Vector Calculus

# 43 Surface Integrals

### Learning Objectives

• Find the parametric representations of a cylinder, a cone, and a sphere.
• Describe the surface integral of a scalar-valued function over a parametric surface.
• Use a surface integral to calculate the area of a given surface.
• Explain the meaning of an oriented surface, giving an example.
• Describe the surface integral of a vector field.
• Use surface integrals to solve applied problems.

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.

A parameterized surface is given by a description of the form

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.

Definition

Given a parameterization of surface the parameter domain of the parameterization is the set of points in the uv-plane that can be substituted into r.

Parameterizing a Cylinder

Describe surface S parameterized by

To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of we can choose any value for u and v and plot the corresponding point. If then so point (1, 0, 0) is on S. Similarly, points and are on S.

Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. To visualize S, we visualize two families of curves that lie on S. In the first family of curves we hold u constant; in the second family of curves we hold v constant. This allows us to build a “skeleton” of the surface, thereby getting an idea of its shape.

First, suppose that u is a constant K. Then the curve traced out by the parameterization is which gives a vertical line that goes through point in the xy-plane.

Now suppose that v is a constant K. Then the curve traced out by the parameterization is which gives a circle in plane with radius 1 and center (0, 0, K).

If u is held constant, then we get vertical lines; if v is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Therefore the surface traced out by the parameterization is cylinder ((Figure)).

(a) Lines for (b) Circles for (c) The lines and circles together. As u and v vary, they describe a cylinder.

Notice that if and then so points from S do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle for some k, and therefore each point on the cylinder is contained in the parameterized surface ((Figure)).

Cylinder has parameterization

Analysis

Notice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to then the surface would be a half-cylinder of height 6.

Describe the surface with parameterization

Cylinder

Hint

Hold u and v constant, and see what kind of curves result.

It follows from (Figure) that we can parameterize all cylinders of the form If S is a cylinder given by equation then a parameterization of S is

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

Describing a Surface

Describe surface S parameterized by

Notice that if u is held constant, then the resulting curve is a circle of radius u in plane Therefore, as u increases, the radius of the resulting circle increases. If v is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that

Therefore, the surface is elliptic paraboloid ((Figure)).

(a) Circles arise from holding u constant; the vertical parabolas arise from holding v constant. (b) An elliptic paraboloid results from all choices of u and v in the parameter domain.

Describe the surface parameterized by

Cone

Hint

Hold u constant and see what kind of curves result. Imagine what happens as u increases or decreases.

Finding a Parameterization

Give a parameterization of the cone lying on or above the plane

The horizontal cross-section of the cone at height is circle Therefore, a point on the cone at height u has coordinates for angle v. Hence, a parameterization of the cone is Since we are not interested in the entire cone, only the portion on or above plane the parameter domain is given by ((Figure)).

Cone has parameterization

Give a parameterization for the portion of cone lying in the first octant.

Hint

Consider the parameter domain for this 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 centered at the origin is given by the parameterization

The idea of this parameterization is that as sweeps downward from the positive z-axis, a circle of radius is traced out by letting run from 0 to To see this, let be fixed. Then

This results in the desired circle ((Figure)).

The sphere of radius has parameterization

Finally, to parameterize the graph of a two-variable function, we first let be a function of two variables. The simplest parameterization of the graph of is where x and y vary over the domain of ((Figure)). For example, the graph of can be parameterized by where the parameters x and y vary over the domain of If we only care about a piece of the graph of —say, the piece of the graph over rectangle —then we can restrict the parameter domain to give this piece of the surface:

Similarly, if S is a surface given by equation or equation then a parameterization of S is

or respectively. For example, the graph of paraboloid can be parameterized by Notice that we do not need to vary over the entire domain of y because x and z are squared.

The simplest parameterization of the graph of a function is

Let’s now generalize the notions of smoothness and regularity to a parametric surface. Recall that curve parameterization is regular if for all t in 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 The image of this parameterization is simply point which is not a curve. Notice also that 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 ((Figure)). How could we avoid parameterizations such as this? Parameterizations that do not give an actual surface? Notice that and and the corresponding cross product is zero. The analog of the condition is that is not zero for point in the parameter domain, which is a regular parameterization.

The image of parameterization is a line.

Definition

Parameterization is a regular parameterization if is not zero for point in the parameter domain.

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

Recall that curve parameterization is smooth if is continuous and for all t in 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.

Definition

A surface parameterization is smooth if vector is not zero for any choice of u and v in the parameter domain.

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

Identifying Smooth and Nonsmooth Surfaces

Which of the figures in (Figure) is smooth?

(a) This surface is smooth. (b) This surface is piecewise smooth.

The surface in (Figure)(a) can be parameterized by

(we can use technology to verify). Notice that vectors

exist for any choice of u and v in the parameter domain, and

The k component of this vector is zero only if or If or then the only choices for u that make the j component zero are or But, these choices of u do not make the i component zero. Therefore, is not zero for any choice of u and v in the parameter domain, and the parameterization is smooth. Notice that the corresponding surface has no sharp corners.

In the pyramid in (Figure)(b), the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, the pyramid has no smooth parameterization. However, the pyramid consists of four smooth faces, and thus this surface is piecewise smooth.

Is the surface parameterization smooth?

Yes

Hint

Investigate the cross product

### 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 over some parameter domain D. We assume here and throughout that the surface parameterization 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 with horizontal width and vertical length 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 Choose point in each piece Point corresponds to point in the parameter domain.

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 Without loss of generality, we assume that is located at the corner of two grid curves, as in (Figure). If we think of r as a mapping from the uv-plane to the grid curves are the image of the grid lines under r. To be precise, consider the grid lines that go through point One line is given by the other is given by In the first grid line, the horizontal component is held constant, yielding a vertical line through In the second grid line, the vertical component is held constant, yielding a horizontal line through The corresponding grid curves are and and these curves intersect at point

Grid lines on a parameter domain correspond to grid curves on a surface.

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

For grid curve the tangent vector at is

If vector exists and is not zero, then the tangent plane at exists ((Figure)). If piece is small enough, then the tangent plane at point is a good approximation of piece

If the cross product of vectors and exists, then there is a tangent plane.

The tangent plane at contains vectors and and therefore the parallelogram spanned by and is in the tangent plane. Since the original rectangle in the uv-plane corresponding to has width and length the parallelogram that we use to approximate is the parallelogram spanned by and In other words, we scale the tangent vectors by the constants and 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 is

Varying point over all pieces and the previous approximation leads to the following definition of surface area of a parametric surface ((Figure)).

The parallelogram spanned by and approximates the piece of surface

Definition

Let with parameter domain D be a smooth parameterization of surface S. Furthermore, assume that S is traced out only once as varies over D. The surface area of S is

where and

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 (Figure), we need a parameterization. We assume this cone is in with its vertex at the origin ((Figure)). To obtain a parameterization, let be the angle that is swept out by starting at the positive z-axis and ending at the cone, and let For a height value v with the radius of the circle formed by intersecting the cone with plane is Therefore, a parameterization of this cone is

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.

The right circular cone with radius r = kh and height h has parameterization

With a parameterization in hand, we can calculate the surface area of the cone using (Figure). The tangent vectors are and Therefore,

The magnitude of this vector is

By (Figure), the surface area of the cone is

Since

Therefore, the lateral surface area of the cone is

Analysis

The surface area of a right circular cone with radius r and height h is usually given as The reason for this is that the circular base is included as part of the cone, and therefore the area of the base is added to the lateral surface area that we found.

Find the surface area of the surface with parameterization

Hint

Use (Figure).

Calculating Surface Area

Show that the surface area of the sphere is

The sphere has parameterization

The tangent vectors are

Therefore,

Now,

Notice that on the parameter domain because and this justifies equation The surface area of the sphere is

We have derived the familiar formula for the surface area of a sphere using surface integrals.

Show that the surface area of cylinder is Notice that this cylinder does not include the top and bottom circles.

With the standard parameterization of a cylinder, (Figure) shows that the surface area is

Hint

Use the standard parameterization of a cylinder and follow the previous example.

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 be a positive single-variable function on the domain and let S be the surface obtained by rotating about the x-axis ((Figure)). Let be the angle of rotation. Then, S can be parameterized with parameters x and by

We can parameterize a surface of revolution by

Calculating Surface Area

Find the area of the surface of revolution obtained by rotating about the x-axis ((Figure)).

A surface integral can be used to calculate the surface area of this solid of revolution.

This surface has parameterization

The tangent vectors are Therefore,

and

The area of the surface of revolution is

Use (Figure) to find the area of the surface of revolution obtained by rotating curve about the x-axis.

Hint

Use the parameterization of surfaces of revolution given before (Figure).

### 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 with parameter domain D and let 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 with horizontal width and vertical length 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 Choose point in each piece evaluate at , and multiply by area to form the Riemann sum

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.

Definition

The surface integral of a scalar-valued function of over a piecewise smooth surface S is

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

From the material we have already studied, we know that

Therefore,

This approximation becomes arbitrarily close to as we increase the number of pieces by letting m and n go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:

(Figure) allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to (Figure) for line integrals:

In this case, vector is perpendicular to the surface, whereas vector is tangent to the curve.

Calculating a Surface Integral

Calculate surface integral where is the surface with parameterization for and

Notice that this parameter domain D is a triangle, and therefore the parameter domain is not rectangular. This is not an issue though, because (Figure) does not place any restrictions on the shape of the parameter domain.

To use (Figure) to calculate the surface integral, we first find vector and Note that and Therefore,

and

By (Figure),

Calculating the Surface Integral of a Cylinder

Calculate surface integral where S is cylinder ((Figure)).

Integrating function over a cylinder.

To calculate the surface integral, we first need a parameterization of the cylinder. Following (Figure), a parameterization is

The tangent vectors are and Then,

and By (Figure),

*** QuickLaTeX cannot compile formula:
\begin{array}{}\\ \\ \\ \phantom{\rule{1.2em}{0ex}}{\iint }_{S}f\left(x,y,z\right)dS={\iint }_{D}f\left(\text{r}\left(u,v\right)\right)‖{\text{t}}_{u}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\text{t}}_{v}‖\phantom{\rule{0.2em}{0ex}}dA\hfill \\ ={\int }_{0}^{3}{\int }_{0}^{2\pi }\left(\text{cos}\phantom{\rule{0.2em}{0ex}}u+{\text{sin}}^{2}u\right)dudv\hfill \\ ={{\int }_{0}^{3}\left[\text{sin}\phantom{\rule{0.2em}{0ex}}u+\frac{u}{2}-\frac{\text{sin}\left(2u\right)}{4}\right]}_{0}^{2\pi }dv={\int }_{0}^{3}\pi dv=3\pi .\hfill \end{array}

*** Error message:
Missing # inserted in alignment preamble.
leading text: $\begin{array}{} Missing$ inserted.
leading text: ...\ \\ \\ \phantom{\rule{1.2em}{0ex}}{\iint }
Extra }, or forgotten $. leading text: ...\ \\ \\ \phantom{\rule{1.2em}{0ex}}{\iint } Missing } inserted. leading text: ...‖\phantom{\rule{0.2em}{0ex}}dA\hfill \\ = Extra }, or forgotten$.
Missing } inserted.
Extra }, or forgotten $. leading text: ...‖\phantom{\rule{0.2em}{0ex}}dA\hfill \\ = Missing } inserted. leading text: ...‖\phantom{\rule{0.2em}{0ex}}dA\hfill \\ = Extra }, or forgotten$.
Missing } inserted.



Calculate where S is the surface with parameterization

24

Hint

Use (Figure).

Calculating the Surface Integral of a Piece of a Sphere

Calculate surface integral where and S is the surface that consists of the piece of sphere that lies on or above plane and the disk that is enclosed by intersection plane and the given sphere ((Figure)).

Calculating a surface integral over surface S.

Notice that S is not smooth but is piecewise smooth; S can be written as the union of its base and its spherical top and both and are smooth. Therefore, to calculate we write this integral as and we calculate integrals and

First, we calculate To calculate this integral we need a parameterization of This surface is a disk in plane centered at To parameterize this disk, we need to know its radius. Since the disk is formed where plane intersects sphere we can substitute into equation

Therefore, the radius of the disk is and a parameterization of is The tangent vectors are and and thus

The magnitude of this vector is u. Therefore,

Now we calculate To calculate this integral, we need a parameterization of The parameterization of full sphere is

Since we are only taking the piece of the sphere on or above plane we have to restrict the domain of To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in (Figure) (the comes from the fact that the base of S is a disk with radius Therefore, the tangent of is which implies that is We now have a parameterization of

The maximum value of has a tangent value of

The tangent vectors are

and thus

The magnitude of this vector is

Therefore,

Since we have

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.

Calculate line integral where S is cylinder including the circular top and bottom.

0

Hint

Break the integral into three separate surface integrals.

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 is then mass m of the sheet is

Calculating the Mass of a Sheet

A flat sheet of metal has the shape of surface that lies above rectangle and If the density of the sheet is given by what is the mass of the sheet?

Let S be the surface that describes the sheet. Then, the mass of the sheet is given by To compute this surface integral, we first need a parameterization of S. Since S is given by the function a parameterization of S is

The tangent vectors are and Therefore, and By (Figure),

A piece of metal has a shape that is modeled by paraboloid and the density of the metal is given by Find the mass of the piece of metal.

Hint

The mass of a sheet is given by A useful parameterization of a paraboloid was given in a previous example.

### 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.

Let S be a smooth surface. For any point on S, we can identify two unit normal vectors and If it is possible to choose a unit normal vector N at every point 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 ((Figure)). We also could choose the inward normal vector at each point to give an “inward” orientation, which is the negative orientation of the surface.

An oriented sphere with positive orientation.

A portion of the graph of any smooth function 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 in the standard way: where x and y vary over the domain of Then, and and therefore the cross product (which is normal to the surface at any point on the surface) is 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.

Let S be a smooth orientable surface with parameterization For each point on the surface, vectors and lie in the tangent plane at that point. Vector is normal to the tangent plane at and is therefore normal to S at that point. Therefore, the choice of unit normal vector

gives an orientation of surface S.

Choosing an Orientation

Give an orientation of cylinder

This surface has parameterization

The tangent vectors are and To get an orientation of the surface, we compute the unit normal vector

In this case, and therefore

An orientation of the cylinder is

Notice that all vectors are parallel to the xy-plane, which should be the case with vectors that are normal to the cylinder. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder ((Figure)).

If all the vectors normal to a cylinder point outward, then this is an outward orientation of the cylinder.

Give the “upward” orientation of the graph of

Hint

Parameterize the surface and use the fact that the surface is the graph of a function.

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 ((Figure)). 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.

The construction of a Möbius strip.

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 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 at the point.

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 Imagine the fluid flows through S, but S is completely permeable so that it does not impede the fluid flow ((Figure)). 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?

Fluid flows across a completely permeable surface S.

The rate of flow, measured in mass per unit time per unit area, is To calculate the mass flux across S, chop S into small pieces If is small enough, then it can be approximated by a tangent plane at some point P in Therefore, the unit normal vector at P can be used to approximate across the entire piece because the normal vector to a plane does not change as we move across the plane. The component of the vector at P in the direction of N is at P. Since is small, the dot product changes very little as we vary across and therefore can be taken as approximately constant across To approximate the mass of fluid per unit time flowing across (and not just locally at point P), we need to multiply by the area of Therefore, the mass of fluid per unit time flowing across in the direction of N can be approximated by where N, and v are all evaluated at P ((Figure)). 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 To approximate the mass flux across S, form the sum As pieces get smaller, the sum gets arbitrarily close to the mass flux. Therefore, the mass flux is

This is a surface integral of a vector field. Letting the vector field be an arbitrary vector field F leads to the following definition.

The mass of fluid per unit time flowing across in the direction of N can be approximated by

Definition

Let F be a continuous vector field with a domain that contains oriented surface S with unit normal vector N. The surface integral of F over S is

Notice the parallel between this definition and the definition of vector line integral 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 is called the flux of F across S, just as integral is the flux of F across curve C. A surface integral over a vector field is also called a flux integral.

Just as with vector line integrals, surface integral is easier to compute after surface S has been parameterized. Let be a parameterization of S with parameter domain D. Then, the unit normal vector is given by and, from (Figure), we have

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

Calculating a Surface Integral

Calculate the surface integral where and S is the surface with parameterization

The tangent vectors are and Therefore,

By (Figure),

Therefore, the flux of F across S is 340.

Calculate surface integral where and S is the portion of the unit sphere in the first octant with outward orientation.

0

Hint

Use (Figure).

Calculating Mass Flow Rate

Let represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m3. Let S be hemisphere with such that S is oriented outward. Find the mass flow rate of the fluid across S.

A parameterization of the surface is

As in (Figure), the tangent vectors are

and their cross product is

Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Therefore we use the orientation for the sphere.

By (Figure),

Therefore, the mass flow rate is

Let m/sec represent a velocity field of a fluid with constant density 100 kg/m3. Let S be the half-cylinder oriented outward. Calculate the mass flux of the fluid across S.

400 kg/sec/m

Hint

Use (Figure).

In (Figure), 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 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 (Figure), we simply remove the density constant, which gives a flow rate of

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 in an object is 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 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 The temperature at point in a region containing the cylinder is 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 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

First we consider the circular bottom of the object, which we denote We can see that is a circle of radius 1 centered at point sitting in plane This surface has parameterization Therefore,

and

Since the surface is oriented outward and is the bottom of the object, it makes sense that this vector points downward. By (Figure), the heat flow across is

Now let’s consider the circular top of the object, which we denote We see that is a circle of radius 1 centered at point sitting in plane This surface has parameterization Therefore,

and

Since the surface is oriented outward and is the top of the object, we instead take vector By (Figure), the heat flow across is

Last, let’s consider the cylindrical side of the object. This surface has parameterization By (Figure), we know that By (Figure),

Therefore, the rate of heat flow across S is

A cast-iron solid ball is given by inequality The temperature at a point in a region containing the ball is Find the heat flow across the boundary of the solid if this boundary is oriented outward.

Hint

### Key Concepts

• Surfaces can be parameterized, just as curves can be parameterized. In general, surfaces must be parameterized with two parameters.
• Surfaces can sometimes be oriented, just as curves can be oriented. Some surfaces, such as a Möbius strip, cannot be oriented.
• A surface integral is like a line integral in one higher dimension. The domain of integration of a surface integral is a surface in a plane or space, rather than a curve in a plane or space.
• The integrand of a surface integral can be a scalar function or a vector field. To calculate a surface integral with an integrand that is a function, use (Figure). To calculate a surface integral with an integrand that is a vector field, use (Figure).
• If S is a surface, then the area of S is

### Key Equations

• Scalar surface integral
• Flux integral

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

If surface S is given by then

True

If surface S is given by then

Surface is the same as surface for

True

Given the standard parameterization of a sphere, normal vectors are outward normal vectors.

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

Plane

for and

Paraboloid for

Plane

for and

The frustum of cone

The portion of cylinder in the first octant, for

for

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

[T] Plane above square

For the following exercises, let S be the hemisphere with and evaluate each surface integral, in the counterclockwise direction.

For the following exercises, evaluate for vector field F, where N is an outward normal vector to surface S.

and S is that part of plane that lies above unit square

and S is hemisphere

and S is the portion of plane that lies inside cylinder

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

[T]S is surface

[T]S is surface

Evaluate where S is the surface of cube in a counterclockwise direction.

Evaluate surface integral where and S is the portion of plane that lies over unit square R:

Evaluate where S is the surface defined parametrically by for

[T] Evaluate where S is the surface defined by

[T] Evaluate where S is the surface defined by for

Evaluate where S is the surface bounded above hemisphere and below by plane

Evaluate where S is the portion of plane that lies inside cylinder

[T] Evaluate where S is the portion of cone that lies between planes and

[T] Evaluate where S is the portion of cylinder that lies in the first octant between planes and

[T] Evaluate where S is the part of the graph of in the first octant between the xz-plane and plane

Evaluate if S is the part of plane 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 in the shape of hemisphere

Compute where and N is an outward normal vector S, where S is the union of two squares and

Compute where and N is an outward normal vector S, where S is the triangular region cut off from plane by the positive coordinate axes.

Compute where and N is an outward normal vector S, where S is the surface of sphere

Compute where and N is an outward normal vector S, where S is the surface of the five faces of the unit cube missing

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

S is the first-octant portion of plane

S is the portion of the graph of bounded by the coordinate planes and plane

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

S is the first-octant portion of plane

S is the portion of the graph of bounded by the coordinate planes and plane

Evaluate surface integral where S is the first-octant part of plane where is a positive constant.

Evaluate surface integral where S is hemisphere

Evaluate surface integral where S is surface

Evaluate surface integral where S is the part of plane that lies above rectangle

Evaluate surface integral where S is plane that lies in the first octant.

Evaluate surface integral where S is the part of plane that lies inside cylinder

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

where S is surface

where S is surface oriented with unit normal vectors pointing outward

where S is disc on plane oriented with unit normal vectors pointing upward

A lamina has the shape of a portion of sphere that lies within cone 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

A lamina has the shape of a portion of sphere that lies within cone 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 Determine the mass of that portion of the shape enclosed in the intersection of S and C. Assume

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

For the following exercises, the heat flow vector field for conducting objects i is the temperature in the object and 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

S consists of the faces of cube

S is sphere

For the following exercises, consider the radial fields where p is a real number. Let S consist of spheres A and B centered at the origin with radii 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.

Find the total flux across S with

Show that for the flux across S is independent of a and b.

The net flux is zero.

### Glossary

flux integral
another name for a surface integral of a vector field; the preferred term in physics and engineering
grid curves
curves on a surface that are parallel to grid lines in a coordinate plane
heat flow
a vector field proportional to the negative temperature gradient in an object
mass flux
the rate of mass flow of a fluid per unit area, measured in mass per unit time per unit area
orientation of a surface
if a surface has an “inner” side and an “outer” side, then an orientation is a choice of the inner or the outer side; the surface could also have “upward” and “downward” orientations
parameter domain (parameter space)
the region of the uv plane over which the parameters u and v vary for parameterization
parameterized surface (parametric surface)
a surface given by a description of the form where the parameters u and v vary over a parameter domain in the uv-plane
regular parameterization
parameterization such that is not zero for point in the parameter domain
surface area
the area of surface S given by the surface integral
surface integral
an integral of a function over a surface
surface integral of a scalar-valued function
a surface integral in which the integrand is a scalar function
surface integral of a vector field
a surface integral in which the integrand is a vector field