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SECTION 13.7 SURFACE INTEGRALS SURFACE INTEGRALS The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length. 13.7 P2 SURFACE INTEGRALS Suppose f is a function of three variables whose domain includes a surface S. We will define the surface integral of f over S such that, in the case where f(x, y, z) = 1, the value of the surface integral is equal to the surface area of S. We start with parametric surfaces. Then, we deal with the special case where S is the graph of a function of two variables. 13.7 P3 PARAMETRIC SURFACES Suppose a surface S has a vector equation r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k (u, v) D We first assume that the parameter domain D is a rectangle and we divide it into subrectangles Rij with dimensions ∆u and ∆v. 13.7 P4 PARAMETRIC SURFACES Then, the surface S is divided into corresponding patches Sij. We evaluate f at a point Pij* in each patch, multiply by the area ∆Sij of the patch, and form the Riemann sum m n * f ( P ij ) Sij i 1 j 1 13.7 P5 SURFACE INTEGRAL Then, we take the limit as the number of patches increases and define the surface integral of f over the surface S as: f ( x, y, z) dS S m lim max ui , v j 0 n f ( P ) S i 1 j 1 * ij ij 13.7 P6 SURFACE INTEGRALS Notice the analogy with: The definition of a line integral (Definition 2 in Section 13.2) The definition of a double integral (Definition 5 in Section 12.1) 13.7 P7 SURFACE INTEGRALS To evaluate the surface integral in Equation 1, we approximate the patch area ∆Sij by the area of an approximating parallelogram in the tangent plane. 13.7 P8 SURFACE INTEGRALS In our discussion of surface area in Section 13.6, we made the approximation ∆Sij ≈ |ru × rv| ∆u ∆v where: x y z x y z ru i j k rv i j k u u u v v v are the tangent vectors at a corner of Sij. 13.7 P9 SURFACE INTEGRALS If the components are continuous and ru and rv are nonzero and nonparallel in the interior of D, it can be shown from Definition 1—even when D is not a rectangle—that: f ( x, y, z) dS f (r(u, v)) | r r | dA u S v D 13.7 P10 SURFACE INTEGRALS This should be compared with the formula for a line integral: C b f ( x, y, z ) ds f (r(t )) | r '(t ) | dt a Observe also that: 1dS | r r u S v | dA A( S ) D 13.7 P11 SURFACE INTEGRALS Formula 2 allows us to compute a surface integral by converting it into a double integral over the parameter domain D. When using this formula, remember that f(r(u, v) is evaluated by writing x = x(u, v), y = y(u, v), z = z(u, v) in the formula for f(x, y, z) 13.7 P12 Example 1 Compute the surface integral x 2 dS , S where S is the unit sphere x2 + y2 + z2 = 1. 13.7 P13 Example 1 SOLUTION As in Example 4 in Section 13.6, we use the parametric representation x = sin f cos , y = sin f sin , z = cos f 0 ≤ f ≤ p, 0 ≤ ≤ 2 p That is, r(f, ) = sin f cos i + sin f sin j + cos f k 13.7 P14 Example 1 SOLUTION As in Example 9 in Section 13.6, we can compute that: |rf × r| = sin f 13.7 P15 Example 1 SOLUTION Therefore, by Formula 2, 2 2 x dS (sin f cos ) | rf r | dA S D 2p 0 p 0 2p sin 2 f cos 2 sin f df d p cos d sin f df 2 0 2p 0 1 2 3 0 1 2 p (1 cos 2 ) d (sin f sin f cos f ) df 2 0 1 2 sin 2 0 2p 4p cos f cos f 0 3 1 3 3 p 13.7 P16 APPLICATIONS Surface integrals have applications similar to those for the integrals we have previously considered. For example, suppose a thin sheet (say, of aluminum foil) has: The shape of a surface S. The density (mass per unit area) at the point (x, y, z) as r(x, y, z). 13.7 P17 MASS Then, the total mass of the sheet is: m r ( x, y, z ) dS S 13.7 P18 CENTER OF MASS The center of mass is x , y , z , where 1 x x r ( x, y, z ) dS m S 1 y y r ( x, y, z ) dS m S 1 z z r ( x, y, z ) dS m S Moments of inertia can also be defined as before. See Exercise 35. 13.7 P19 GRAPHS Any surface S with equation z = g(x, y) can be regarded as a parametric surface with parametric equations x=x y=y z = g(x, y) So, we have: g rx i k x g ry j k y 13.7 P20 GRAPHS Thus, g g rx ry i jk x x and 2 z z | rx ry | 1 x y 2 13.7 P21 GRAPHS Therefore, in this case, Formula 2 becomes: f ( x, y, z) dS S D 2 z z f ( x, y, g ( x, y )) 1 dA x y 2 13.7 P22 GRAPHS Similar formulas apply when it is more convenient to project S onto the yz-plane or xyplane. 13.7 P23 GRAPHS For instance, if S is a surface with equation y = h(x, z) and D is its projection on the xz-plane, then f ( x, y, z ) dS S D y y f ( x, h( x, z ), z ) 1 dA x z 2 2 13.7 P24 Example 2 Evaluate y dS where S is the surface S z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 See Figure 2. SOLUTION z z 1 and 2y x y 13.7 P25 Example 2 SOLUTION So, Formula 4 gives: 2 z z S y dS D y 1 x y dA 2 1 2 0 0 y 1 1 4 y 2 dy dx 1 2 0 0 dx 2 y 1 2 y 2 dy 2 1 4 2 3 (1 2 y ) 2 3/ 2 13 2 0 3 2 13.7 P26 GRAPHS If S is a piecewise-smooth surface—a finite union of smooth surfaces S1, S2, …, Sn that intersect only along their boundaries—then the surface integral of f over S is defined by: f ( x, y, z) dS S f ( x, y, z ) dS S1 f ( x, y, z ) dS Sn 13.7 P27 Example 3 Evaluate z dS , where S is the surface whose: S Sides S1 are given by the cylinder x2 + y2 = 1. Bottom S2 is the disk x2 + y2 ≤ 1 in the plane z = 0. Top S3 is the part of the plane z = 1 + x that lies above S2. 13.7 P28 Example 3 SOLUTION The surface S is shown in Figure 3. We have changed the usual position of the axes to get a better look at S. 13.7 P29 Example 3 SOLUTION For S1, we use and z as parameters (Example 5 in Section 13.6) and write its parametric equations as: x = cos y = sin z=z where: 0 ≤ ≤ 2p 0 ≤ z ≤ 1 + x = 1 + cos 13.7 P30 Example 3 SOLUTION Therefore, i r rz sin 0 j cos 0 k 0 cos i sin j 1 and | r rz | cos 2 sin 2 1 13.7 P31 Example 3 SOLUTION Thus, the surface integral over S1 is: z dS z | r r z S1 | dA D 2p 2p 0 0 1 2 1 cos 0 1 2 z dz d (1 cos ) 2 d 2p 1 1 2 cos 2 (1 cos 2 ) d 0 3p 2sin sin 2 2 1 2 3 2 1 4 2p 0 13.7 P32 Example 3 SOLUTION Since S2 lies in the plane z = 0, we have: z dS 0 dS 0 S2 S2 S3 lies above the unit disk D and is part of the plane z = 1 + x. So, taking g(x, y) = 1 + x in Formula 4 and converting to polar coordinates, we have the following result. 13.7 P33 Example 3 SOLUTION 2 z z S z dS D (1 x) 1 x y dA 3 2 2p 0 1 (1 r cos ) 0 2 2p 2 2p 0 0 1 (r r 0 2 1 1 0 r dr d cos ) dr d 12 13 cos d 2p sin 2 2p 3 0 2 13.7 P34 Example 3 SOLUTION Therefore, z dS z dS z dS z dS S S1 S2 S3 3p 0 2p 2 3 2 2 p 13.7 P35 ORIENTED SURFACES To define surface integrals of vector fields, we need to rule out nonorientable surfaces such as the Möbius strip shown in Figure 4. It is named after the German geometer August Möbius (1790–1868). 13.7 P36 MOBIUS STRIP You can construct one for yourself by: 1. Taking a long rectangular strip of paper. 2. Giving it a half-twist. 3. Taping the short edges together. 13.7 P37 MOBIUS STRIP If an ant were to crawl along the Möbius strip starting at a point P, it would end up on the “other side” of the strip—that is, with its upper side pointing in the opposite direction. 13.7 P38 MOBIUS STRIP Then, if it continued to crawl in the same direction, it would end up back at the same point P without ever having crossed an edge. If you have constructed a Möbius strip, try drawing a pencil line down the middle. 13.7 P39 MOBIUS STRIP Therefore, a Möbius strip really has only one side. You can graph the Möbius strip using the parametric equations in Exercise 28 in Section 13.6. 13.7 P40 ORIENTED SURFACES From now on, we consider only orientable (twosided) surfaces. 13.7 P41 ORIENTED SURFACES We start with a surface S that has a tangent plane at every point (x, y, z) on S (except at any boundary point). There are two unit normal vectors n1 and n2 = –n1 at (x, y, z). See Figure 6. 13.7 P42 ORIENTED SURFACE & ORIENTATION If it is possible to choose a unit normal vector n at every such point (x, y, z) so that n varies continuously over S, then S is called an oriented surface. The given choice of n provides S with an orientation. 13.7 P43 POSSIBLE ORIENTATIONS There are two possible orientations for any orientable surface. 13.7 P44 UPWARD ORIENTATION For a surface z = g(x, y) given as the graph of g, we use Equation 3 to associate with the surface a natural orientation given by the unit normal vector g g n x i y g g 1 x y 2 jk 2 As the k-component is positive, this gives the upward orientation of the surface. 13.7 P45 ORIENTATION If S is a smooth orientable surface given in parametric form by a vector function r(u, v), then it is automatically supplied with the orientation of the unit normal vector ru rv n | ru rv | The opposite orientation is given by –n. 13.7 P46 ORIENTATION For instance, in Example 4 in Section 13.6, we found the parametric representation r(f, ) = a sin f cos i + a sin f sin j + a cos f k for the sphere x2 + y2 + z2 = a2 13.7 P47 ORIENTATION Then, in Example 9 in Section 13.6, we found that: rf × r = a2 sin2 f cos i + a2 sin2 f sin j + a2 sin f cos f k and |rf × r | = a2 sin f 13.7 P48 ORIENTATION So, the orientation induced by r(f, ) is defined by the unit normal vector n rf r | rf r | sin f cos i sin f sin j cos f k 1 r (f , ) a 13.7 P49 POSITIVE ORIENTATION Observe that n points in the same direction as the position vector—that is, outward from the sphere. See Figure 8. 13.7 P50 Fig. 17.7.8, p. 1122 NEGATIVE ORIENTATION The opposite (inward) orientation would have been obtained (see Figure 9) if we had reversed the order of the parameters because r × rf = –rf × r 13.7 P51 CLOSED SURFACES For a closed surface—a surface that is the boundary of a solid region E—the convention is that: The positive orientation is the one for which the normal vectors point outward from E. Inward-pointing normals give the negative orientation. 13.7 P52 SURFACE INTEGRALS OF VECTOR FIELDS Suppose that S is an oriented surface with unit normal vector n. Then, imagine a fluid with density r(x, y, z) and velocity field v(x, y, z) flowing through S. Think of S as an imaginary surface that doesn’t impede the fluid flow—like a fishing net across a stream. 13.7 P53 SURFACE INTEGRALS OF VECTOR FIELDS Then, the rate of flow (mass per unit time) per unit area is r v. If we divide S into small patches Sij , as in Figure 10 (compare with Figure 1), then Sij is nearly planar. 13.7 P54 SURFACE INTEGRALS OF VECTOR FIELDS So, we can approximate the mass of fluid crossing Sij in the direction of the normal n per unit time by the quantity (r v · n)A(Sij) where r, v, and n are evaluated at some point on Sij. Recall that the component of the vector r v in the direction of the unit vector n is r v · n. 13.7 P55 VECTOR FIELDS Summing these quantities and taking the limit, we get, according to Definition 1, the surface integral of the function r v · n over S: r v n dS S r ( x, y, z ) v( x, y, z ) n( x, y, z) dS S This is interpreted physically as the rate of flow through S. 13.7 P56 VECTOR FIELDS If we write F = r v, then F is also a vector field on 3. Then, the integral in Equation 7 becomes: F n dS S 13.7 P57 FLUX INTEGRAL A surface integral of this form occurs frequently in physics—even when F is not r v. It is called the surface integral (or flux integral) of F over S. 13.7 P58 Definition 8 If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is F dS F n dS S S This integral is also called the flux of F across S. 13.7 P59 FLUX INTEGRAL In words, Definition 8 says that: The surface integral of a vector field over S is equal to the surface integral of its normal component over S (as previously defined). 13.7 P60 FLUX INTEGRAL If S is given by a vector function r(u, v), then n is given by Equation 6. Then, from Definition 8 and Equation 2, we have: ru rv S F dS S F ru rv dS ru rv F(r (u, v)) ru rv dA ru rv D where D is the parameter domain. 13.7 P61 FLUX INTEGRAL Thus, we have: F dS F (r r ) dA u S v D 13.7 P62 Example 4 Find the flux of the vector field F(x, y, z) = z i + y j + x k across the unit sphere x 2 + y 2 + z2 = 1 13.7 P63 Example 4 SOLUTION Using the parametric representation r(f, ) = sin f cos i + sin f sin j + cos f k 0≤f≤p 0 ≤ ≤ 2p we have: F(r(f, )) = cos f i + sin f sin j + sin f cos k 13.7 P64 Example 4 SOLUTION From Example 9 in Section 13.6, r f × r = sin2 f cos i + sin2 f sin j + sin f cos f k Therefore, F(r(f, )) · (rf × r ) = cos f sin2 f cos + sin3 f sin2 + sin2 f cos f cos 13.7 P65 Example 4 SOLUTION Then, by Formula 9, the flux is: F dS S F (rf r ) dA D 2p 0 p 0 (2sin 2 f cos f cos sin 3 f sin 2 ) df d 13.7 P66 Example 4 SOLUTION p 2p 2 sin f cos f df cos d 2 0 0 p 2p sin f df sin 2 d 3 0 p 0 2p 0 sin f df sin 2 d 0 3 0 4p 3 This is by the same calculation as in Example 1. 13.7 P67 FLUX INTEGRALS Figure 11 shows the vector field F in Example 4 at points on the unit sphere. 13.7 P68 VECTOR FIELDS If, for instance, the vector field in Example 4 is a velocity field describing the flow of a fluid with density 1, then the answer, 4p/3, represents: The rate of flow through the unit sphere in units of mass per unit time. 13.7 P69 VECTOR FIELDS In the case of a surface S given by a graph z = g(x, y), we can think of x and y as parameters and use Equation 3 to write: g g F (rx ry ) ( P i Q j R k ) i jk y x 13.7 P70 VECTOR FIELDS Thus, Formula 9 becomes: g g S F dS D P x Q y R dA This formula assumes the upward orientation of S. For a downward orientation, we multiply by –1. Similar formulas can be worked out if S is given by y = h(x, z) or x = k(y, z). See Exercises 31 and 32. 13.7 P71 Example 5 Evaluate where: F dS S F(x, y, z) = y i + x j + z k S is the boundary of the solid region E enclosed by the paraboloid z = 1 – x2 – y2 and the plane z = 0. 13.7 P72 Example 5 SOLUTION S consists of: A parabolic top surface S1. A circular bottom surface S2. See Figure 12. 13.7 P73 Example 5 SOLUTION Since S is a closed surface, we use the convention of positive (outward) orientation. This means that S1 is oriented upward. So, we can use Equation 10 with D being the projection of S1 on the xy-plane, namely, the disk x2 + y2 ≤ 1. 13.7 P74 Example 5 SOLUTION On S1, P(x, y, z) = y Q(x, y, z) = x R(x, y, z) = z = 1 – x2 – y2 Also, g 2x x g 2 y y 13.7 P75 Example 5 SOLUTION So, we have: F dS S1 g g P Q R dA x y D [ y (2 x) x(2 y ) 1 x 2 y 2 ] dA D (1 4 xy x y ) dA 2 2 D 13.7 P76 Example 5 SOLUTION 2p 1 0 0 2 1 0 0 p (r r 4r cos sin ) dr d p ( cos sin ) d 2 0 (1 4r 2 cos sin r 2 ) r dr d 3 3 1 4 14 (2p ) 0 p 2 13.7 P77 Example 5 SOLUTION The disk S2 is oriented downward. So, its unit normal vector is n = –k and we have: F dS F (k ) dS ( z) dA 0 dA 0 S2 S2 D D since z = 0 on S2. 13.7 P78 Example 5 SOLUTION Finally, we compute, by definition, F dS as the sum of the surface integrals S of F over the pieces S1 and S2: p p F dS F dS F dS 2 0 2 S S1 S2 13.7 P79 APPLICATIONS Although we motivated the surface integral of a vector field using the example of fluid flow, this concept also arises in other physical situations. 13.7 P80 ELECTRIC FLUX For instance, if E is an electric field (Example 5 in Section 13.1), the surface integral E dS S is called the electric flux of E through the surface S. 13.7 P81 GAUSS’S LAW One of the important laws of electrostatics is Gauss’s Law, which says that the net charge enclosed by a closed surface S is: Q e 0 E dS S where e0 is a constant (called the permittivity of free space) that depends on the units used. In the SI system, e0 ≈ 8.8542 × 10–12 C2/N · m2 13.7 P82 GAUSS’S LAW Thus, if the vector field F in Example 4 represents an electric field, we can conclude that the charge enclosed by S is: Q = 4pe0/3 13.7 P83 HEAT FLOW Another application occurs in the study of heat flow. Suppose the temperature at a point (x, y, z) in a body is u(x, y, z). Then, the heat flow is defined as the vector field F = –K ∇u where K is an experimentally determined constant called the conductivity of the substance. 13.7 P84 HEAT FLOW Then, the rate of heat flow across the surface S in the body is given by the surface integral F dS K u dS S S 13.7 P85 Example 6 The temperature u in a metal ball is proportional to the square of the distance from the center of the ball. Find the rate of heat flow across a sphere S of radius a with center at the center of the ball. 13.7 P86 Example 6 SOLUTION Taking the center of the ball to be at the origin, we have: u(x, y, z) = C(x2 + y2 + z2) where C is the proportionality constant. Then, the heat flow is F(x, y, z) = –K ∇u = –KC(2x i + 2y j + 2z k) where K is the conductivity of the metal. 13.7 P87 Example 6 SOLUTION Instead of using the usual parametrization of the sphere as in Example 4, we observe that the outward unit normal to the spherex2 + y2 + z2 = a2 at the point (x, y, z) is: n = 1/a (x i + y j + z k) Thus, 2 KC 2 2 2 F n (x y z ) a 13.7 P88 Example 6 SOLUTION However, on S, we have: x 2 + y 2 + z2 = a 2 Thus, F · n = –2aKC 13.7 P89 Example 6 SOLUTION Thus, the rate of heat flow across S is: F dS= F n dS = 2aKC dS S S S = 2aKCA( S )= 2aKC (4p a 2 ) = 8KCp a 3 13.7 P90