VECTOR CALCULUS 16.7 Surface Integrals In this section, we will learn about: Integration of different types of surfaces. 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. •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 mn * f() Pij S ij ij11 SURFACE INTEGRAL Equation 1 Then, we take the limit as the number of patches increases and define the surface integral of f over the surface S as: mn * f( x , y , z ) dS lim f ( Pij ) S ij mn, S ij11 . Analogues to: The definition of a line integral (Definition 2 in Section 16.2);The definition of a double integral (Definition 5 in Section 15.1) . 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 x y z ruv i j k r i j k u u u v v v are the tangent vectors at a corner of Sij. SURFACE INTEGRALS Formula 2 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: fxyzdS(,,) f ((,))|r uv r r | dA uv SD SURFACE INTEGRALS This should be compared with the formula for a line integral: b fxyzds(,,) f (())|'()|rr t tdt Ca Observe also that: 1dS |rr | dA A ( S ) uv SD SURFACE INTEGRALS Example 1 Compute the surface integral x2 dS , where S is the unit sphere S 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 . we can compute: |rΦ x rθ| = sin Φ SURFACE INTEGRALS Example 1 Therefore, by Formula 2, x2 dS S (sin cos )2 |rr |dA D 2 (sin22 cos sin dd 00 2 cos23dd sin 00 2 1 (1 cos 2 )dd (sin sin cos2 ) 002 2 1 1sin 2 cos 1 cos3 2 20 3 0 4 3 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). CENTER OF MASS Then, the total mass of the sheet is: m (,,) x y z dS S The center of mass is: x,, y z 1 x x(,,) x y z dS m S 1 where y y(,,) x y z dS m S 1 z z(,,) x y z dS m S 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: gg rxy i k r j k xy GRAPHS Equation 3 •Thus, gg r r i j k xy xx 2 2 zz and |rrxy | 1 xy •Formula 2 becomes: f(,,) x y z dS S 2 2 zz f( x , y , g ( x , y )) 1 dA D xy GRAPHS Example 2 Evaluate y dS where S is the surface S z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 z . 1 x and z 2y y GRAPHS Example 2 So, Formula 4 gives: 2 2 zz y dS y1 dA SDxy 12 y1 1 4 y2 dy dx 00 12 dx2 y 1 2 y2 dy 00 2 13 2 12 2 3/ 2 2 43 (1 2y ) 0 3 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 fxyzdS(,,)(,,) fxyzdS SS1 n GRAPHS Example 3 Evaluate z dS , where S is the surfaceS whose: 2 2 . Sides S1 are given by the cylinder x + y = 1. 2 2 . Bottom S2 is the disk x + y ≤ 1 in the plane z = 0. Top S3 is the part of the plane z = 1 + x that lies above S2. 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 GRAPHS Example 3 Therefore, i j k r rz sin cos 0 cos i sin j 0 0 1 and 22 |rr z | cos sin 1 GRAPHS Example 3 Thus, the surface integral over S1 is: z dS z||rr dA z SD1 2 1 cos z dz d 00 2 1 (1 cos )2 d 0 2 2 111 2cos (1 cos 2 ) d 220 3 113 2sin sin 2 2 2 2 4 0 2 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. GRAPHS Example 3 2 2 zz z dS(1 x ) 1 dA xy SD3 21 (1 r cos ) 1 1 0 r dr d 00 21 2 (r r2 cos ) dr d 00 2 2 11 cosd 0 23 2 sin 22 230 GRAPHS Example 3 Therefore, z dS z dS z dS z dS SSSS1 2 3 3 02 2 3 2 2 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. 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: vn dS S (,,)(,,)(,,)xyzvn xyz xyzdS 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 R3. Then, the integral in Equation 7 becomes: Fn dS S 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. FLUX INTEGRAL 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: Fd S F n dS SS . This integral is also called the flux of F across S. 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 (D is the parameters’ domain): rr FSFd uv dS SSrruv rruv F( r (u , v )) ruv r dA rr D uv . So, Fd S F () r r dA uv SD FLUX INTEGRALS Example 4 •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 •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 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, 2 2 rΦ x rθ = sin Φ cos θ i + sin Φ sin θ j + sin Φ cos Φ k 2 Therefore, F(r(Φ, θ)) · (rΦ x rθ) = cos Φ sin Φ cos θ + sin3 Φ sin2 θ + sin2 Φ cos Φ cos θ Then, by Formula 9, the flux is: FSd S F () r r dA D 2 (2sin2 cos cos sin 3 sin 2 ) dd 00 FLUX INTEGRALS Example 4 2 2 sin2 cos dd cos 00 2 sin32dd sin 00 2 0 sin32dd sin 00 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.