9.13 Surface Integrals

INTRODUCTION

In the xy-plane, the length of an arc of the graph of y = f(x) from x = a to x = b is given by the definite integral

(1)

The problem in three dimensions, which is the counterpart of the arc length problem, is to find the area A(s) of that portion of the surface S given by a function z = f(x, y) having continuous first partial derivatives on a closed region R in the xy-plane. Such a surface is said to be smooth.

Surface Area

Suppose, as shown in FIGURE 9.13.1(a), that an inner partition P of R is formed using lines parallel to the x- and y-axes. P then consists of n rectangular elements Rk of area ΔAk = Δxk Δyk that lie entirely within R. Let (xk, yk, 0) denote any point in an Rk. As we see in Figure 9.13.1(a), by projecting the sides of Rk upward, we determine two quantities: a portion or patch Sk of the surface and a portion Tk of a tangent plane at (xk, yk, f(xk, yk)). It seems reasonable to assume that when Rk is small, the area ΔTk of Tk is approximately the same as the area ΔSk of Sk.

A graph and a visual representation. Graph a. A three dimensional surface, a closed curve, and a rectangular prism, are graphed on a three dimensional x y z coordinate system. The surface is labeled z = f(x y) over R. A closed curve is graphed on the x y plane. Several equally spaced out lines, parallel to the x and y axes, partition the region inside the closed curve. All the rectangles in the grid that lie entirely inside the closed curve, are shaded and labeled R. The closed curve is projected vertically up on the surface. A rectangular prism, with one of the rectangles as its base, and its top face on the surface, connects the surface and the region R. The top face of the rectangular prism on the surface is labeled T subscript k, and contains a marked point (x subscript k, y subscript k, f(x subscript k, y subscript k)). The base of the rectangular prism on the x y plane, is highlighted and contains a second marked point (x subscript k, y subscript k, 0). A dashed vertical line connecting the 2 marked points, passes through the central axis of the rectangular prism. A patch of the surface on the projection of the rectangle R subscript k, is labeled S subscript k. Visual representation b. Caption. Enlargement of R subscript k, S subscript k, and T subscript k. Figure. The rectangular prism in the graph is enlarged. The base of the rectangular prism is shaded and labeled R subscript k. The left side of the base at the bottom is labeled delta x subscript k, and the side at the front is labeled delta y subscript k. The bottom left vertex at the back of the base is at the marked point (x subscript k, y subscript k, 0). A vector, labeled u, begins at the top left vertex of the base at the back, and goes along the left side of the top base. A vector, labeled v, begins at the top left vertex of the base at the back, and goes along the side at the back of the top base. The top base is shaded and labeled T subscript k. A patch of the surface on the sides at the top of the rectangular prism is labeled S subscript k.

FIGURE 9.13.1 What is the area of the surface above R?

To find the area of Tk let us choose (xk, yk, 0) at a corner of Rk as shown in Figure 9.13.1(b). The indicated vectors u and v, which form two sides of Tk, are given by

where fx(xk, yk) and fy(xk, yk) are the slopes of the lines containing u and v, respectively. Now from (16) of Section 7.4 we know that ΔTk = u ×v where

In other words,

Consequently, the area A is approximately

Taking the limit of the foregoing sum as → 0 leads us to the next definition.

DEFINITION 9.13.1 Surface Area

Let f be a function for which the first partial derivatives fx and fy are continuous on a closed region R. Then the area of the surface over R is given by

.(2)

One could have almost guessed the form of (2) by naturally extending the one-variable structure of (1) to two variables.

EXAMPLE 1 Surface Area

Find the surface area of that portion of the sphere x2 + y2 + z2 = a2 that is above the xy-plane and within the cylinder x2 + y2 = b2, 0 < b < a.

SOLUTION

If we define z = f(x, y) by f(x, y) = then

and so 1 + [fx(x, y)]2 + [fy(x, y)]2 =

Hence, (2) is A(S) =

where R is indicated in FIGURE 9.13.2. To evaluate this double integral, we change to polar coordinates:

≡

A graph. A hemisphere and right circular cylinder is graphed on a three dimensional x y z coordinate system. The hemisphere is labeled x^2 + y^2 + z^2 = a^2, and its center is at the origin. The circular base of the cylinder, graphed inside the base of the hemisphere, is labeled x^2 + y^2 = b^2, and its center is also at the origin. The base of the cylinder is shaded and labeled R. The top base of the cylinder is above the surface of the hemisphere. The surface of the hemisphere inside the cylinder is shaded.

FIGURE 9.13.2 Portion of a sphere in Example 1

Differential of Surface Area

The function

(3)

is called the differential of the surface area. We will use this function in the discussion that follows.

Surface Integral

As we have seen, a double integral

is a generalization of the definite integral f(x) dx. The surface area integral (2) is a generalization of the arc length integral (1). We are now going to consider a generalization of the line integral G(x, y) ds. This generalization is called a surface integral.

  1. Let w = G(x, y, z) be defined in a region of 3-space that contains a surface S, which is the graph of a function z = f(x, y). Let the projection R of the surface onto the xy-plane be either a Type I or a Type II region.
  2. Divide the surface S into n patches Sk with areas ΔSk that correspond to a partition P of R into n rectangles Rk with areas ΔAk.
  3. Let be the norm of the partition or the length of the longest diagonal of the Rk.
  4. Choose a sample point (, , ) on each patch Sk of surface area. See FIGURE 9.13.3.
  5. Form the sum G(, , ) ΔSk.
A graph. A three dimensional surface, a closed curve is graphed on a three dimensional x y z coordinate system. The surface is labeled S. A closed curve is graphed on the x y plane. Several equally spaced out lines, parallel to the x and y axes, partition the region inside the closed curve. All the rectangles in the grid that lie entirely inside the closed curve, are shaded and labeled R. One of the rectangles on the grid is projected vertically up on the surface. The projected patch on the surface is shaded and labeled delta S subscript k. The point (x subscript k asterisk, y subscript k asterisk, z subscript k asterisk) is graphed inside the projected patch.

FIGURE 9.13.3 Sample point on kth patch

DEFINITION 9.13.2 Surface Integral

Let G be a function of three variables defined over a region of 3-space containing the surface S. Then the surface integral of G over S is given by

.(4)

Method of Evaluation

If G, f, fx, and fy are continuous throughout a region containing S, we can evaluate (4) by means of a double integral. From (3) the left side of (4) becomes

(5)

Note that when G = 1, (5) reduces to formula (2) for surface area.

Projection of S into Other Planes

If y = g(x, z) is the equation of a surface S that projects onto a region R of the xz-plane, then

(6)

Similarly, if x = h(y, z) is the equation of a surface that projects onto the yz-plane, then the analogue of (5) is

(7)

Mass of a Surface

Suppose ρ(x, y, z) represents the density of a surface at any point, or mass per unit surface area; then the mass m of the surface is

(8)

EXAMPLE 2 Mass of a Surface

Find the mass of the surface of the paraboloid z = 1 + x2 + y2 in the first octant for 1 ≤ z ≤ 5 if the density at a point P on the surface is directly proportional to its distance from the xy-plane.

SOLUTION

The surface in question and its projection onto the xy-plane are shown in FIGURE 9.13.4. Now, since ρ(x, y, z) = kz and z = 1 + x2 + y2, (8) and (5) give

A graph. A portion of a paraboloid, a plane and a region are graphed on a three dimensional x y z coordinate system. The paraboloid is graphed in the first octant. The low point of the paraboloid is on the z axis, above the x y plane. A horizontal plane at z = 5 intersects the paraboloid along a curve. The intersection curve is a quarter circle, with its center at the point (0, 0, 5). The curve begins at the point (0, 2, 5) and ends at the point (2, 0, 5). The surface of the paraboloid is shaded. The paraboloid is projected vertically down on the x y plane. The projected curve on the x y plane is a quarter circle labeled x^2 + y^2 = 4 or r = 2, and its center is at the origin. The curve begins at the point (0, 2, 0) and ends at the point (2, 0, 0). The region inside the projected curve is shaded and labeled R.

FIGURE 9.13.4 Surface in Example 2

By changing to polar coordinates, we obtain

≡

EXAMPLE 3 Evaluating a Surface Integral

Evaluate xz2 dS, where S is that portion of the cylinder y = 2x2 + 1 in the first octant bounded by x = 0, x = 2, z = 4, and z = 8.

SOLUTION

We shall use (6) with g(x, z) = 2x2 + 1 and R the rectangular region in the xz-plane shown in FIGURE 9.13.5. Since gx(x, z) = 4x and gz(x, z) = 0, it follows that

≡

A graph. A portion of a cylinder and a region is graphed on a three dimensional x y z coordinate system. A portion of the lateral surface of the cylinder begins on the y z plane between the points (0, 1, 4) and (0, 1, 8) and ends between the points (2, 8, 4) and (2, 8, 8). The surface is projected vertically down on the x y plane. The projected curve on the x y plane is labeled y = 2 x^2 + 1. The curve begins at the point (0, 1, 0) and ends at the point (2, 8, 0). The cylinder surface is projected horizontally on the x z plane. The projected region on the x z plane is a rectangle. The vertices of the rectangle are at the following points: (0, 0, 4), (2, 0, 4), (2, 0, 8), (0, 0, 8). The region inside the rectangle is shaded and labeled R.

FIGURE 9.13.5 Surface in Example 3

Orientable Surfaces

In Example 5, we are going to evaluate a surface integral of a vector field. In order to do this we need to examine the concept of an orientable surface. Roughly, an orientable surface S, such as that given in FIGURE 9.13.6(a), has two sides that could be painted different colors. The Möbius strip* shown in Figure 9.13.6(b) is not an orientable surface and is one-sided. A person who starts to paint the surface of a Möbius strip at a point will paint the entire surface and return to the starting point.

A graph and a visual representation. Graph a. Caption. Two-sided surface. Graph. A three dimensional convex surface is graphed on a three dimensional x y z coordinate system. The top and bottom sides of the surface are painted in different colors. The top side is labeled S. Visual representation b. Caption. One-sided surface. Figure. A surface in the form of a ring. The surface of the ring is twisted thrice, exposing the top and bottom sides of the surface.

FIGURE 9.13.6 Oriented surface in (a); non-oriented surface in (b)

Specifically, we say a smooth surface S is orientable or is an oriented surface if there exists a continuous unit normal vector function n defined at each point (x, y, z) on the surface. The vector field n(x, y, z) is called the orientation of S. But since a unit normal to the surface S at (x, y, z) can be either n(x, y, z) or –n(x, y, z), an orientable surface has two orientations. See FIGURE 9.13.7(a)–(c). The Möbius strip shown again in Figure 9.13.7(d) is not an oriented surface, since if a unit normal n starts at P on the surface and moves once around the strip on the curve C, it ends up on the “opposite side” of the strip at P and so points in the opposite direction. A surface S defined by z = f(x, y) has an upward orientation (Figure 9.13.7(b)) when the unit normals are directed upward—that is, have positive k components, and it has a downward orientation (Figure 9.13.7(c)) when the unit normals are directed downward—that is, have negative k components.

4 visual representations. Visual representation a. A convex surface labeled S. The surface is curved upward and contains a point (x y z). A vector labeled n points vertically up and an identical opposite vector labeled negative n points vertically down. Visual representation b. A convex surface labeled S. The surface is curved upward. Several vectors on the upper surface point outward at different angles. Visual representation c. A convex surface labeled S. The surface is curved upward. Several vectors on the lower surface point downward at different angles. Visual representation d. A surface in the form of a ring. The surface of the ring is twisted thrice, exposing the top and bottom sides of the surface. A dashed line passes through the center of the strip, and contains a point labeled P. A vector labeled n is perpendicular to the strip at the point P on the upper surface and points outward. A second identical opposite vector labeled negative n is perpendicular to the strip at the point P on the lower surface and points in the opposite direction.

FIGURE 9.13.7 Upward orientation in (b); downward orientation in (c)

If a smooth surface S is defined by g(x, y, z) = 0, then recall that a unit normal is

(9)

where ∇g = is the gradient of g. If S is defined by z = f(x, y), then we can use g(x, y, z) = z – f(x, y) = 0 or g(x, y, z) = f(x, y) – z = 0 depending on the orientation of S.

As we shall see in the next example, the two orientations of an orientable closed surface are outward and inward.

EXAMPLE 4 Orientations of a Surface

Consider the sphere of radius a > 0: x2 + y2 + z2 = a2. If we define g(x, y, z) = x2 + y2 + z2 – a2, then

Then the two orientations of the surface are

The vector field n defines an outward orientation, whereas n1 = –n defines an inward orientation. See FIGURE 9.13.8. ≡

2 graphs. Graph a. A sphere is graphed on a three dimensional x y z coordinate system. The center of the sphere is at the origin. Several vectors of similar magnitudes begin at various points on the surface of the sphere and point outward. Graph b. A sphere is graphed on a three dimensional x y z coordinate system. The center of the sphere is at the origin. Several vectors of similar magnitudes begin at various points on surface of the sphere and point inward toward the center.

FIGURE 9.13.8 Sphere in Example 4

Integrals of Vector Fields

If F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k is the velocity field of a fluid, then, as we saw in Figure 9.7.3, the volume of the fluid flowing through an element of surface area ΔS per unit time is approximated by

(height)(area of base) = (compn F) ΔS = (F · n) ΔS,

where n is a unit normal to the surface. See FIGURE 9.13.9. The total volume of a fluid passing through S per unit time is called the flux of F through S and is given by

(10)

In the case of a closed surface S, if n is the outer (inner) normal, then (10) gives the volume of fluid flowing out (in) through S per unit time.

A graph. A convex surface is graphed on a three dimensional x y z coordinate system. A patch of the surface, labeled delta S, contains a point. A vector labeled n is normal to the surface at the point. A second vector at the point labeled F points upward to the left. Several vector around the convex surface point up and to the left. The surface is projected vertically down on the x y plane. The projected region is shaded and labeled R.

FIGURE 9.13.9 Surface S in (10)

EXAMPLE 5 Flux Through a Surface

Let F(x, y, z) = zj + zk represent the flow of a liquid. Find the flux of F through the surface S given by that portion of the plane z = 6 – 3x – 2y in the first octant oriented upward.

SOLUTION

The vector field and the surface are illustrated in FIGURE 9.13.10. By defining the plane by g(x, y, z) = 3x + 2y + z – 6 = 0, we see that a unit normal with a positive k component is

Hence,

With R the projection of the surface onto the xy-plane, we find from (10) that

≡

A graph. A surface given by a portion of a plane is graphed on a three dimensional x y z coordinate system. The portion of the plane is formed by 3 lines. The first line labeled 3 x + 2 y = 6, graphed on the x y plane, begins at the point (2, 0, 0), and ends at the point (0, 3, 0). The second line begins at the point (2, 0, 0) and ends at the point (0, 0, 6). The third line begins at the point (0, 3, 0) and ends at the point (0, 0, 6). The vertical projection of the surface on the x y plane is a triangular region formed by the line 3 x + 2 y = 6 and the x and y axes. The projected region is shaded and labeled R. Several parallel vectors of different magnitudes begin on the surface and point upward.

FIGURE 9.13.10 Surface in Example 5

Depending on the nature of the vector field, the integral in (10) can represent other kinds of flux. For example, (10) could also give electric flux, magnetic flux, flux of heat, and so on.

REMARKS

If the surface S is piecewise defined, we express a surface integral over S as the sum of the surface integrals over the various pieces of the surface. For example, suppose S is the orientable piecewise-smooth closed surface bounded by the paraboloid z = x2 + y2 (S1) and the plane z = 1 (S2). Then the flux of a vector field F out of the surface S is

where we take S1 oriented downward and S2 oriented upward. See FIGURE 9.13.11 and Problem 35 in Exercises 9.13.

A graph. A piecewise-defined surface bound by a paraboloid and a plane is graphed on a three dimensional x y z coordinate system. The paraboloid graphed around the vertical z axis, and reaches a low point at the origin. The plane, represented by a disk, is graphed above the paraboloid and is parallel to the x y plane. The outer surface of the paraboloid is labeled S subscript 1. The upper surface of the disk is labeled S subscript 2. Several vectors on the surface S subscript 2 point vertically up. Several vectors on the paraboloid surface point downward at different angles.

FIGURE 9.13.11 Piecewise-defined surface

9.13 Exercises Answers to selected odd-numbered problems begin on page ANS-26.

  1. Find the surface area of that portion of the plane 2x + 3y + 4z = 12 that is bounded by the coordinate planes in the first octant.
  2. Find the surface area of that portion of the plane 2x + 3y + 4z = 12 that is above the region in the first quadrant bounded by the graph r = sin 2θ.
  3. Find the surface area of that portion of the cylinder x2 + z2 = 16 that is above the region in the first quadrant bounded on the graphs of x = 0, x = 2, y = 0, y = 5.
  4. Find the surface area of that portion of the paraboloid z = x2 + y2 that is below the plane z = 2.
  5. Find the surface area of that portion of the paraboloid z = 4 – x2 – y2 that is above the xy-plane.
  6. Find the surface area of those portions of the sphere x2 + y2 + z2 = 2 that are within the cone z2 = x2 + y2.
  7. Find the surface area of the portion of the sphere x2 + y2 + z2 = 25 that is above the region in the first quadrant bounded by the graphs of x = 0, y = 0, 4x2 + y2 = 25. [Hint: Integrate first with respect to x.]
  8. Find the surface area of that portion of the graph of z = x2 – y2 that is in the first octant within the cylinder x2 + y2 = 4.
  9. Find the surface area of the portions of the sphere x2 + y2 + z2 = a2 that are within the cylinder x2 + y2 = ay.
  10. Find the surface area of the portions of the cone z2 = (x2 + y2) that are within the cylinder (x – 1)2 + y2 = 1.
  11. Find the surface area of the portions of the cylinder y2 + z2 = a2 that are within the cylinder x2 + y2 = a2. [Hint: See Figure 9.10.12.]
  12. Use the result given in Example 1 to prove that the surface area of a sphere of radius a is 4πa2. [Hint: Consider a limit as b → a.]
  13. Find the surface area of that portion of the sphere x2 + y2 + z2 = a2 that is bounded between y = c1 and y = c2, 0 < c1 < c2 < a. [Hint: Use polar coordinates in the xz-plane.]
  14. Show that the area found in Problem 13 is the same as the surface area of the cylinder x2 + z2 = a2 between y = c1 and y = c2.

In Problems 15–24, evaluate the surface integral G(x, y, z) dS.

  1. G(x, y, z) = x; S the portion of the cylinder z = 2 – x2 in the first octant bounded by x = 0, y = 0, y = 4, z = 0
  2. G(x, y, z) = xy(9 – 4z); same surface as in Problem 15
  3. G(x, y, z) = xz3; S the cone z = inside the cylinder x2 + y2 = 1
  4. G(x, y, z) = x + y + z; S the cone z = between z = 1 and z = 4
  5. G(x, y, z) = (x2 + y2)z; S that portion of the sphere x2 + y2 + z2 = 36 in the first octant
  6. G(x, y, z) = z2; S that portion of the plane z = x + 1 within the cylinder y = 1 – x2, 0 ≤ y ≤ 1
  7. G(x, y, z) = xy; S that portion of the paraboloid 2z = 4 – x2 – y2 within 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
  8. G(x, y, z) = 2z; S that portion of the paraboloid 2z = 1 + x2 + y2 in the first octant bounded by x = 0, y = , z = 1
  9. G(x, y, z) = 24z; S that portion of the cylinder y = x2 in the first octant bounded by y = 0, y = 4, z = 0, z = 3
  10. G(x, y, z) = (1 + 4y2 + 4z2)1/2; S that portion of the paraboloid x = 4 – y2 – z2 in the first octant outside the cylinder y2 + z2 = 1

In Problems 25 and 26, evaluate (3z2 + 4yz) dS, where S is that portion of the plane x + 2y + 3z = 6 in the first octant. Use the projection of S onto the coordinate plane indicated in the given figure.

  1. A graph. A surface given by a portion of a plane is graphed on a three dimensional x y z coordinate system. The portion of the plane is formed by 3 lines. The first line, graphed on the x y plane, begins at the point (6, 0, 0), and ends at the point (0, 3, 0). The second line begins at the point (6, 0, 0) and ends at the point (0, 0, 2). The third line begins at the point (0, 3, 0) and ends at the point (0, 0, 2). The surface, projected horizontally on the x z plane is a triangular region formed by the second line and the x and z axes. The projected region is shaded and labeled R.

    FIGURE 9.13.12 Region R for Problem 25

  2. A graph. A surface given by a portion of a plane is graphed on a three dimensional x y z coordinate system. The portion of the plane is formed by 3 lines. The first line, graphed on the x y plane, begins at the point (6, 0, 0), and ends at the point (0, 3, 0). The second line begins at the point (6, 0, 0) and ends at the point (0, 0, 2). The third line begins at the point (0, 3, 0) and ends at the point (0, 0, 2). The surface, projected horizontally on the y z plane is a triangular region formed by the third line and the y and z axes. The projected region is shaded and labeled R.

    FIGURE 9.13.13 Region R for Problem 26

In Problems 27 and 28, find the mass of the given surface with the indicated density function.

  1. S that portion of the plane x + y + z = 1 in the first octant; density at a point P directly proportional to the square of the distance from the yz-plane
  2. S the hemisphere z =

In Problems 29–34, let F be a vector field. Find the flux of F through the given surface. Assume the surface S is oriented upward.

  1. F = xi + 2zj + yk; S that portion of the cylinder y2 + z2 = 4 in the first octant bounded by x = 0, x = 3, y = 0, z = 0
  2. F = zk; S that part of the paraboloid z = 5 – x2 – y2 inside the cylinder x2 + y2 = 4
  3. F = xi + yj + zk; same surface S as in Problem 30
  4. F = –x3yi + yz3j + xy3k; S that portion of the plane z = x + 3 in the first octant within the cylinder x2 + y2 = 2x
  5. F = x2i + y2j + zk; S that portion of the paraboloid z = 4 – x2 – y2 for 0 ≤ z ≤ 4
  6. F = eyi + exj + 18yk; S that portion of the plane x + y + z = 6 in the first octant
  7. Find the flux of F = y2i + x2j + 5zk out of the closed surface S given in Figure 9.13.11.
  8. Find the flux of F = –yi + xj + 6z2k out of the closed surface S bounded by the paraboloids z = 4 – x2 – y2 and z = x2 + y2.
  9. Let T(x, y, z) = x2 + y2 + z2 represent temperature and let the “flow” of heat be given by the vector field F = −∇T. Find the flux of heat out of the sphere x2 + y2 + z2 = a2. [Hint: The surface area of a sphere of radius a is 4πa2.]
  10. Find the flux of F = xi + yj + zk out of the unit cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1. See FIGURE 9.13.14. Use the fact that the flux out of the cube is the sum of the fluxes out of the sides.
    A graph. A cube is graphed on a three dimensional x y z coordinate system. The bottom left vertex at the back of the cube is at the origin. The left, back and bottom faces of the cube are on the x y, y z and x y planes respectively. The surface of the cube is labeled S. The right face of the cube is shaded and labeled D. A vector labeled n subscript 1 on the top face, points vertically up. A vector labeled n subscript 4 on the bottom face, points vertically down. A vector labeled n subscript 2 on the right face, points up and to the right. A vector labeled n subscript 5 on the front face, points downward in the opposite direction as the vector n subscript 2. A vector labeled n subscript 3 on the right face, points horizontally to the right. A vector labeled n subscript 6 on the left face, points horizontally to the left.

    FIGURE 9.13.14 Cube in Problem 38

  11. Coulomb’s law states that the electric field E due to a point charge q at the origin is given by E = kqr/r3, where k is a constant and r = xi + yj + zk. Determine the flux out of a sphere x2 + y2 + z2 = a2.
  12. If σ(x, y, z) is charge density in an electrostatic field, then the total charge on a surface S is Find the total charge on that part of the hemisphere z = that is inside the cylinder x2 + y2 = 9 if the charge density at a point P on the surface is directly proportional to distance from the xy-plane.
  13. The coordinates of the centroid of a surface are given by

    where A(S) is the area of the surface. Find the centroid of that portion of the plane 2x + 3y + z = 6 in the first octant.
  14. Use the information in Problem 41 to find the centroid of the hemisphere z = .
  15. Let z = f(x, y) be the equation of a surface S and F be the vector field F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k. Show that (F · n) dS equals

 

* To construct a Möbius strip cut out a long strip of paper, give one end a half-twist, and then attach the ends with tape.