–278, 281–283, and 284–286.