Surface Integrals

We have seen that a line integral is an integral over a path in a plane or in space. However, if we wish to integrate over a surface (a two-dimensional object) rather than a path (a one-dimensional object) in space, then we need a new kind of integral that can handle integration over objects in higher dimensions. We can extend the concept of a line integral to a surface integral to allow us to perform this integration.

Surface integrals are important for the same reasons that line integrals are important. They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. In particular, surface integrals allow us to generalize Green’s theorem to higher dimensions, and they appear in some important theorems we discuss in later sections.

Parametric Surfaces

A surface integral is similar to a line integral, except the integration is done over a surface rather than a path. In this sense, surface integrals expand on our study of line integrals. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field.

However, before we can integrate over a surface, we need to consider the surface itself. Recall that to calculate a scalar or vector line integral over curve C, we first need to parameterize C. In a similar way, to calculate a surface integral over surface S, we need to parameterize S. That is, we need a working concept of a parameterized surface (or a parametric surface), in the same way that we already have a concept of a parameterized curve.

Notice that this parameterization involves two parameters, u and v, because a surface is two-dimensional, and therefore two variables are needed to trace out the surface. The parameters u and v vary over a region called the parameter domain, or parameter space—the set of points in the uv-plane that can be substituted into r. Each choice of u and v in the parameter domain gives a point on the surface, just as each choice of a parameter t gives a point on a parameterized curve. The entire surface is created by making all possible choices of u and v over the parameter domain.

It follows from [link] that we can parameterize all cylinders of the form

x^{2} + y^{2} = R^{2} .

We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface.

We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions. To parameterize a sphere, it is easiest to use spherical coordinates. The sphere of radius

ρ

Finally, to parameterize the graph of a two-variable function, we first let

z = f (x, y)

be a function of two variables. The simplest parameterization of the graph of

f

can be parameterized by

r (x, z) = 〈 x, \frac{x^{2} + z^{2}}{2}, z 〉, 0 \leq x < \infty, 0 \leq z < \infty .

Notice that we do not need to vary over the entire domain of y because x and z are squared.

Let’s now generalize the notions of smoothness and regularity to a parametric surface. Recall that curve parameterization

r (t), a \leq t \leq b

For a curve, this condition ensures that the image of r really is a curve, and not just a point. For example, consider curve parameterization

r (t) = 〈 1, 2 〉, 0 \leq t \leq 5 .

The fact that the derivative is zero indicates we are not actually looking at a curve.

Analogously, we would like a notion of regularity for surfaces so that a surface parameterization really does trace out a surface. To motivate the definition of regularity of a surface parameterization, consider parameterization

Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line ([link]). How could we avoid parameterizations such as this? Parameterizations that do not give an actual surface? Notice that

r_{u} = 〈 0, 0, 0 〉

and the corresponding cross product is zero. The analog of the condition

r ′ (t) = 0

If parameterization r is regular, then the image of r is a two-dimensional object, as a surface should be. Throughout this chapter, parameterizations

r (u, v) = 〈 x (u, v), y (u, v), z (u, v) 〉

Informally, a curve parameterization is smooth if the resulting curve has no sharp corners. The definition of a smooth surface parameterization is similar. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners.

A surface may also be piecewise smooth if it has smooth faces but also has locations where the directional derivatives do not exist.

Surface Area of a Parametric Surface

Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. The second step is to define the surface area of a parametric surface. The notation needed to develop this definition is used throughout the rest of this chapter.

Let S be a surface with parameterization

r (u, v) = 〈 x (u, v), y (u, v), z (u, v) 〉

over some parameter domain D. We assume here and throughout that the surface parameterization

r (u, v) = 〈 x (u, v), y (u, v), z (u, v) 〉

is continuously differentiable—meaning, each component function has continuous partial derivatives. Assume for the sake of simplicity that D is a rectangle (although the following material can be extended to handle nonrectangular parameter domains). Divide rectangle D into subrectangles

D_{i j}

Suppose that i ranges from 1 to m and j ranges from 1 to n so that D is subdivided into mn rectangles. This division of D into subrectangles gives a corresponding division of surface S into pieces

S_{i j} .

Note that we can form a grid with lines that are parallel to the u-axis and the v-axis in the uv-plane. These grid lines correspond to a set of grid curves on surface S that is parameterized by

r (u, v) .

is located at the corner of two grid curves, as in [link]. If we think of r as a mapping from the uv-plane to

ℝ^{3},

the grid curves are the image of the grid lines under r. To be precise, consider the grid lines that go through point

(u_{i}, v_{j}) .

In the first grid line, the horizontal component is held constant, yielding a vertical line through

(u_{i}, v_{j}) .

In the second grid line, the vertical component is held constant, yielding a horizontal line through

(u_{i}, v_{j}) .

Now consider the vectors that are tangent to these grid curves. For grid curve

r (u_{i}, v),

is in the tangent plane. Since the original rectangle in the uv-plane corresponding to

S_{i j}

to match the scale of the original division of rectangles in the parameter domain. Therefore, the area of the parallelogram used to approximate the area of

S_{i j}

and the previous approximation leads to the following definition of surface area of a parametric surface ([link]).

Calculating Surface Area

Calculate the lateral surface area (the area of the “side,” not including the base) of the right circular cone with height h and radius r.

Before calculating the surface area of this cone using [link], we need a parameterization. We assume this cone is in $ℝ^{3}$

with its vertex at the origin ([link]). To obtain a parameterization, let $α$

be the angle that is swept out by starting at the positive z-axis and ending at the cone, and let $k = tan α .$

For a height value v with $0 \leq v \leq h,$

the radius of the circle formed by intersecting the cone with plane $z = v$

is $k v .$

Therefore, a parameterization of this cone is

s (u, v) = 〈 k v cos u, k v sin u, v 〉, 0 \leq u < 2 π, 0 \leq v \leq h .

The idea behind this parameterization is that for a fixed v value, the circle swept out by letting u vary is the circle at height v and radius kv. As v increases, the parameterization sweeps out a “stack” of circles, resulting in the desired cone.

A right circular cone in three dimensions, opening upwards on the z axis. It has radius r = kh and height h with the given parameterization. Alpha is the angle that is swept out by starting at the positive z-axis and ending at the cone. It is noted that k is equal to the tangent of alpha.

With a parameterization in hand, we can calculate the surface area of the cone using [link]. The tangent vectors are $t_{u} = 〈 − k v sin u, k v cos u, 0 〉$

and $t_{v} = 〈 k cos u, k sin u, 1 〉 .$

Therefore,

\begin{matrix} t_{u} \times t_{v} & = \| \begin{matrix} i & j & k \\ − k v sin u & k v cos u & 0 \\ k cos u & k sin u & 1 \end{matrix} \| \\ = 〈 k v cos u, k v sin u, − k^{2} v {sin}^{2} u - k^{2} v {cos}^{2} u 〉 \\ = 〈 k v cos u, k v sin u, − k^{2} v 〉 . \end{matrix}

The magnitude of this vector is

\begin{matrix} ‖ 〈 k v cos u, k v sin u, − k^{2} v 〉 ‖ & = \sqrt{k^{2} v^{2} {cos}^{2} u + k^{2} v^{2} {sin}^{2} u + k^{4} v^{2}} \\ = \sqrt{k^{2} v^{2} + k^{4} v^{2}} \\ = k v \sqrt{1 + k^{2}} . \end{matrix}

By [link], the surface area of the cone is

\begin{matrix} \iint_{D} ‖ t_{u} \times t_{v} ‖ d A & = \int_{0}^{h} \int_{0}^{2 π} k v \sqrt{1 + k^{2}} d u d v \\ = 2 π k \sqrt{1 + k^{2}} \int_{0}^{h} v d v \\ = 2 π k \sqrt{1 + k^{2}} {[\frac{v^{2}}{2}]}_{0}^{h} \\ = π k h^{2} \sqrt{1 + k^{2}} . \end{matrix}

Since $k = tan α = r / h,$

\begin{matrix} π k h^{2} \sqrt{1 + k^{2}} & = π \frac{r}{h} h^{2} \sqrt{1 + \frac{r^{2}}{h^{2}}} \\ = π r h \sqrt{1 + \frac{r^{2}}{h^{2}}} \\ = π r \sqrt{h^{2} + h^{2} (\frac{r^{2}}{h^{2}})} \\ = π r \sqrt{h^{2} + r^{2}} . \end{matrix}

Therefore, the lateral surface area of the cone is $π r \sqrt{h^{2} + r^{2}} .$

Analysis

The surface area of a right circular cone with radius r and height h is usually given as $π r^{2} + π r \sqrt{h^{2} + r^{2}} .$

The reason for this is that the circular base is included as part of the cone, and therefore the area of the base $π r^{2}$

is added to the lateral surface area $π r \sqrt{h^{2} + r^{2}}$

that we found.

In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. Therefore, we can calculate the surface area of a surface of revolution by using the same techniques. Let

y = f (x) \geq 0

be the angle of rotation. Then, S can be parameterized with parameters x and

θ

Surface Integral of a Scalar-Valued Function

Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. First, let’s look at the surface integral of a scalar-valued function. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). Therefore, the definition of a surface integral follows the definition of a line integral quite closely. For scalar line integrals, we chopped the domain curve into tiny pieces, chose a point in each piece, computed the function at that point, and took a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain region (no longer a curve) into tiny pieces and proceed in the same fashion.

Let S be a piecewise smooth surface with parameterization

r (u, v) = 〈 x (u, v), y (u, v), z (u, v) 〉

be a function with a domain that contains S. For now, assume the parameter domain D is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation more manageable). Divide rectangle D into subrectangles

D_{i j}

Suppose that i ranges from 1 to m and j ranges from 1 to n so that D is subdivided into mn rectangles. This division of D into subrectangles gives a corresponding division of S into pieces

S_{i j} .

To define a surface integral of a scalar-valued function, we let the areas of the pieces of S shrink to zero by taking a limit.

Again, notice the similarities between this definition and the definition of a scalar line integral. In the definition of a line integral we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. In the definition of a surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. Thus, a surface integral is similar to a line integral but in one higher dimension.

The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero.

Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To develop a method that makes surface integrals easier to compute, we approximate surface areas

Δ S_{i j}

with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors

t_{u}

This approximation becomes arbitrarily close to

lim_{m, n \to \infty} \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (P_{i j}) Δ S_{i j}

by letting m and n go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:

[link] allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to [link] for line integrals:

Calculating the Surface Integral of a Piece of a Sphere

Calculate surface integral $\iint_{S} f (x, y, z) d S,$

where $f (x, y, z) = z^{2}$

and S is the surface that consists of the piece of sphere $x^{2} + y^{2} + z^{2} = 4$

that lies on or above plane $z = 1$

and the disk that is enclosed by intersection plane $z = 1$

and the given sphere ([link]).

A diagram in three dimensions of the upper half of a sphere. The center is at the origin, and the radius is 2. The top part above the plane z=1 is cut off and shaded; the rest is simply an outline of the hemisphere. The top section has center at (0,0,1) and radius of radical three.

Notice that S is not smooth but is piecewise smooth; S can be written as the union of its base $S_{1}$

and its spherical top $S_{2},$

and both $S_{1}$

and $S_{2}$

are smooth. Therefore, to calculate $\iint_{S} z^{2} d S,$

we write this integral as $\iint_{S_{1}} z^{2} d S + \iint_{S_{2}} z^{2} d S$

and we calculate integrals $\iint_{S_{1}} z^{2} d S$

and $\iint_{S_{2}} z^{2} d S .$

First, we calculate $\iint_{S_{1}} z^{2} d S .$

To calculate this integral we need a parameterization of $S_{1} .$

This surface is a disk in plane $z = 1$

centered at $(0, 0, 1) .$

To parameterize this disk, we need to know its radius. Since the disk is formed where plane $z = 1$

intersects sphere $x^{2} + y^{2} + z^{2} = 4,$

we can substitute $z = 1$

into equation $x^{2} + y^{2} + z^{2} = 4 :$

x^{2} + y^{2} + 1 = 4 \Rightarrow x^{2} + y^{2} = 3 .

Therefore, the radius of the disk is $\sqrt{3}$

and a parameterization of $S_{1}$

is $r (u, v) = 〈 u cos v, u sin v, 1 〉, 0 \leq u \leq \sqrt{3}, 0 \leq v \leq 2 π .$

The tangent vectors are $t_{u} = 〈 cos v, sin v, 0 〉$

and $t_{v} = 〈 − u sin v, u co s v, 0 〉,$

and thus

t_{u} \times t_{v} = \| \begin{matrix} i & j & k \\ cos v & sin v & 0 \\ − u sin v & u cos v & 0 \end{matrix} \| = 〈 0, 0, u {cos}^{2} v + u {sin}^{2} v 〉 = 〈 0, 0, u 〉 .

The magnitude of this vector is u. Therefore,

\begin{matrix} \iint_{S_{1}} z^{2} d S & = \int_{0}^{\sqrt{3}} \int_{0}^{2 π} f (r (u, v)) ‖ t_{u} \times t_{v} ‖ d v d u \\ = \int_{0}^{\sqrt{3}} \int_{0}^{2 π} u d v d u \\ = 2 π \int_{0}^{\sqrt{3}} u d u \\ = 2 π \sqrt{3} . \end{matrix}

Now we calculate $\iint_{S_{2}} d S .$

To calculate this integral, we need a parameterization of $S_{2} .$

The parameterization of full sphere $x^{2} + y^{2} + z^{2} = 4$

r (ϕ, θ) = 〈 2 cos θ sin ϕ, 2 sin θ sin ϕ, 2 cos ϕ 〉, 0 \leq θ \leq 2 π, 0 \leq ϕ \leq π .

Since we are only taking the piece of the sphere on or above plane $z = 1,$

we have to restrict the domain of $ϕ .$

To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in [link] (the $\sqrt{3}$

comes from the fact that the base of S is a disk with radius $\sqrt{3}) .$

Therefore, the tangent of $ϕ$

is $\sqrt{3},$

which implies that $ϕ$

is $π / 6 .$

We now have a parameterization of $S_{2} :$

r (ϕ, θ) = 〈 2 cos θ sin ϕ, 2 sin θ sin ϕ, 2 cos ϕ 〉, 0 \leq θ \leq 2 π, 0 \leq ϕ \leq π / 3 .

The tangent vectors are

t_{ϕ} = 〈 2 cos θ cos ϕ, 2 sin θ cos ϕ, −2 sin ϕ 〉 and t_{θ} = 〈 −2 sin θ sin ϕ, u cos θ sin ϕ, 0 〉,

and thus

\begin{matrix} t_{ϕ} \times t_{θ} & = \| \begin{matrix} i & j & k \\ 2 cos θ cos ϕ & 2 sin θ cos ϕ & −2 sin ϕ \\ −2 sin θ sin ϕ & 2 cos θ sin ϕ & 0 \end{matrix} \| \\ = 〈 4 cos θ {sin}^{2} ϕ, 4 sin θ {sin}^{2} ϕ, 4 {cos}^{2} θ cos ϕ sin ϕ + 4 {sin}^{2} θ cos ϕ sin ϕ 〉 \\ = 〈 4 cos θ {sin}^{2} ϕ, 4 sin θ {sin}^{2} ϕ, 4 cos ϕ sin ϕ 〉 . \end{matrix}

The magnitude of this vector is

\begin{matrix} ‖ t_{ϕ} \times t_{θ} ‖ & = \sqrt{16 {cos}^{2} θ {sin}^{4} ϕ + 16 {sin}^{2} θ {sin}^{4} ϕ + 16 {cos}^{2} ϕ {sin}^{2} ϕ} \\ = 4 \sqrt{{sin}^{4} ϕ + {cos}^{2} ϕ {sin}^{2} ϕ} . \end{matrix}

Therefore,

\begin{matrix} \iint_{S_{2}} z d S & = \int_{0}^{π / 6} \int_{0}^{2 π} f (r (ϕ, θ)) ‖ t_{ϕ} \times t_{θ} ‖ d θ d ϕ \\ = \int_{0}^{π / 6} \int_{0}^{2 π} 16 {cos}^{2} ϕ \sqrt{{sin}^{4} ϕ + {cos}^{2} ϕ {sin}^{2} ϕ} d θ d ϕ \\ = 32 π \int_{0}^{π / 6} {cos}^{2} ϕ \sqrt{{sin}^{4} ϕ + {cos}^{2} ϕ {sin}^{2} ϕ} d ϕ \\ = 32 π \int_{0}^{π / 6} {cos}^{2} ϕ sin ϕ \sqrt{{sin}^{2} ϕ + {cos}^{2} ϕ} d ϕ \\ = 32 π \int_{0}^{π / 6} {cos}^{2} ϕ sin ϕ d ϕ \\ = 32 π {[- \frac{{cos}^{3} ϕ}{3}]}_{0}^{π / 6} = 32 π [\frac{1}{3} - \frac{\sqrt{3}}{8}] = \frac{32 π}{3} - 4 \sqrt{3} . \end{matrix}

Since $\iint_{S} z^{2} d S = \iint_{S_{1}} z^{2} d S + \iint_{S_{2}} z^{2} d S,$

we have $\iint_{S} z^{2} d S = (2 π - 4) \sqrt{3} + \frac{32 π}{3} .$

Analysis

In this example we broke a surface integral over a piecewise surface into the addition of surface integrals over smooth subsurfaces. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces.

Scalar surface integrals have several real-world applications. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. If a thin sheet of metal has the shape of surface S and the density of the sheet at point

(x, y, z)

Orientation of a Surface

Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. The same was true for scalar surface integrals: we did not need to worry about an “orientation” of the surface of integration.

On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. That is, we needed the notion of an oriented curve to define a vector line integral without ambiguity. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. An oriented surface is given an “upward” or “downward” orientation or, in the case of surfaces such as a sphere or cylinder, an “outward” or “inward” orientation.

on S so that N varies continuously over S, then S is “orientable.” Such a choice of unit normal vector at each point gives the orientation of a surface S. If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the “negative” side and the side of the surface at which the water flows away is the “positive” side. Informally, a choice of orientation gives S an “outer” side and an “inner” side (or an “upward” side and a “downward” side), just as a choice of orientation of a curve gives the curve “forward” and “backward” directions.

Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. This is called the positive orientation of the closed surface ([link]). We also could choose the inward normal vector at each point to give an “inward” orientation, which is the negative orientation of the surface.

is also orientable. If we choose the unit normal vector that points “above” the surface at each point, then the unit normal vectors vary continuously over the surface. We could also choose the unit normal vector that points “below” the surface at each point. To get such an orientation, we parameterize the graph of

f

(which is normal to the surface at any point on the surface) is

〈 − f_{x}, − f_{y}, 1 〉 .

Since the z component of this vector is one, the corresponding unit normal vector points “upward,” and the upward side of the surface is chosen to be the “positive” side.

and is therefore normal to S at that point. Therefore, the choice of unit normal vector

Since every curve has a “forward” and “backward” direction (or, in the case of a closed curve, a clockwise and counterclockwise direction), it is possible to give an orientation to any curve. Hence, it is possible to think of every curve as an oriented curve. This is not the case with surfaces, however. Some surfaces cannot be oriented; such surfaces are called nonorientable. Essentially, a surface can be oriented if the surface has an “inner” side and an “outer” side, or an “upward” side and a “downward” side. Some surfaces are twisted in such a fashion that there is no well-defined notion of an “inner” or “outer” side.

The classic example of a nonorientable surface is the Möbius strip. To create a Möbius strip, take a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together ([link]). Because of the half-twist in the strip, the surface has no “outer” side or “inner” side. If you imagine placing a normal vector at a point on the strip and having the vector travel all the way around the band, then (because of the half-twist) the vector points in the opposite direction when it gets back to its original position. Therefore, the strip really only has one side.

Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve.

Surface Integral of a Vector Field

With the idea of orientable surfaces in place, we are now ready to define a surface integral of a vector field. The definition is analogous to the definition of the flux of a vector field along a plane curve. Recall that if F is a two-dimensional vector field and C is a plane curve, then the definition of the flux of F along C involved chopping C into small pieces, choosing a point inside each piece, and calculating

F \cdot N

at the point (where N is the unit normal vector at the point). The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface S into small pieces, choose a point in the small (two-dimensional) piece, and calculate

F \cdot N

To place this definition in a real-world setting, let S be an oriented surface with unit normal vector N. Let v be a velocity field of a fluid flowing through S, and suppose the fluid has density

ρ (x, y, z) .

Imagine the fluid flows through S, but S is completely permeable so that it does not impede the fluid flow ([link]). The mass flux of the fluid is the rate of mass flow per unit area. The mass flux is measured in mass per unit time per unit area. How could we calculate the mass flux of the fluid across S?

is small enough, then it can be approximated by a tangent plane at some point P in

S_{i j} .

because the normal vector to a plane does not change as we move across the plane. The component of the vector

ρ v

and v are all evaluated at P ([link]). This is analogous to the flux of two-dimensional vector field F across plane curve C, in which we approximated flux across a small piece of C with the expression

(F \cdot N) Δ s .

To approximate the mass flux across S, form the sum

\sum_{i = 1}^{m} \sum_{j = 1}^{n} (ρ v \cdot N) Δ S_{i j} .

get smaller, the sum

\sum_{i = 1}^{m} \sum_{j = 1}^{n} (ρ v \cdot N) Δ S_{i j}

Notice the parallel between this definition and the definition of vector line integral

\int_{C} F \cdot N d s .

A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (a two-dimensional object) rather than a curve (a one-dimensional object). Integral

\iint_{S} F \cdot N d S

is the flux of F across curve C. A surface integral over a vector field is also called a flux integral.

be a parameterization of S with parameter domain D. Then, the unit normal vector is given by

N = \frac{t_{u} \times t_{v}}{‖ t_{u} \times t_{v} ‖}

Therefore, to compute a surface integral over a vector field we can use the equation

In [link], we computed the mass flux, which is the rate of mass flow per unit area. If we want to find the flow rate (measured in volume per time) instead, we can use flux integral

\int \int_{S} v • N d S,

which leaves out the density. Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization of the flow rate of a fluid with velocity v across a surface S:

To compute the flow rate of the fluid in [link], we simply remove the density constant, which gives a flow rate of

90 π m^{3} / sec .

Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.

In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point

(x, y, z)

Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the heat flow is defined as vector field

F = − k \nabla T,

where the constant k is the thermal conductivity of the substance from which the object is made (this constant is determined experimentally). The rate of heat flow across surface S in the object is given by the flux integral

Calculating Heat Flow

A cast-iron solid cylinder is given by inequalities $x^{2} + y^{2} \leq 1,$

1 \leq z \leq 4 .

The temperature at point $(x, y, z)$

in a region containing the cylinder is $T (x, y, z) = (x^{2} + y^{2}) z .$

Given that the thermal conductivity of cast iron is 55, find the heat flow across the boundary of the solid if this boundary is oriented outward.

Let S denote the boundary of the object. To find the heat flow, we need to calculate flux integral $\iint_{S} − k \nabla T \cdot d S .$

Notice that S is not a smooth surface but is piecewise smooth, since S is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of S. Before calculating any integrals, note that the gradient of the temperature is $\nabla T = 〈 2 x z, 2 y z, x^{2} + y^{2} 〉 .$

First we consider the circular bottom of the object, which we denote $S_{1} .$

We can see that $S_{1}$

is a circle of radius 1 centered at point $(0, 0, 1),$

sitting in plane $z = 1 .$

This surface has parameterization $r (u, v) = 〈 v cos u, v sin u, 1 〉, 0 \leq u < 2 π, 0 \leq v \leq 1 .$

Therefore,

t_{u} = 〈 − v sin u, v cos u, 0 〉 and t_{v} = 〈 cos u, v sin u, 0 〉,

and

t_{u} \times t_{v} = 〈 0, 0, − v {sin}^{2} u - v {cos}^{2} u 〉 = 〈 0, 0, − v 〉 .

Since the surface is oriented outward and $S_{1}$

is the bottom of the object, it makes sense that this vector points downward. By [link], the heat flow across $S_{1}$

\begin{matrix} \iint_{S_{1}} − k \nabla T \cdot d S & = −55 \int_{0}^{2 π} \int_{0}^{1} \nabla T (u, v) \cdot (t_{u} \times t_{v}) d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} 〈 2 v cos u, 2 v sin u, v^{2} {cos}^{2} u + v^{2} {sin}^{2} u 〉 \cdot 〈 0, 0, − v 〉 d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} 〈 2 v cos u, 2 v sin u, v^{2} 〉 \cdot 〈 0, 0, − v 〉 d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} − v^{3} d v d u = −55 \int_{0}^{2 π} - \frac{1}{4} d u = \frac{55 π}{2} . \end{matrix}

Now let’s consider the circular top of the object, which we denote $S_{2} .$

We see that $S_{2}$

is a circle of radius 1 centered at point $(0, 0, 4),$

sitting in plane $z = 4 .$

This surface has parameterization $r (u, v) = 〈 v cos u, v sin u, 4 〉, 0 \leq u < 2 π, 0 \leq v \leq 1 .$

Therefore,

t_{u} = 〈 − v sin u, v cos u, 0 〉 and t_{v} = 〈 cos u, v sin u, 0 〉,

and

t_{u} \times t_{v} = 〈 0, 0, − v {sin}^{2} u - v {cos}^{2} u 〉 = 〈 0, 0, − v 〉 .

Since the surface is oriented outward and $S_{1}$

is the top of the object, we instead take vector $t_{v} \times t_{u} = 〈 0, 0, v 〉 .$

By [link], the heat flow across $S_{1}$

\begin{matrix} \int \int_{S_{2}} − k \nabla T • d S & = −55 \int_{0}^{2 π} \int_{0}^{1} \nabla T (u, v) • (t_{v} \times t_{u}) d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} 〈 8 v cos u, 8 v sin u, v^{2} {cos}^{2} u + v^{2} {sin}^{2} u 〉 • 〈 0, 0, v 〉 d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} 〈 8 v cos u, 8 v sin u, v^{2} 〉 • 〈 0, 0, v 〉 d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} v^{3} d v d u = - \frac{55 π}{2} . \end{matrix}

Last, let’s consider the cylindrical side of the object. This surface has parameterization $r (u, v) = 〈 cos u, sin u, v 〉, 0 \leq u < 2 π, 1 \leq v \leq 4 .$

By [link], we know that $t_{u} \times t_{v} = 〈 cos u, sin u, 0 〉 .$

By [link],

\begin{matrix} \iint_{S_{3}} − k \nabla T • d S & = −55 \int_{0}^{2 π} \int_{1}^{4} \nabla T (u, v) • (t_{v} \times t_{u}) d v d u \\ = −55 \int_{0}^{2 π} \int_{1}^{4} 〈 2 v cos u, 2 v sin u, {cos}^{2} u + {sin}^{2} u 〉 • 〈 cos u, sin u, 0 〉 d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} 〈 2 v cos u, 2 v sin u, 1 〉 • 〈 cos u, sin u, 0 〉 d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} (2 v {cos}^{2} u + 2 v {sin}^{2} u) d v d u \\ = −55 \int_{0}^{2 π} \int_{0}^{1} 2 v d v d u = −55 \int_{0}^{2 π} d u = −110 π . \end{matrix}

Therefore, the rate of heat flow across S is $\frac{55 π}{2} - \frac{55 π}{2} - 110 π = −110 π .$

Key Concepts

Key Equations

For the following exercises, determine whether the statements are true or false.

If surface S is given by ${(x, y, z) : 0 \leq x \leq 1, 0 \leq y \leq 1, z = 10},$

then $\iint_{S} f (x, y, z) d S = \int_{0}^{1} \int_{0}^{1} f (x, y, 10) d x d y .$

True

If surface S is given by ${(x, y, z) : 0 \leq x \leq 1, 0 \leq y \leq 1, z = x},$

then $\iint_{S} f (x, y, z) d S = \int_{0}^{1} \int_{0}^{1} f (x, y, x) d x d y .$

Surface $r = 〈 v cos u, v sin u, v^{2} 〉, for 0 \leq u \leq π, 0 \leq v \leq 2,$

is the same as surface $r = 〈 \sqrt{v} cos 2 u, \sqrt{v} sin 2 u, v 〉,$

for $0 \leq u \leq \frac{π}{2}, 0 \leq v \leq 4 .$

True

Given the standard parameterization of a sphere, normal vectors $t_{u}^{} \times t_{v}$

are outward normal vectors.

For the following exercises, find parametric descriptions for the following surfaces.

Plane $3 x - 2 y + z = 2$

r (u, v) = 〈 u, v, 2 - 3 u + 2 v 〉

for $− \infty \leq u < \infty$

and $− \infty \leq v < \infty .$

Paraboloid $z = x^{2} + y^{2},$

for $0 \leq z \leq 9 .$

Plane $2 x - 4 y + 3 z = 16$

r (u, v) = 〈 u, v, \frac{1}{3} (16 - 2 u + 4 v) 〉

for $\| u \| < \infty$

and $\| v \| < \infty .$

The frustum of cone $z^{2} = x^{2} + y^{2}, for 2 \leq z \leq 8$

The portion of cylinder $x^{2} + y^{2} = 9$

in the first octant, for $0 \leq z \leq 3$

A diagram in three dimensions of a section of a cylinder with radius 3. The center of its circular top is (0,0,3). The section exists for x, y, and z between 0 and 3.

r (u, v) = 〈 3 cos u, 3 sin u, v 〉

for $0 \leq u \leq \frac{π}{2}, 0 \leq v \leq 3$

A cone with base radius r and height h, where r and h are positive constants

For the following exercises, use a computer algebra system to approximate the area of the following surfaces using a parametric description of the surface.

[T] Half cylinder ${(r, θ, z) : r = 4, 0 \leq θ \leq π, 0 \leq z \leq 7}$

A = 87.9646

[T] Plane $z = 10 - x - y$

above square $\| x \| \leq 2, \| y \| \leq 2$

For the following exercises, let S be the hemisphere $x^{2} + y^{2} + z^{2} = 4,$

with $z \geq 0,$

and evaluate each surface integral, in the counterclockwise direction.

\iint_{S} z d S

\iint_{S} z d S = 8 π

\iint_{S} (x - 2 y) d S

\iint_{S} (x^{2} + y^{2}) z d S

\iint_{S} (x^{2} + y^{2}) z d S = 16 π

For the following exercises, evaluate $\int \int_{S} F \cdot N d s$

for vector field F, where N is an outward normal vector to surface S.

F (x, y, z) = x i + 2 y j - 3 z k,

and S is that part of plane $15 x - 12 y + 3 z = 6$

that lies above unit square $0 \leq x \leq 1, 0 \leq y \leq 1 .$

F (x, y, z) = x i + y j,

and S is hemisphere $z = \sqrt{1 - x^{2} - y^{2}} .$

\iint_{S} F \cdot N d S = \frac{4 π}{3}

F (x, y, z) = x^{2} i + y^{2} j + z^{2} k,

and S is the portion of plane $z = y + 1$

that lies inside cylinder $x^{2} + y^{2} = 1 .$

A cylinder and an intersecting plane shown in three-dimensions. S is the portion of the plane z = y + 1 inside the cylinder x^2 + y ^2 = 1.

For the following exercises, approximate the mass of the homogeneous lamina that has the shape of given surface S. Round to four decimal places.

[T] S is surface $z = 4 - x - 2 y, with z \geq 0, x \geq 0, y \geq 0; ξ = x .$

m \approx 13.0639

[T] S is surface $z = x^{2} + y^{2}, with z \leq 1; ξ = z .$

[T] S is surface $x^{2} + y^{2} + x^{2} = 5, with z \geq 1; ξ = θ^{2} .$

m \approx 228.5313

Evaluate $\iint_{S} (y^{2} z i + y^{3} j + x z k) \cdot d S,$

where S is the surface of cube $−1 \leq x \leq 1, −1 \leq y \leq 1, and 0 \leq z \leq 2 .$

in a counterclockwise direction.

Evaluate surface integral $\iint_{S} g d S,$

where $g (x, y, z) = x z + 2 x^{2} - 3 x y$

and S is the portion of plane $2 x - 3 y + z = 6$

that lies over unit square R: $0 \leq x \leq 1, 0 \leq y \leq 1 .$

\iint_{S} g d S = 3 \sqrt{4}

Evaluate $\iint_{S} (x + y + z) d S,$

where S is the surface defined parametrically by $R (u, v) = (2 u + v) i + (u - 2 v) j + (u + 3 v) k$

for $0 \leq u \leq 1, and 0 \leq v \leq 2 .$

A three-dimensional diagram of the given surface, which appears to be a steeply sloped plane stretching through the (x,y) plane.

[T] Evaluate $\iint_{S} (x - y^{2} + z) d S,$

where S is the surface defined by $R (u, v) = u^{2} i + v j + u k, 0 \leq u \leq 1, 0 \leq v \leq 1 .$

A three-dimensional diagram of the given surface, which appears to be a curve with edges parallel to the y-axis. It increases in x components and decreases in z components the further it is from the y axis.

\iint_{S} (x^{2} + y - z) d S \approx 0.9617

[T] Evaluate where S is the surface defined by $R (u, v) = u i - u^{2} j + v k, 0 \leq u \leq 2, 0 \leq v \leq 1 .$

for $0 \leq u \leq 1, 0 \leq v \leq 2 .$

Evaluate $\iint_{S} (x^{2} + y^{2}) d S,$

where S is the surface bounded above hemisphere $z = \sqrt{1 - x^{2} - y^{2}},$

and below by plane $z = 0 .$

\iint_{S} (x^{2} + y^{2}) d S = \frac{4 π}{3}

Evaluate $\iint_{S} (x^{2} + y^{2} + z^{2}) d S,$

where S is the portion of plane $z = x + 1$

that lies inside cylinder $x^{2} + y^{2} = 1 .$

[T] Evaluate $\iint_{S} x^{2} z d S,$

where S is the portion of cone $z^{2} = x^{2} + y^{2}$

that lies between planes $z = 1$

and $z = 4 .$

A diagram of the given upward opening cone in three dimensions. The cone is cut by planes z=1 and z=4.

\iint_{S} x^{2} z d S = \frac{1023 \sqrt{2 π}}{5}

[T] Evaluate $\iint_{S} (x z / y) d S,$

where S is the portion of cylinder $x = y^{2}$

that lies in the first octant between planes $z = 0, z = 5, y = 1,$

and $y = 4 .$

A diagram of the given cylinder in three-dimensions. It is cut by the planes z=0, z=5, y=1, and y=4.

[T] Evaluate $\iint_{S} (z + y) d S,$

where S is the part of the graph of $z = \sqrt{1 - x^{2}}$

in the first octant between the xz-plane and plane $y = 3 .$

A diagram of the given surface in three dimensions in the first octant between the xz-plane and plane y=3. The given graph of z= the square root of (1-x^2) stretches down in a concave down curve from along (0,y,1) to along (1,y,0). It looks like a portion of a horizontal cylinder with base along the xz-plane and height along the y axis.

\iint_{S} (z + y) d S \approx 10.1

Evaluate $\iint_{S} x y z d S$

if S is the part of plane $z = x + y$

that lies over the triangular region in the xy-plane with vertices (0, 0, 0), (1, 0, 0), and (0, 2, 0).

Find the mass of a lamina of density $ξ (x, y, z) = z$

in the shape of hemisphere $z = {(a^{2} - x^{2} - y^{2})}^{1 / 2} .$

m = π a^{3}

Compute $\int \int_{S} F \cdot N d S,$

where $F (x, y, z) = x i - 5 y j + 4 z k$

and N is an outward normal vector S, where S is the union of two squares $S_{1} : x = 0, 0 \leq y \leq 1, 0 \leq z \leq 1$

and $S_{2} : z = 1, 0 \leq x \leq 1, 0 \leq y \leq 1 .$

A diagram in three dimensions. It shows the square formed by the components x=0, 0 <= y <= 1, and 0 <= z <= 1. It also shows the square formed by the components z=1, 0 <= x <= 1, and 0 <= y <= 1.

Compute $\int \int_{S} F \cdot N d S,$

where $F (x, y, z) = x y i + z j + (x + y) k$

and N is an outward normal vector S, where S is the triangular region cut off from plane $x + y + z = 1$

by the positive coordinate axes.

\iint_{S} F \cdot N d S = \frac{13}{24}

Compute $\int \int_{S} F \cdot N d S,$

where $F (x, y, z) = 2 y z i + ({tan}^{−1} x z) j + e^{x y} k$

and N is an outward normal vector S, where S is the surface of sphere $x^{2} + y^{2} + z^{2} = 1 .$

Compute $\int \int_{S} F \cdot N d S,$

where $F (x, y, z) = x y z i + x y z j + x y z k$

and N is an outward normal vector S, where S is the surface of the five faces of the unit cube $0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq z \leq 1$

missing $z = 0 .$

\iint_{S} F \cdot N d S = \frac{3}{4}

For the following exercises, express the surface integral as an iterated double integral by using a projection on S on the yz-plane.

\iint_{S} x y^{2} z^{3} d S;

S is the first-octant portion of plane $2 x + 3 y + 4 z = 12 .$

\iint_{S} (x^{2} - 2 y + z) d S;

S is the portion of the graph of $4 x + y = 8$

bounded by the coordinate planes and plane $z = 6 .$

\int_{0}^{8} \int_{0}^{6} (4 - 3 y + \frac{1}{16} y^{2} + z) (\frac{1}{4} \sqrt{17}) d z d y

For the following exercises, express the surface integral as an iterated double integral by using a projection on S on the xz-plane

\iint_{S} x y^{2} z^{3} d S;

S is the first-octant portion of plane $2 x + 3 y + 4 z = 12 .$

\iint_{S} (x^{2} - 2 y + z) d S;

S is the portion of the graph of $4 x + y = 8$

bounded by the coordinate planes and plane $z = 6 .$

\int_{0}^{2} \int_{0}^{6} [x^{2} - 2 (8 - 4 x) + z] \sqrt{17} d z d x

Evaluate surface integral $\iint_{S} y z d S,$

where S is the first-octant part of plane $x + y + z = λ,$

where $λ$

is a positive constant.

Evaluate surface integral $\iint_{S} (x^{2} z + y^{2} z) d S,$

where S is hemisphere $x^{2} + y^{2} + z^{2} = a^{2}, z \geq 0 .$

\iint_{S} (x^{2} z + y^{2} z) d S = \frac{π a^{5}}{2}

Evaluate surface integral $\iint_{S} z d A,$

where S is surface $z = \sqrt{x^{2} + y^{2}}, 0 \leq z \leq 2 .$

Evaluate surface integral $\iint_{S} x^{2} y z d S,$

where S is the part of plane $z = 1 + 2 x + 3 y$

that lies above rectangle $0 \leq x \leq 3 and 0 \leq y \leq 2 .$

\iint_{S} x^{2} y z d S = 171 \sqrt{14}

Evaluate surface integral $\iint_{S} y z d S,$

where S is plane $x + y + z = 1$

that lies in the first octant.

Evaluate surface integral $\iint_{S} y z d S,$

where S is the part of plane $z = y + 3$

that lies inside cylinder $x^{2} + y^{2} = 1 .$

\iint_{S} y z d S = \frac{\sqrt{2} π}{4}

For the following exercises, use geometric reasoning to evaluate the given surface integrals.

\iint_{S} \sqrt{x^{2} + y^{2} + z^{2}} d S,

where S is surface $x^{2} + y^{2} + z^{2} = 4, z \geq 0$

\iint_{S} (x i + y j) \cdot d S,

where S is surface $x^{2} + y^{2} = 4, 1 \leq z \leq 3,$

oriented with unit normal vectors pointing outward

\iint_{S} (x i + y j) \cdot d S = 16 π

\iint_{S} (z k) \cdot d S,

where S is disc $x^{2} + y^{2} \leq 9$

on plane $z = 4,$

oriented with unit normal vectors pointing upward

A lamina has the shape of a portion of sphere $x^{2} + y^{2} + z^{2} = a^{2}$

that lies within cone $z = \sqrt{x^{2} + y^{2}} .$

Let S be the spherical shell centered at the origin with radius a, and let C be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the z-axis. Determine the mass of the lamina if $δ (x, y, z) = x^{2} y^{2} z .$

A diagram in three dimensions. A cone opens upward with point at the origin and an asic of symmetry that coincides with the z-axis. The upper half of a hemisphere with center at the origin opens downward and is cut off by the xy-plane.

m = \frac{π a^{7}}{192}

A lamina has the shape of a portion of sphere $x^{2} + y^{2} + z^{2} = a^{2}$

that lies within cone $z = \sqrt{x^{2} + y^{2}} .$

Let S be the spherical shell centered at the origin with radius a, and let C be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the z-axis. Suppose the vertex angle of the cone is $ϕ_{0}, with 0 \leq ϕ_{0} < \frac{π}{2} .$

Determine the mass of that portion of the shape enclosed in the intersection of S and C. Assume $δ (x, y, z) = x^{2} y^{2} z .$

A paper cup has the shape of an inverted right circular cone of height 6 in. and radius of top 3 in. If the cup is full of water weighing $62.5 lb / {ft}^{3},$

find the total force exerted by the water on the inside surface of the cup.

F \approx 4.57 lb .

For the following exercises, the heat flow vector field for conducting objects i $F = − k \nabla T, where T (x, y, z)$

is the temperature in the object and $k > 0$

is a constant that depends on the material. Find the outward flux of F across the following surfaces S for the given temperature distributions and assume $k = 1 .$

T (x, y, z) = 100 e^{− x - y};

S consists of the faces of cube $\| x \| \leq 1, \| y \| \leq 1, \| z \| \leq 1 .$

T (x, y, z) = − ln (x^{2} + y^{2} + z^{2});

S is sphere $x^{2} + y^{2} + z^{2} = a^{2} .$

8 π a

For the following exercises, consider the radial fields $F = \frac{〈 x, y, z 〉}{{(x^{2} + y^{2} + z^{2})}^{\frac{p}{2}}} = \frac{r}{{\| r \|}^{p}},$

where p is a real number. Let S consist of spheres A and B centered at the origin with radii $0 < a < b .$

The total outward flux across S consists of the outward flux across the outer sphere B less the flux into S across inner sphere A.

A diagram in three dimensions of two spheres, one contained completely inside the other. Their centers are both at the origin. Arrows point in toward the origin from outside both spheres.

Find the total flux across S with $p = 0 .$

Show that for $p = 3$

the flux across S is independent of a and b.

The net flux is zero.

Parametric Surfaces

Surface Area of a Parametric Surface

Surface Integral of a Scalar-Valued Function

Orientation of a Surface

Surface Integral of a Vector Field

Key Concepts

Key Equations

Glossary