14.6 surface integrals

Few architectural structures are as impressive as domes. Whether it is the ceiling of the Sistine Chapel, the dome of a college library or the massive roof of the Toronto SkyDome, looking up at a domed ceiling can be awe-inspiring. One impressive aspect of a dome is its lack of visible support. As you might imagine, this is also a feature of domes that worries architects, who must be certain that the weight of the dome is properly supported. A critical part of an architect's calculation is the mass of the dome.

How would you compute the mass of a dome? You have already seen how to use double integrals to compute the mass of a two-dimensional lamina and triple integrals to find the mass of a three-dimensional solid. However, a dome is a three-dimensional structure more like a thin shell (a surface) than a solid. We hope you're one step ahead of us on this one: if you don't know how to find the mass of the dome exactly, you can try to approximate its mass by slicing it into a number of small sections and estimating the mass of each section. In Figure 14.39, we show a curved surface that has been divided into a number of pieces. If the pieces are small enough, notice that the density of each piece will be approximately constant.



Figure 14.39
Partition of a surface.

So, first subdivide (partition) the surface into n smaller pieces, S1, S2, , Sn. Next, let (x, y, z) be the density function (measured in units of mass per unit area), and for each i = 1, 2, , n, let (xi, yi, zi) be a point on the section Si and let Si be the surface area of Si . The mass of the section Si is then given approximately by (xi, yi, zi) Si . The total mass m of the surface is given approximately by the sum of these approximate masses,

 
You should expect that the exact mass is given by the limit of these sums as the size of the pieces gets smaller and smaller. We define the diameter of a section Si as the maximum distance between any two points on Si and the norm of the partition as the maximum of the diameters of the Si 's. Then we have that
 
This limit is an example of a new type of integral, the surface integral, which is the focus of this section.

6.1   
 
The surface integral of a function g(x, y, z) over a surface S 3, written is given by
 
provided the limit exists and is the same for all choices of the evaluation points (xi, yi, zi) .
 



Notice how our development of the surface integral parallels our development of the line integral. Whereas the line integral extended a single integral over an interval to an integral over a curve in three dimensions, the surface integral extends a double integral over a two-dimensional region to an integral over a two-dimensional surface in three dimensions. In both cases, we are “curving'' our domain into three dimensions.

Now that we have defined the surface integral, how can we calculate one? The basic idea is to rewrite a surface integral as a double integral and then evaluate the double integral using existing techniques. To convert a given surface integral into a double integral, you will have two main tasks:

1. Write the integrand g(x, y, z) as a function of two variables.
2. Write the surface area element dS in terms of the area element dA.

We will develop a general rule for step (2) before considering specific examples.

Consider a surface such as the one pictured in Figure 14.39. For the sake of simplicity, we assume that the surface is the graph of the equation z = f (x, y) , where f  has continuous first partial derivatives in some region R in the xy - plane. Notice that for an inner partition R1, R2, , Rn of R, if we take the point (xi, yi, 0) as the point in Ri closest to the origin, then the portion of the surface Si lying above Ri will differ very little from the portion Ti of the tangent plane to the surface at (xi, yi, f (xi, yi)) lying above Ri . More to the point, the surface area of Si will be approximately the same as the area of the parallelogram Ti . In Figure 14.40, we have indicated the portion Ti of the tangent plane lying above Ri.



Figure 14.40
Portion of the tangent plane lying above Ri .

Let the vectors and form two adjacent sides of the parallelogram Ti, as indicated in Figure 14.40. Notice that since ui and vi lie in the tangent plane, n i = ui vi = is a normal vector to the tangent plane. We saw in Chapter 10 that the area of the parallelogram can be written as

 
We further observe that the area of Ri is given by Ai = | ac | and ni k = -ac, so that . We can now write
 
since ac 0. The corresponding expression relating the surface area element dS and the area element dA is then

In the exercises, we will ask you to derive similar formulas for the cases where the surface S is written as a function of x and z or as a function of y and z.

We will consider two main cases of surface integrals. In the first, the surface is defined by a function z = f (x, y) . In the second, the surface is defined by parametric equations x = x(u, v) , y = y(u, v) and z = z(u, v) . In each case, your primary task will be to determine a normal vector to use in the general conversion formula for dS.

If S is the surface z = f (x, y) , recall from our discussion in section 12.4 that a normal vector to S is given by n = f x, f y, -1 . This is a convenient normal vector for our purposes, since . With , we have the following result.

6.1    (Evaluation Theorem)
 
If the surface S is given by z = f (x, y) for (x, y) in the region R 2, where f  has continuous first partial derivatives, then
 

   
 
From the definition of surface integral in Definition 6.1, we have
        
        
        
        
as desired.
 

Think about this theorem carefully! It tells us that we can evaluate a surface integral by evaluating a related double integral. To convert the surface integral into a double integral, you must first substitute for z in the function g(x, y, z) . This is done by substituting in the equation of the surface z = f (x, y) . Also, you must replace the surface area element dS with , which for the surface z = f (x, y) is given by
 
(6.1)

6.1   
Evaluating a Surface Integral
 
Evaluate where the surface S is the portion of the plane 2x+y+z = 2 lying in the first octant.
 
 
On S, we have z = 2-2x-y , so we must evaluate Note that a normal vector to the plane 2x+y+z = 2 is , so that in this case, the element of surface area is given by (6.1) to be
 
From the Evaluation Theorem, we then have
 
where R is the projection of the surface onto the xy - plane. A graph of the surface S is shown in Figure 14.41a. In this case, notice that R is the triangle indicated in Figure 14.41b. The triangle is bounded by x = 0 , y = 0 and the line 2x+y = 2 (the intersection of the plane 2x+y+z = 2 with the plane z = 0 ). If we integrate with respect to y first, we have inside integration limits of y = 0 and y = 2-2x , with x ranging from 0 to 1. This gives us
 
 
  = 2,
where we leave the details of the integration as an exercise.



Figure 14.41a
z = 2-2x-y.



Figure 14.41b
The projection R of the surface S onto the xy - plane.
 

In the following example, we will need to rewrite the double integral using polar coordinates.

6.2   
Evaluating a Surface Integral Using Polar Coordinates
 
Evaluate where the surface S is the portion of the paraboloid z = 4-x2-y2 lying above the xy - plane.
 
 
Substituting z = 4-x2-y2 , we have
 
In this case, a normal vector to the surface z = 4-x2-y2 is n = -2x, -2y, -1, so that
 
This gives us
 
Here, the region R is the projection of the paraboloid onto the xy - plane, which is the circle x2+y2 = 4 (see Figure 14.42). With a circular region of integration and the term x2+y2 appearing (twice!) in the integrand, you had better be thinking about polar coordinates. We have 4-x2-y2 = 4-r2 , and dA = r dr d . For the circle x2+y2 = 4 , r ranges from 0 to 2 and ranges from 0 to 2 . Then, we have
 
 
  = - ,
where we leave the details of the final integration to you.



Figure 14.42
z = 4-x2-y2.
 

Parametric Representation of Surfaces

In the remainder of this section, we study parametric representations of surface integrals. Before applying parametric equations to the computation of surface integrals, we need a better understanding of surfaces that have been defined parametrically. You have already seen many parametric surfaces, even if you didn't think of them in parametric terms. For instance, you can think of the equation as a parametric representation of a cone with parameters x and y . Recall that we can write this cone in cylindrical coordinates as z = r, 0 2 , where the parameters are r and . Similarly, the equation = 4 , 0 2 and , is a parametric representation of the sphere x2+y2+z2 = 16, with parameters . It will be helpful to review these graphs as well as to look at some new ones.

6.3   
Sketching a Surface Defined Parametrically
 
Sketch the surface defined parametrically by , and , 0 u 2 and - < v < .
 
 
A sketch such as the one we show in Figure 14.43 can be obtained from a computer algebra system. Notice that this looks like a hyperboloid of one sheet wrapped around the z - axis. To verify that this is correct, observe that
 
 
 
where we have used the identities cos 2u+sin 2u = 1 and . Recall that the graph of x2+y2-z2 = 4 is indeed a hyperboloid of one sheet.



Figure 14.43
x2+y2-z2 = 4.
 

Most of the time when we use parametric representations of surfaces, the task is the opposite of that in example 6.3. Given a particular surface, we may need to find a convenient parametric representation of the surface. The general form for parametric equations representing a surface in three dimensions is x = x(u, v) , y = y(u, v) and z = z(u, v) for u1 u u2 and v1 v v2 . The parameters u and v can correspond to familiar coordinates ( x and y , or r and , for instance), or less familiar expressions. Keep in mind that to fully describe a surface, you will need to define two parameters.

6.4   
Finding Parametric Representations of a Surface
 
Find simple parametric representations for (a) the portion of the cone inside the cylinder x2+y2 = 4 and (b) the portion of the sphere x2+y2+z2 = 16 inside of the cone .
 
 
It is important to realize that both parts (a) and (b) have numerous solutions. (In fact, every surface can be represented parametrically in an infinite number of ways.) The solutions we show below are among the simplest and most useful, but they are not the only reasonable solutions. In (a), the repeated appearance of the term x2+y2 suggests that cylindrical coordinates (r, , z) might be convenient. A sketch of the surface is shown in Figure 14.44a. Notice that the cone becomes z = r in cylindrical coordinates. Recall that in cylindrical coordinates, x = rcos and y = rsin . Notice that the parameters r and have ranges determined by the cylinder x2+y2 = 4 , so that 0 r 2 and 0 2 . Parametric equations for the cone are then x = rcos , y = rsin , z = r with 0 r 2 and 0 2.



Figure 14.44a
The cone and the cylinder x2+y2 = 4.

The surface in part (b) is a portion of a sphere, which suggests (what else?) spherical coordinates. Recall that we convert from spherical coordinates to rectangular co-ordinates by using , and . We only want two parameters, so one of the spherical coordinates must be eliminated. Notice that on the sphere x2+y2+z2 = 16, the distance from the origin is constant and so, an equation of the sphere is = 4 . Using this, we have the parametric representation of the sphere: , and , where 0 2 and . To find the portion of the sphere inside the cone, we must observe that the cone can be described in spherical coordinates as Referring to Figure 14.44b, note that the portion of the sphere inside the cone is then described by , and where 0 2 and .



Figure 14.44b
The portion of the sphere inside the cone.
 

Suppose that we have a parametric representation for the surface S : x = x(u, v) , y = y(u, v) and z = z(u, v), defined on the rectangle R = { (u, v) | a u b and c v d} in the uv - plane. It is often convenient to use parametric equations to evaluate the surface integral Of course, to do this, we must substitute for x, y and z to rewrite the integrand in terms of the parameters u and v, as
  g(u, v) = f (x(u, v), y(u, v), z(u, v)).
However, this is only a small piece of the puzzle, as we must also write the surface area element dS in terms of the area element dA for the uv plane. Unfortunately, we can't use (6.1) here, since this only holds for the case of a surface written in the form z = f (x, y). Instead, we'll need to back up just a bit.

First, notice that the position vector for points on the surface S is r(u, v) = x(u, v), y(u, v), z(u, v) . We define the vectors ru and rv (the subscripts denote partial derivatives) by:

 
and
r v(u, v) = xv(u, v), yv(u, v), zv(u, v).
Notice that for any fixed (u, v), both of the vectors ru(u, v) and rv(u, v) lie in the tangent plane to S at the point (x(u, v), y(u, v), z(u, v)). So, unless these two vectors are parallel, we can find a normal vector to the tangent plane by computing their cross product. That is, n = ru rv is a normal vector to the surface at the point (x(u, v), y(u, v), z(u, v)). We say that the surface S is smooth if ru and rv are continuous and ru rv 0, for all (u, v) R. (This says that the surface will not have any corners.) We say that S is piecewise-smooth if we can write S = S1 S2 Sn, for some smooth surfaces S1, S2, , Sn.

As we have done many times now, we partition the rectangle R in the uv - plane. For each rectangle Ri in the partition, let (ui, vi) be the closest point in Ri to the origin, as indicated in Figure 14.45a. Notice that each of the sides of Ri gets mapped to a curve in xyz - space, so that Ri gets mapped to a curvilinear region Si in xyz - space, as indicated in Figure 14.45b (on the following page). Observe that if we locate their initial points at the point Pi(x(ui, vi), y(ui, vi), z(ui, vi)), the vectors ru(ui, vi) and rv(ui, vi) lie tangent to two adjacent curved sides of Si. So, we can approximate the area Si of Si by the area of the parallelogram Ti whose sides are formed by the vectors ui ru(ui, vi) and vi rv(ui, vi) (see Figure 14.45c).



Figure 14.45a
Partition of parameter domain(u - plane).



Figure 14.45b
Curvilinear region Si.



Figure 14.45c
The parallelogram Ti.

As we know, the area of the parallelogram is given by the magnitude of the cross product
 
 
where Ai is the area of the rectangle Ri. We then have that
 
and it follows that the element of surface area can be written as
 
(6.2)
Notice that this corresponds closely to (6.1), as ru rv is a normal vector to S . Finally, we developed (6.2) in the comparatively simple case where the parameter domain R (that is, the domain in the uv - plane) was a rectangle. If the parameter domain is not a rectangle, you should recognize that we can do the same thing by constructing an inner partition of the region. We can now evaluate surface integrals using parametric equations, as in the following example.

6.5   
Evaluating a Surface Integral Using Spherical Coordinates
 
Evaluate where S is the sphere x2+y2+z2 = 4 .
 
 
Since the surface is a sphere and the integrand contains the term x2 + y2+z2, spherical coordinates are indicated. Notice that the sphere is described by = 2 and so, on the surface of the sphere, the integrand becomes 3(x2+y2+z2) = 12 . Further, we can describe the sphere = 2 with the parametric equations , and for 0 2 and . This says that the sphere is traced out by the endpoint of the vector-valued function
 
We then have the partial derivatives
 
and
 
We leave it as an exercise to show that a normal vector to the surface is given by
 
so that . Equation (6.2) now gives us so that
 
 
    = 192,
where we used the fact that for , to replace by .
 



 If you did the calculation of dS in example 6.5, you may not think that parametric equations lead to simple solutions. (That's why we didn't show all of the details!) However, recall that for changing a triple integral from rectangular to spherical coordinates, you replace dx dy dz by . In example 6.5, we have 2 = 4 and dS = . Looks familiar now, doesn't it? This shortcut is valuable, since surface integrals over spheres are reasonably common.

So, when you evaluate a surface integral, what have you computed? We close the section with two examples. The first is familiar: observe that the surface integral of the function f (x, y, z) = 1 over the surface S is simply the surface area of S . That is,

 
The proof of this follows directly from the definition of the surface integral and is left as an exercise.

6.6   
Using a Surface Integral to Compute Surface Area
 
Compute the surface area of the portion of the hyperboloid x2+y2-z2 = 4 between z = 0 and z = 2 .
 
 
We need to evaluate . Recall from example 6.3 that we can write the hyperboloid parametrically by , and . (If you didn't remember this, you could always derive parametric equations in the following way. To get a circular cross section of radius 2 in the xy - plane, start with x = 2cos u and y = 2sin u . To get a hyperbola in the xz - or yz - plane, multiply x and y by and set .) We have 0 u 2 to get the circular cross sections and ( 0.88 ). The hyperboloid is traced out by the endpoint of the vector-valued function
 
so that
 
and
 
This gives us the normal vector
 
where We now have
 
 
    31.95,
where we evaluated the final integral numerically.
 

Our final example of a surface integral requires some preliminary discussion. We must first define an oriented (or two-sided) surface. We say that a surface S is oriented (or orientable) if it has a unit normal vector n at each point (x, y, z) not on the boundary of the surface and if n is a continuous function of (x, y, z) . Further, we assume that S has two identifiable sides (a top and a bottom or an inside and an outside). To orient such a surface, we choose a consistent direction for all normal vectors to point. For instance, a sphere is a two-sided surface; the two sides of the surface are the inside and the outside. Notice that you can't get from the inside to the outside without passing through the sphere. The positive orientation for the sphere (or for any other closed surface) is to choose outward normal vectors (normal vectors pointing away from the interior).

All of the surfaces we have seen so far in this course are two-sided, but it's not difficult to construct a one-sided surface. Perhaps the most famous example of a one-sided surface is the Möbius strip, named after the German mathematician A. F. Möbius. You can easily construct a Möbius strip by taking a long rectangular strip of paper, giving it a half-twist and then taping the short edges together, as illustrated in Figures 14.46a through 14.46c. Notice that if you started painting the strip, you would eventually return to your starting point, having painted both “sides'' of the strip, but without having crossed any edges. This says that the Möbius strip has no inside and no outside and is therefore not orientable.

One reason we need to be able to orient a surface is to compute the flux of a vector field. It's easiest to visualize the flux for a vector field representing the velocity field for a fluid in motion. In this context, the flux measures the net flow rate of the fluid across the surface, that is, from the inside to the outside. (Notice that for this to make sense, the surface must have two identifiable sides. That is, the surface must be orientable.) The orientation of the surface lets us distinguish one direction from the other. In general, we have the following definition.



Figure 14.46a
A long, thin strip.



Figure 14.46b
Make one half-twist.



Figure 14.46c
A Möbius strip.

6.2   
 
Let F(x, y, z) be a continuous vector field defined on an oriented surface S with unit normal vector n . The surface integral of F over S (or the flux of F over S ) is given by .
 

Think carefully about the role of the unit normal vector in this definition. Notice that since n is a unit vector, the integrand F n gives (at any given point on S ) the component of F in the direction of n . So, if F represents the velocity field for a fluid in motion, F n corresponds to the component of the velocity that moves the fluid across the surface (from one side to the other). Also, note that F n can be positive or negative, depending on which normal vector we have chosen. (Keep in mind that at each point on a surface, there are two unit normal vectors; one pointing toward each side of the surface.) You should recognize that this is why we need to have an oriented surface.

6.7   
Computing the Flux of a Vector Field
 
Compute the flux of the vector field over the portion of the paraboloid z = x2+y2 below z = 4 (oriented with upward-pointing normal vectors).
 
 
First, observe that at any given point, the normal vectors for the paraboloid z = x2+y2 are 2x, 2y, -1 . For the normal vector to point upward, we need a positive z - component. In this case,
  m = - 2x, 2y, -1 = -2x, -2y, 1
is such a normal vector. However, the definition of flux requires a unit normal vector. A unit vector pointing in the same direction as m is
 
Before computing F n , we use the normal vector m to write the surface area increment dS in terms of dA. From (6.1), we have
 
Putting this all together gives us
 
 
 
where the region R is the projection of the portion of the paraboloid under consideration onto the xy - plane. Note how the square roots arising from the calculation of and dS cancelled one another out. Look at the graph in Figure 14.47 and recognize that this projection is bounded by the circle x2+y2 = 4 . You should quickly realize that the double integral should be set up in polar coordinates. We have
 
 



Figure 14.47
z = x2+y2.
 

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