Surface Integrals - Math.upenn.edu
SurfaceIntegralsMath 240ScalarintegralsSurface areaVectorintegralsChangingorientationSurface IntegralsMath 240 — Calculus IIISummer 2013, Session IIWednesday, July 3, 2013
SurfaceIntegralsAgendaMath 240ScalarintegralsSurface areaVectorintegralsChangingorientation1. Scalar surface integralsSurface area2. Vector surface integrals3. Changing orientation
SurfaceIntegralsMath 240Now, we use the notion of a parametrized surface to calculate the surface area of asmooth surface. In the discussion that follows, we take S X(D) to be a smoothparametrized surface, where D is the union of finitely many elementary regionsin R2 and X: D R3 is of class C 1 and one-one except possibly along D.Scalar surface ngorientationS X(D)XΔtSurface areaX(s0, t0)Tt(s0, t0)TssyxFigure 7.13 The image of the s t rectangle in D is approximately aDefinitionparallelogram spanned by T (s , t ) s and T (s , t ) t.s00t00Let X : D R2 R3 be a smooth parameterized surface. LetThe key geometric observation is as follows: Consider a small rectangulart0 ) D andincludeswhose widthof D whose lowerleft corneris at thewhosepoint (s0 ,domainf be asubsetcontinuousscalarfunctionand height are s and t, respectively. The image of this rectangle under X is aS X(D).scalarsurfaceintegral ofalong Xwithis apiece of theTheunderlyingsurfaceS that is approximatelythe fparallelogramcornerZatZX(s0 , t0 ) and spannedZ Z by the vectors Ts (s0 , t0 ) s and Tt (s0 , t0 ) t. (SeeFigure 7.13.) The area A of this piece isf dS f (X(s, t)) kTs Tt k ds dt A X Ts (s0 , t0 ) s TDt (s0 , t0 ) t Ts (s0 , t0 ) Tt (s0 , t0 ) s t.ZZNow, suppose D [a, b] [c, d]; that is, suppose D itself is a rectangle.Partition D into n 2 subrectangles fvia(X(s, t)) kN(s, t)k ds dt.a s0 s1 · · · sDn bandc t0 t1 · · · tn d.
coll50424 ch07SurfaceIntegralsScalar surface integralsMath 240ScalarintegralsSurface ey472Chapter 7zExampleLet S be the closed cylinder of radius 3 withaxis along the z-axis, top face at z 15,RR andbottom face at z 0. Let’s calculate S z dS.Denote the lateral cylindrical face of S by S1and the bottom and top faces by S2 and S3 ,respectively.Surface IntegS3: z 15S 1: x 2 y 2 9xS2: z 0Figure 7.15 The closedcylinder of radius 3 andheight 15 of Example 2.We computeZZZZZZz dS 675π,z dS 0, andz dS 135π.S1S2Therefore,ZZZZz dS SS1S3ZZz dS ZZz dS S2z dS 810π.S3y
is only piecewise smooth.SurfaceIntegralsMath 240ScalarintegralsArea of a Smooth Parametrized SurfaceNow, we use the notion of a parametrized surface to calculate the surface area of asmooth surface. In the discussion that follows, we take S X(D) to be a smoothparametrized surface, where D is the union of finitely many elementary regionsin R2 and X: D R3 is of class C 1 and one-one except possibly along D.VectorintegralsChangingorientationztSurface areaSurface areaS X(D)XΔtΔsX(s0, t0)Tt(s0, t0)TssyFigure: The quantitykTs Tt k is thearea of the graysquare on the right.xFigure 7.13 The image of the s t rectangle in D is approximately aparallelogram spanned by Ts (s0 , t0 ) s and Tt (s0 , t0 ) t.FactThe key geometric observation is as follows: Consider a small rectangularsubset of D whose lower left corner is at the point (s0 , t0 ) D and whose widthand height are s and t, respectively. The image of this rectangle under X is apiece of the underlying surface S that is approximately the parallelogram with acorner at X(s0 , t0 ) and spanned by the vectors Ts (s0 , t0 ) s and Tt (s0 , t0 ) t. (SeeFigure 7.13.) The area A of this piece isIf S is a smooth surface parameterized by X : D R2 R3then the surface area of S is given byZZZZZZ A Ts (s0 , t0 ) s Tt (s0 , t0 ) t Ts (s0 , t0 ) Tt (s0 , t0 ) s t.kNk ds dt kTs Tt k ds dt 1 dS.Now, suppose D [a, b] [c, d]; that is, suppose D itself is a rectangle.D2subrectangles viaPartition D into nDa s 0 s1 · · · s n bandc 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 turnpartitioned into pieces, each of which is approximately a parallelogram, assuming si and t j are small for i, j 1, . . . , n. If Ai j denotes the area of the pieceX
SurfaceIntegralsSurface areaMath 240ScalarintegralsSurface l our parameterization of a sphere:X(s, t) r(cos s)(sin t) i r(sin s)(sin t) j r(cos t) k.We calculateTs 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 kNk r2 sin t.Therefore, the surface area of the sphere isZ π Z 2πZ π2r sin t ds dt 2πr2 sin t dt 4πr2 .000
SurfaceIntegralsVector surface integralsMath 240ScalarintegralsSurface t X : D R2 R3 be a smooth parameterized surface. LetF be a continuous vector field whose domain includesS X(D). The vector surface integral of F along X isZZZZF · dS F(X(s, t)) · N(s, t) ds dt.XDIn physical terms, we can interpret F as the flow of some kindof fluid. Then the vector surface integral measures the volumeof fluid that flows through S per unit time. This is called theflux of F across S.
SurfaceIntegralsMath 240Scalarintegralsis a continuous function of the entries of H, it thus cannot change sign.) Hence,the standard normal NY either always points in the same direction as NX or elsealways points in the opposite direction (Figure 7.20). Under these assumptions, wesay that both H and Y are orientation-preserving if the Jacobian (u, v)/ (s, t)is positive, orientation-reversing if (u, v)/ (s, t) is negative.Changing orientationvD1Surface areauVectorintegralsChangingorientationtNXXD2NYsS X(D1) Y(D2)YFigure: X and Yparameterize thesame surface withopposite normaldirections.Figure 7.20 If Y is an orientation-reversing reparametrization of X, then NYZZpoints opposite to NX .ZZf dS 1.4, Chapter 6,fshowsdS thatThe following result, a close analogue of TheoremY value of a scalarZlinesmooth reparametrization has no effectZ Z on theZX integral.THEOREM 2.4 Let X: D1 R3 be a smoothandanyF · parametrizeddS surfaceF ·f dS3is any smoothcontinuous function whose domain includesY X(D1 ). If Y: D2 R Xreparametrization of X, then This can be achievedbyexchangings and t:f dS f d S.YXTt Ts (Ts Tt ) .
Surface area Vector integrals Changing orientation Changing orientation P1: OSO coll50424úch07 PEAR591-Colley July 29, 2011 13:58 478 Chapter 7 Surface Integrals and Vector Analysis The parametrized surface Y is the same as X, except that the standard nor-mal vector arising from Y points in the opposite direction to the one arising
Peter-Michael Osera posera@cis.upenn.edu Richard Eisenberg eir@cis.upenn.edu Christian DeLozier delozier@cis.upenn.edu Santosh Nagarakatte santoshn@cis.upenn.edu Milo M. K. Martin milom@cis.upenn.edu Steve Zdancewic stevez@cis.upenn.edu August 5, 2013 Core Ironclad is a c
Iterated Double Integrals Double Integrals over General Regions Changing the Order of Integration Triple Integrals Triple Integrals over General Regions Iterated Double Integrals Let f be a real-valued function of two variables x,
Nathan Close, Amalia Hawkins, Sworupini Sureshkumar, Peter-Michael Osera, Lyle Ungar, and Steve Zdancewic University of Pennsylvania - Philadelphia, PA {closen, hawka, sures}@seas.upenn.edu, {posera, ungar, stevez}@cis.upenn.edu ABSTRACT The Judgement of Code Style (JOCS) system is an auto-
Peter-Michael Osera posera@cis.upenn.edu Vilhelm Sj oberg vilhelm@cis.upenn.edu Steve Zdancewic stevez@cis.upenn.edu March 29, 2012. Abstract In this paper we study the problem of interoperability combining constructs from two separate programming languages within one program
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S8: Double integrals in polar co–ordinates. Sometimes we can reduce a very difficult double integral to a simple one via a substitution. You will have seen this general technique for single integrals. However, for double integrals, we can make a transformation tha
1. Double Integrals 2. Triple Integrals 1. Double Integrals R f(x,y) dx dy is called the double integral where R ϵ [(x1, x2)(y1, y2)] Methods to Evaluate Double Integrals Method 1: If y1, y2 are functions of x only and x1, x2 are co
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