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16 VECTOR CALCULUS VECTOR CALCULUS 16.7 Surface Integrals In this section, we will learn about: Integration of different types of surfaces. SURFACE INTEGRALS The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length. 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. SURFACE INTEGRALS We start with parametric surfaces. Then, we deal with the special case where S is the graph of a function of two variables. 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 PARAMETRIC SURFACES We first assume that the parameter domain D is a rectangle and we divide it into subrectangles Rij with dimensions ∆u and ∆v. PARAMETRIC SURFACES Then, the surface S is divided into corresponding patches Sij. PARAMETRIC SURFACES 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 ) S i 1 j 1 * ij ij Equation 1 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 m, n n f (P ) S i 1 j 1 * ij ij SURFACE INTEGRALS Notice the analogy with: The definition of a line integral (Definition 2 in Section 16.2) The definition of a double integral (Definition 5 in Section 15.1) 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. SURFACE INTEGRALS In our discussion of surface area in Section 16.6, we made the approximation ∆Sij ≈ |ru x rv| ∆u ∆v where: x y z ru i j k u u u x y z rv i j k v v v are the tangent vectors at a corner of Sij. Formula 2 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 u S D rv | dA 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: 1 dS | r u S D rv | dA A( S ) 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) SURFACE INTEGRALS Example 1 Compute the surface integral x dS , S where S is the unit sphere 2 x2 + y2 + z2 = 1. SURFACE INTEGRALS Example 1 As in Example 4 in Section 16.6, we use the parametric representation x = sin Φ cos θ, y = sin Φ sin θ, z = cos Φ 0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π That is, r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k SURFACE INTEGRALS Example 1 As in Example 10 in Section 16.6, we can compute that: |rΦ x rθ| = sin Φ Example 1 SURFACE INTEGRALS Therefore, by Formula 2, x dS 2 S (sin cos ) | r r | dA 2 D 2 0 0 (sin cos sin d d 2 2 Example 1 SURFACE INTEGRALS 2 cos d sin d 2 0 2 0 1 2 0 1 2 (1 cos 2 ) d (sin sin cos ) d 4 3 3 2 0 1 2 sin 2 0 cos cos 0 2 1 3 3 APPLICATIONS Surface integrals have applications similar to those for the integrals we have previously considered. APPLICATIONS 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 ρ(x, y, z). MASS Then, the total mass of the sheet is: m ( x, y , z ) dS S CENTER OF MASS The center of mass is: x, y , z where 1 x x ( x, y , z ) dS m S 1 y y ( x, y , z ) dS m S 1 z z ( x, y , z ) dS m S MOMENTS OF INERTIA Moments of inertia can also be defined as before. See Exercise 39. 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 Equation 3 GRAPHS Thus, g g rx ry i jk x x and 2 z z | rx ry | 1 x y 2 Formula 4 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 GRAPHS Similar formulas apply when it is more convenient to project S onto the yz-plane or xy-plane. 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 Example 2 GRAPHS Evaluate y dS where S is the surface S z 1 x and z 2y y z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 Example 2 GRAPHS 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 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 f ( x, y, z ) dS S1 Sn GRAPHS Evaluate Example 3 z dS , where S is S the surface whose: 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. GRAPHS Example 3 The surface S is shown. We have changed the usual position of the axes to get a better look at S. GRAPHS Example 3 For S1, we use θ and z as parameters (Example 5 in Section 16.6) and write its parametric equations as: x = cos θ y = sin θ z=z where: 0 ≤ θ ≤ 2π 0 ≤ z ≤ 1 + x = 1 + cos θ Example 3 GRAPHS Therefore, i r rz sin 0 j cos 0 k 0 cos i sin j 1 and | r rz | cos sin 1 2 2 Example 3 GRAPHS Thus, the surface integral over S1 is: z dS z | r r z S1 | dA D 2 2 0 0 1 2 0 1 2 1 cos (1 cos ) d 2 0 z dz d 2 1 1 2 cos 2 (1 cos 2 ) d 3 2sin sin 2 2 1 2 3 2 1 4 2 0 GRAPHS Example 3 Since S2 lies in the plane z = 0, we have: z dS S2 0 dS S2 0 GRAPHS Example 3 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. Example 3 GRAPHS 2 z z S z dS D (1 x) 1 x y dA 3 2 2 1 0 0 (1 r cos ) 1 1 0 r dr d 2 (r r cos ) dr d 2 1 0 0 2 2 0 1 2 2 13 cos d 2 sin 2 2 2 3 0 Example 3 GRAPHS Therefore, z dS z dS z dS z dS S2 S1 S 3 0 2 2 3 2 2 S3 ORIENTED SURFACES To define surface integrals of vector fields, we need to rule out nonorientable surfaces such as the Möbius strip shown. It is named after the German geometer August Möbius (1790– 1868). 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. 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. 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. MOBIUS STRIP Therefore, a Möbius strip really has only one side. You can graph the Möbius strip using the parametric equations in Exercise 32 in Section 16.6. ORIENTED SURFACES From now on, we consider only orientable (two-sided) surfaces. 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). 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. POSSIBLE ORIENTATIONS There are two possible orientations for any orientable surface. UPWARD ORIENTATION Equation 5 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 g g the unit normal vector i jk x y n 2 2 g g 1 x y As the k-component is positive, this gives the upward orientation of the surface. Equation 6 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. ORIENTATION For instance, in Example 4 in Section 16.6, we found the parametric representation r(Φ, θ) = a sin Φ cos θ i + a sin Φ sin θ j + a cos Φ k for the sphere x2 + y2 + z2 = a2 ORIENTATION Then, in Example 10 in Section 16.6, we found that: rΦ x rθ = a2 sin2 Φ cos θ i + a2 sin2 Φ sin θ j + a2 sin Φ cos Φ k and |rΦ x rθ| = a2 sin Φ ORIENTATION So, the orientation induced by r(Φ, θ) is defined by the unit normal vector n r r | r r | sin cos i sin sin j cos k 1 r ( , ) a POSITIVE ORIENTATION Observe that n points in the same direction as the position vector—that is, outward from the sphere. NEGATIVE ORIENTATION The opposite (inward) orientation would have been obtained if we had reversed the order of the parameters because rθ x rΦ = –rΦ x rθ 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. SURFACE INTEGRALS OF VECTOR FIELDS Suppose that S is an oriented surface with unit normal vector n. Then, imagine a fluid with density ρ(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. SURFACE INTEGRALS OF VECTOR FIELDS Then, the rate of flow (mass per unit time) per unit area is ρv. SURFACE INTEGRALS OF VECTOR FIELDS If we divide S into small patches Sij , then Sij is nearly planar. 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 (ρv · n)A(Sij) where ρ, v, and n are evaluated at some point on Sij. Recall that the component of the vector ρv in the direction of the unit vector n is ρv · n. VECTOR FIELDS Equation 7 Summing these quantities and taking the limit, we get, according to Definition 1, the surface integral of the function ρv · n over S: v n dS S ( x, y, z ) v( x, y, z ) n( x, y, z ) dS S This is interpreted physically as the rate of flow through S. VECTOR FIELDS If we write F = ρv, then F is also a vector field on ° 3. Then, the integral in Equation 7 becomes: F n dS S FLUX INTEGRAL A surface integral of this form occurs frequently in physics—even when F is not ρv. It is called the surface integral (or flux integral) of F over S. Definition 8 FLUX INTEGRAL 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. 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). 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. Formula 9 FLUX INTEGRAL Thus, we have: F d S F ( r r ) dA u v S D Example 4 FLUX INTEGRALS Find the flux of the vector field F(x, y, z) = z i + y j + x k across the unit sphere x2 + y2 + z2 = 1 FLUX INTEGRALS Example 4 Using the parametric representation r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k 0≤Φ≤π 0 ≤ θ ≤ 2π we have: F(r(Φ, θ)) = cos Φ i + sin Φ sin θ j + sin Φ cos θ k FLUX INTEGRALS Example 4 From Example 10 in Section 16.6, r Φ x rθ = sin2 Φ cos θ i + sin2 Φ sin θ j + sin Φ cos Φ k FLUX INTEGRALS Example 4 Therefore, F(r(Φ, θ)) · (rΦ x rθ) = cos Φ sin2 Φ cos θ + sin3 Φ sin2 θ + sin2 Φ cos Φ cos θ FLUX INTEGRALS Example 4 Then, by Formula 9, the flux is: F dS S F (r r ) dA D 2 0 0 (2sin cos cos sin sin ) d d 2 3 2 Example 4 FLUX INTEGRALS 2 2 sin cos d cos d 2 0 0 2 sin d sin d 3 0 2 2 0 0 sin d sin d 0 3 2 0 4 3 This is by the same calculation as in Example 1. FLUX INTEGRALS The figure shows the vector field F in Example 4 at points on the unit sphere. 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, 4π/3, represents: The rate of flow through the unit sphere in units of mass per unit time. 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 VECTOR FIELDS Formula 10 Thus, Formula 9 becomes: g g F d S P Q R dA S D x y This formula assumes the upward orientation of S. For a downward orientation, we multiply by –1. VECTOR FIELDS Similar formulas can be worked out if S is given by y = h(x, z) or x = k(y, z). See Exercises 35 and 36. VECTOR FIELDS Evaluate Example 5 F d S S where: 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. VECTOR FIELDS S consists of: A parabolic top surface S1. A circular bottom surface S2. Example 5 VECTOR FIELDS Example 5 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. Example 5 VECTOR FIELDS 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 Example 5 VECTOR FIELDS So, we have: F dS S1 g g P Q R dA x y D [ y (2 x) x(2 y ) 1 x y ] dA 2 D (1 4 xy x y ) dA 2 D 2 2 Example 5 VECTOR FIELDS 2 1 0 0 2 1 0 0 (1 4r cos sin r ) r dr d (r r 4r cos sin ) dr d ( cos sin ) d 2 0 1 4 14 (2 ) 0 2 2 3 2 3 Example 5 VECTOR FIELDS The disk S2 is oriented downward. So, its unit normal vector is n = –k and we have: F dS F (k ) dS ( z) dA S2 S2 D 0 dA 0 D since z = 0 on S2. Example 5 VECTOR FIELDS Finally, we compute, by definition, F dS as the sum of the surface integrals S of F over the pieces S1 and S2: F dS F dS F dS S S1 S2 2 0 2 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. ELECTRIC FLUX For instance, if E is an electric field (Example 5 in Section 16.1), the surface integral E d S S is called the electric flux of E through the surface S. GAUSS’S LAW Equation 11 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 0 E dS S where ε0 is a constant (called the permittivity of free space) that depends on the units used. In the SI system, ε0 ≈ 8.8542 x 10–12 C2/N · m2 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 = 4πε0/3 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). HEAT FLOW 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. HEAT FLOW Then, the rate of heat flow across the surface S in the body is given by the surface integral F d S K u d S S S HEAT FLOW 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. Example 6 HEAT FLOW 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. Example 6 HEAT FLOW 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. HEAT FLOW Example 6 Instead of using the usual parametrization of the sphere as in Example 4, we observe that the outward unit normal to the sphere x2 + 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 Example 6 HEAT FLOW However, on S, we have: x2 + y2 + z2 = a2 Thus, F · n = –2aKC Example 6 HEAT FLOW Thus, the rate of heat flow across S is: F dS F n dS 2aKC dS S S S 2aKCA( S ) 2aKC (4 a ) 2 8KC a 3