Surface Integrals Math 240 Scalar integrals Surface area Vector integrals Changing Surface Integrals orientation Math 240 | Calculus III Summer 2013, Session II Wednesday, July 3, 2013 Surface Integrals Agenda Math 240 Scalar integrals Surface area Vector integrals Changing 1. Scalar surface integrals orientation Surface area 2. Vector surface integrals 3. Changing orientation P1: OSO coll50424˙ch07 PEAR591-Colley July 29, 2011 13:58 464 Chapter 7 Surface Integrals and Vector Analysis 3 the cube’s faces by Xi :[0, 1] × [0, 1] → R , i = 1,...,6, where X1(s, t) = (0, s, t); X2(s, t) = (1, s, t); X3(s, t) = (s, 0, t); X4(s, t) = (s, 1, t); X5(s, t) = (s, t, 0); X6(s, t) = (s, t, 1). 1 Each map Xi is clearly of class C and one-one. In addition, the faces have well-defined nonzero normal vectors. For example, for both X1 and X2, N1 = N2 = Ts × Tt = j × k = i. Similarly, N3 = N4 = i × k =−j and N5 = N6 = i × j = k. None of these vectors vanishes. There is no consistent way to define normal vectors along the edges of the cube (where two faces meet). That is why the cube is only piecewise smooth. ◆ Area of a Smooth Parametrized Surface Now, we use the notion of a parametrized surface to calculate the surface area of a = Surface smooth surface. In the discussion that follows, we take S X(D) to be a smooth Integrals parametrized surface, where D is the union ofScalar finitely many surface elementary regions integrals in R2 and X: D → R3 is of class C1 and one-one except possibly along ∂ D. Math 240 t z Scalar integrals S = X(D) Surface area Δt X Vector Δs X s t integrals ( 0, 0) (s , t ) T Changing 0 0 t T orientation s s y x DefinitionFigure 7.13 The image of the s × t rectangle in D is approximately a parallelogram spanned by Ts (s0, t0)s and Tt (s0, t0)t. 2 3 Let X : DThe keyR geometricR observationbe a smooth is as follows: parameterized Consider a small rectangular surface. Let ⊆ ! , ∈ f be asubset continuous of D whose lower scalar left corner function is at the whose point (s0 domaint0) D and includes whose width and height are s and t, respectively. The image of this rectangle under X is a S = Xpiece(D of). the The underlyingscalar surface surfaceS that is approximately integral of thef parallelogramalong X withis a corner at X(s0, t0) and spanned by the vectors Ts(s0, t0)s and Tt (s0, t0)t. (See ZZ ZZ Figure 7.13.) Thef dS area= A of thisf piece(X( iss; t)) T T ds dt ≈ , × , = , s × t, . A X Ts(s0 t0) s TDt (s0 t0) t Ts(ks0 t0) ×Tt (s0 kt0) s t Now, suppose D = [aZZ, b] × [c, d]; that is, suppose D itself is a rectangle. Partition D into n2 subrectangles= f via(X(s; t)) N(s; t) ds dt: D k k a = s0 < s1 < ···< sn = b and c = t0 < t1 < ···< tn = d. Let si = si − si−1 and t j = t j − t j−1 for i, j = 1,...,n. Then S is in turn partitioned into pieces, each of which is approximately a parallelogram, assuming si and t j are small for i, j = 1,...,n.IfAij denotes the area of the piece P1: OSO coll50424˙ch07 PEAR591-Colley July 29, 2011 13:58 Surface Integrals Scalar surface integrals Math 240 472 Chapter 7 Surface Integrals and Vector Analysis Scalar z To define and evaluate scalar surface integrals over piecewise smooth para- Example S : z = 15 integrals 3 metrized surfaces, simply calculate the surface integral over each smooth piece Surface area Let S be the closed cylinder of radius 3 with and add the results. Vector EXAMPLE 2 Let S be the closed cylinder of radius 3 with axis along the z-axis, integrals axis along the z-axis, top face at z = 15, and top face at z = 15, and bottom face at z = 0, as shown in Figure 7.15. Then S S x2 + y2 = 1: 9 is a piecewise smooth surface; it is the union of the three smooth parametrized Changing bottom face at z = 0. Let's calculate RR z dS. orientation S surfaces S1, S2, and S3 described next. We calculate S zdS. The three smooth pieces may be parametrized as follows: Denote the lateral cylindrical face of S by S1 ⎧ y ⎨x = 3 cos s S : z = 0 = ≤ ≤ π ≤ ≤ and the bottom and top faces by S2 and S3, x 2 S1 (lateral cylindrical surface): ⎩y 3 sin s 0 s 2 ,0 t 15, z = t respectively. Figure 7.15 The closed ⎧ cylinder of radius 3 and ⎨x = s cos t height 15 of Example 2. = ≤ ≤ ≤ ≤ π We compute S2 (bottom disk): ⎩y s sin t 0 s 3, 0 t 2 , z = 0 ZZ ZZ ZZ and z dS = 675π; z dS = 0; and z dS = 135π: ⎧ S S S ⎨x = s cos t 1 2 3 = ≤ ≤ ≤ ≤ π S3 (top disk): ⎩y s sin t 0 s 3, 0 t 2 . Therefore, z = 15 Using Definition 2.1, we have ZZ ZZ ZZ ZZ 15 2π z dS = z dS + z dS + z dS = 810π: zdS= t (−3 sin s i + 3 cos s j) × k ds dt S1 0 0 S S1 S2 S3 15 2π = t 3 sin s j + 3 cos s i ds dt 0 0 15 2π 15 = = π = π 215 = π. 3tdsdt 6 tdt 3 t 0 675 0 0 0 Now, zdS= 0, since z vanishes along the bottom of S.ForS3,wehave S2 zdS= 15 dS = 15 · 1 dS S3 S3 S3 = 15 · area of disk = 15 · (9π) = 135π. Therefore, zdS= zdS+ zdS+ zdS S S1 S2 S3 ◆ = 675π + 0 + 135π = 810π. If a surface S is given by the graph of z = g(x, y), where g is of class C1 on some region D in R2, then S is parametrized by X(x, y) = (x, y, g(x, y)) with (x, y) ∈ D. (See Example 4 of 7.1.) Then, from Example 13 in 7.1, § § N(x, y) =−gx i − gy j + k, so that = , , , 2 + 2 + . fdS f (x y g(x y)) gx gy 1 dx dy (4) X D P1: OSO coll50424˙ch07 PEAR591-Colley July 29, 2011 13:58 464 Chapter 7 Surface Integrals and Vector Analysis 3 the cube’s faces by Xi :[0, 1] × [0, 1] → R , i = 1,...,6, where X1(s, t) = (0, s, t); X2(s, t) = (1, s, t); X3(s, t) = (s, 0, t); X4(s, t) = (s, 1, t); X5(s, t) = (s, t, 0); X6(s, t) = (s, t, 1). 1 Each map Xi is clearly of class C and one-one. In addition, the faces have well-defined nonzero normal vectors. For example, for both X1 and X2, N1 = N2 = Ts × Tt = j × k = i. Similarly, N3 = N4 = i × k =−j and N5 = N6 = i × j = k. None of these vectors vanishes. There is no consistent way to define normal vectors along the edges of the cube (where two faces meet). That is why the cube is only piecewise smooth. ◆ Area of a Smooth Parametrized Surface Now, we use the notion of a parametrized surface to calculate the surface area of a = Surface smooth surface. In the discussion that follows, we take S X(D) to be a smooth Integrals parametrized surface, where D is the union of finitely many elementarySurface regions area in R2 and X: D → R3 is of class C1 and one-one except possibly along ∂ D. Math 240 t z Scalar integrals S = X(D) Surface area Δt X Vector Δs integrals X s t Figure:( 0, 0)The quantity (s , t ) T Changing 0 0 t Ts Tt is the orientation Ts s areak × of thek gray square on they right. x Figure 7.13 The image of the s × t rectangle in D is approximately a parallelogram spanned by Ts (s0, t0)s and Tt (s0, t0)t. Fact The key geometric observation is as follows: Consider a small rectangular , ∈ 2 3 If S issubset a smooth of D whose surface lower left parameterizedcorner is at the point ( bys0 t0X) :DDand whoseR widthR and height are s and t, respectively. The image of this rectangle⊆ under X!is a then thepiece surface of the underlying area surface of S Sisthat given is approximately by the parallelogram with a corner at X(s0, t0) and spanned by the vectors Ts(s0, t0)s and Tt (s0, t0)t. (See FigureZZ 7.13.) The area A ofZZ this piece is ZZ ≈ N ds, dt=× , Ts = Tt ds, dt× = , 1dS:. DAk kTs(s0 t0) s Tt (sD0 kt0) t × Ts(ks0 t0) Tt (s0 t0) Xs t Now, suppose D = [a, b] × [c, d]; that is, suppose D itself is a rectangle. Partition D into n2 subrectangles via a = s0 < s1 < ···< sn = b and c = t0 < t1 < ···< tn = d. Let si = si − si−1 and t j = t j − t j−1 for i, j = 1,...,n. Then S is in turn partitioned into pieces, each of which is approximately a parallelogram, assuming si and t j are small for i, j = 1,...,n.IfAij denotes the area of the piece Surface Integrals Surface area Math 240 Scalar integrals Example Surface area Recall our parameterization of a sphere: Vector integrals X(s; t) = r(cos s)(sin t) i + r(sin s)(sin t) j + r(cos t) k: Changing orientation We calculate Ts = r sin s sin t i + r cos s sin t j; − Tt = r cos s cos t i + r sin s cos t j r sin t k; − N = r2 cos s sin2 t i r2 sin s sin2 t j r2 sin t cos t k; − − − and N = r2 sin t: k k Therefore, the surface area of the sphere is Z π Z 2π Z π r2 sin t ds dt = 2πr2 sin t dt = 4πr2: 0 0 0 Surface Integrals Vector surface integrals Math 240 Scalar integrals Surface area Vector Definition integrals 2 3 Let X : D R R be a smooth parameterized surface.