DOCSLIB.ORG

Surface Integrals

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Surface Integrals 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.
Load more

Recommended publications
  • Line, Surface and Volume Integrals
    Line, Surface and Volume Integrals 1 Line integrals Z Z Z Á dr; a ¢ dr; a £ dr (1) C C C (Á is a scalar ¯eld and a is a vector ¯eld) We divide the path C joining the points A and B into N small line elements ¢rp, p = 1;:::;N. If (xp; yp; zp) is any point on the line element ¢rp, then the second type of line integral in Eq. (1) is de¯ned as Z XN a ¢ dr = lim a(xp; yp; zp) ¢ rp N!1 C p=1 where it is assumed that all j¢rpj ! 0 as N ! 1. 2 Evaluating line integrals The ¯rst type of line integral in Eq. (1) can be written as Z Z Z Á dr = i Á(x; y; z) dx + j Á(x; y; z) dy C CZ C +k Á(x; y; z) dz C The three integrals on the RHS are ordinary scalar integrals. The second and third line integrals in Eq. (1) can also be reduced to a set of scalar integrals by writing the vector ¯eld a in terms of its Cartesian components as a = axi + ayj + azk. Thus, Z Z a ¢ dr = (axi + ayj + azk) ¢ (dxi + dyj + dzk) C ZC = (axdx + aydy + azdz) ZC Z Z = axdx + aydy + azdz C C C 3 Some useful properties about line integrals: 1. Reversing the path of integration changes the sign of the integral. That is, Z B Z A a ¢ dr = ¡ a ¢ dr A B 2. If the path of integration is subdivided into smaller segments, then the sum of the separate line integrals along each segment is equal to the line integral along the whole path.
    [Show full text]
  • Chapter 8 Change of Variables, Parametrizations, Surface Integrals
    Chapter 8 Change of Variables, Parametrizations, Surface Integrals x0. The transformation formula In evaluating any integral, if the integral depends on an auxiliary function of the variables involved, it is often a good idea to change variables and try to simplify the integral. The formula which allows one to pass from the original integral to the new one is called the transformation formula (or change of variables formula). It should be noted that certain conditions need to be met before one can achieve this, and we begin by reviewing the one variable situation. Let D be an open interval, say (a; b); in R , and let ' : D! R be a 1-1 , C1 mapping (function) such that '0 6= 0 on D: Put D¤ = '(D): By the hypothesis on '; it's either increasing or decreasing everywhere on D: In the former case D¤ = ('(a);'(b)); and in the latter case, D¤ = ('(b);'(a)): Now suppose we have to evaluate the integral Zb I = f('(u))'0(u) du; a for a nice function f: Now put x = '(u); so that dx = '0(u) du: This change of variable allows us to express the integral as Z'(b) Z I = f(x) dx = sgn('0) f(x) dx; '(a) D¤ where sgn('0) denotes the sign of '0 on D: We then get the transformation formula Z Z f('(u))j'0(u)j du = f(x) dx D D¤ This generalizes to higher dimensions as follows: Theorem Let D be a bounded open set in Rn;' : D! Rn a C1, 1-1 mapping whose Jacobian determinant det(D') is everywhere non-vanishing on D; D¤ = '(D); and f an integrable function on D¤: Then we have the transformation formula Z Z Z Z ¢ ¢ ¢ f('(u))j det D'(u)j du1::: dun = ¢ ¢ ¢ f(x) dx1::: dxn: D D¤ 1 Of course, when n = 1; det D'(u) is simply '0(u); and we recover the old formula.
    [Show full text]
  • An Introduction to Space–Time Exterior Calculus
    mathematics Article An Introduction to Space–Time Exterior Calculus Ivano Colombaro * , Josep Font-Segura and Alfonso Martinez Department of Information and Communication Technologies, Universitat Pompeu Fabra, 08018 Barcelona, Spain; [email protected] (J.F.-S.); [email protected] (A.M.) * Correspondence: [email protected]; Tel.: +34-93-542-1496 Received: 21 May 2019; Accepted: 18 June 2019; Published: 21 June 2019 Abstract: The basic concepts of exterior calculus for space–time multivectors are presented: Interior and exterior products, interior and exterior derivatives, oriented integrals over hypersurfaces, circulation and flux of multivector fields. Two Stokes theorems relating the exterior and interior derivatives with circulation and flux, respectively, are derived. As an application, it is shown how the exterior-calculus space–time formulation of the electromagnetic Maxwell equations and Lorentz force recovers the standard vector-calculus formulations, in both differential and integral forms. Keywords: exterior calculus; exterior algebra; electromagnetism; Maxwell equations; differential forms; tensor calculus 1. Introduction Vector calculus has, since its introduction by J. W. Gibbs [1] and Heaviside, been the tool of choice to represent many physical phenomena. In mechanics, hydrodynamics and electromagnetism, quantities such as forces, velocities and currents are modeled as vector fields in space, while flux, circulation, divergence or curl describe operations on the vector fields themselves. With relativity theory, it was observed that space and time are not independent but just coordinates in space–time [2] (pp. 111–120). Tensors like the Faraday tensor in electromagnetism were quickly adopted as a natural representation of fields in space–time [3] (pp. 135–144). In parallel, mathematicians such as Cartan generalized the fundamental theorems of vector calculus, i.e., Gauss, Green, and Stokes, by means of differential forms [4].
    [Show full text]
  • Arxiv:2109.03250V1 [Astro-Ph.EP] 7 Sep 2021 by TESS
    Draft version September 9, 2021 Typeset using LATEX modern style in AASTeX62 Efficient and precise transit light curves for rapidly-rotating, oblate stars Shashank Dholakia,1 Rodrigo Luger,2 and Shishir Dholakia1 1Department of Astronomy, University of California, Berkeley, CA 2Center for Computational Astrophysics, Flatiron Institute, New York, NY Submitted to ApJ ABSTRACT We derive solutions to transit light curves of exoplanets orbiting rapidly-rotating stars. These stars exhibit significant oblateness and gravity darkening, a phenomenon where the poles of the star have a higher temperature and luminosity than the equator. Light curves for exoplanets transiting these stars can exhibit deviations from those of slowly-rotating stars, even displaying significantly asymmetric transits depending on the system’s spin-orbit angle. As such, these phenomena can be used as a protractor to measure the spin-orbit alignment of the system. In this paper, we introduce a novel semi-analytic method for generating model light curves for gravity-darkened and oblate stars with transiting exoplanets. We implement the model within the code package starry and demonstrate several orders of magnitude improvement in speed and precision over existing methods. We test the model on a TESS light curve of WASP-33, whose host star displays rapid rotation (v sin i∗ = 86:4 km/s). We subtract the host’s δ-Scuti pulsations from the light curve, finding an asymmetric transit characteristic of gravity darkening. We find the projected spin orbit angle is consistent with Doppler +19:0 ◦ tomography and constrain the true spin-orbit angle of the system as ' = 108:3−15:4 .
    [Show full text]
  • 16.6 Parametric Surfaces and Their Areas
    CHAPTER 16. VECTOR CALCULUS 239 16.6 Parametric Surfaces and Their Areas Comments. On the next page we summarize some ways of looking at graphs in Calc I, Calc II, and Calc III, and pose a question about what comes next. 2 3 2 Definition. Let r : R ! R and let D ⊆ R . A parametric surface S is the set of all points hx; y; zi such that hx; y; zi = r(u; v) for some hu; vi 2 D. In other words, it's the set of points you get using x = some function of u; v y = some function of u; v z = some function of u; v Example 1. (a) Describe the surface z = −x2 − y2 + 2 over the region D : −1 ≤ x ≤ 1; −1 ≤ y ≤ 1 as a parametric surface. Graph it in MATLAB. (b) Describe the surface z = −x2 − y2 + 2 over the region D : x2 + y2 ≤ 1 as a parametric surface. Graph it in MATLAB. Solution: (a) In this case we \cheat": we already have z as a function of x and y, so anything we do to get rid of x and y will work x = u y = v z = −u2 − v2 + 2 −1 ≤ u ≤ 1; −1 ≤ v ≤ 1 [u,v]=meshgrid(linspace(-1,1,35)); x=u; y=v; z=2-u.^2-v.^2; surf(x,y,z) CHAPTER 16. VECTOR CALCULUS 240 How graphs look in different contexts: Pre-Calc, and Calc I graphs: Calc II parametric graphs: y = f(x) x = f(t), y = g(t) • 1 number gets plugged in (usually x) • 1 number gets plugged in (usually t) • 1 number comes out (usually y) • 2 numbers come out (x and y) • Graph is essentially one dimensional (if you zoom in enough it looks like a line, plus you only need • graph is essentially one dimensional (but we one number, x, to specify any location on the don't picture t at all) graph) • We need to picture the graph in two dimensional • We need to picture the graph in a two dimen- world.
    [Show full text]
  • Polynomial Curves and Surfaces
    Polynomial Curves and Surfaces Chandrajit Bajaj and Andrew Gillette September 8, 2010 Contents 1 What is an Algebraic Curve or Surface? 2 1.1 Algebraic Curves . .3 1.2 Algebraic Surfaces . .3 2 Singularities and Extreme Points 4 2.1 Singularities and Genus . .4 2.2 Parameterizing with a Pencil of Lines . .6 2.3 Parameterizing with a Pencil of Curves . .7 2.4 Algebraic Space Curves . .8 2.5 Faithful Parameterizations . .9 3 Triangulation and Display 10 4 Polynomial and Power Basis 10 5 Power Series and Puiseux Expansions 11 5.1 Weierstrass Factorization . 11 5.2 Hensel Lifting . 11 6 Derivatives, Tangents, Curvatures 12 6.1 Curvature Computations . 12 6.1.1 Curvature Formulas . 12 6.1.2 Derivation . 13 7 Converting Between Implicit and Parametric Forms 20 7.1 Parameterization of Curves . 21 7.1.1 Parameterizing with lines . 24 7.1.2 Parameterizing with Higher Degree Curves . 26 7.1.3 Parameterization of conic, cubic plane curves . 30 7.2 Parameterization of Algebraic Space Curves . 30 7.3 Automatic Parametrization of Degree 2 Curves and Surfaces . 33 7.3.1 Conics . 34 7.3.2 Rational Fields . 36 7.4 Automatic Parametrization of Degree 3 Curves and Surfaces . 37 7.4.1 Cubics . 38 7.4.2 Cubicoids . 40 7.5 Parameterizations of Real Cubic Surfaces . 42 7.5.1 Real and Rational Points on Cubic Surfaces . 44 7.5.2 Algebraic Reduction . 45 1 7.5.3 Parameterizations without Real Skew Lines . 49 7.5.4 Classification and Straight Lines from Parametric Equations . 52 7.5.5 Parameterization of general algebraic plane curves by A-splines .
    [Show full text]
  • Vector Calculus and Differential Forms with Applications To
    Vector Calculus and Differential Forms with Applications to Electromagnetism Sean Roberson May 7, 2015 PREFACE This paper is written as a final project for a course in vector analysis, taught at Texas A&M University - San Antonio in the spring of 2015 as an independent study course. Students in mathematics, physics, engineering, and the sciences usually go through a sequence of three calculus courses before go- ing on to differential equations, real analysis, and linear algebra. In the third course, traditionally reserved for multivariable calculus, stu- dents usually learn how to differentiate functions of several variable and integrate over general domains in space. Very rarely, as was my case, will professors have time to cover the important integral theo- rems using vector functions: Green’s Theorem, Stokes’ Theorem, etc. In some universities, such as UCSD and Cornell, honors students are able to take an accelerated calculus sequence using the text Vector Cal- culus, Linear Algebra, and Differential Forms by John Hamal Hubbard and Barbara Burke Hubbard. Here, students learn multivariable cal- culus using linear algebra and real analysis, and then they generalize familiar integral theorems using the language of differential forms. This paper was written over the course of one semester, where the majority of the book was covered. Some details, such as orientation of manifolds, topology, and the foundation of the integral were skipped to save length. The paper should still be readable by a student with at least three semesters of calculus, one course in linear algebra, and one course in real analysis - all at the undergraduate level.
    [Show full text]
  • Surface Integrals
    V9. Surface Integrals Surface integrals are a natural generalization of line integrals: instead of integrating over a curve, we integrate over a surface in 3-space. Such integrals are important in any of the subjects that deal with continuous media (solids, fluids, gases), as well as subjects that deal with force fields, like electromagnetic or gravitational fields. Though most of our work will be spent seeing how surface integrals can be calculated and what they are used for, we first want to indicate briefly how they are defined. The surface integral of the (continuous) function f(x,y,z) over the surface S is denoted by (1) f(x,y,z) dS . ZZS You can think of dS as the area of an infinitesimal piece of the surface S. To define the integral (1), we subdivide the surface S into small pieces having area ∆Si, pick a point (xi,yi,zi) in the i-th piece, and form the Riemann sum (2) f(xi,yi,zi)∆Si . X As the subdivision of S gets finer and finer, the corresponding sums (2) approach a limit which does not depend on the choice of the points or how the surface was subdivided. The surface integral (1) is defined to be this limit. (The surface has to be smooth and not infinite in extent, and the subdivisions have to be made reasonably, otherwise the limit may not exist, or it may not be unique.) 1. The surface integral for flux. The most important type of surface integral is the one which calculates the flux of a vector field across S.
    [Show full text]
  • Surface Integrals, Stokes's Theorem and Gauss's Theorem
    Surface Integrals, Stokes's Theorem and Gauss's Theorem Specifying a surface: ² Parametric form: X(s; t) = (x(s; t); y(s; t); z(s; t)) in some region D in R2. ² Explicit function: z = g(x; y) where g is a function in some region D in R2. ² Implicit function: g(x; y; z) = 0, for some function g of three variables. Normal vector and unit normal vector: Normal vectors and unit normal vectors to a surface play an important role in surface integrals. Here are some ways to obtain these vectors. ² Geometric reasoning. For surfaces of simple shape (e.g., planes or spherical surfaces), the easiest way to get normal vectors is by geometric arguments. For example, if the surface is spherical and centered at the origin, then x is a normal vector. If the surface is horizontal or vertical, one can take one of the unit basis vectors (or its negative) as normal vectors. Such geometric reasoning requires virtually no calculation, and is usually the quickest way to evaluate a surface integral in situations where it can be applied. ² Surfaces given by an equation g(x; y; z) = 0. In this case, the gradient of g, rg, is normal to the surface. Normalizing this vector, we get a unit normal vector, n = rg=krgk. Similarly, if the surface is given by an equation z = g(x; y), the vector (gx; gy; 1) is normal to the surface. ² Surfaces given by a parametric representation X(s; t). In this case, the vector N = Ts£ Tt is a normal vector, called the standard normal vector of the given parametrization.
    [Show full text]
  • Stokes' Theorem
    V13.3 Stokes’ Theorem 3. Proof of Stokes’ Theorem. We will prove Stokes’ theorem for a vector field of the form P (x, y, z) k . That is, we will show, with the usual notations, (3) P (x, y, z) dz = curl (P k ) · n dS . � C � �S We assume S is given as the graph of z = f(x, y) over a region R of the xy-plane; we let C be the boundary of S, and C ′ the boundary of R. We take n on S to be pointing generally upwards, so that |n · k | = n · k . To prove (3), we turn the left side into a line integral around C ′, and the right side into a double integral over R, both in the xy-plane. Then we show that these two integrals are equal by Green’s theorem. To calculate the line integrals around C and C ′, we parametrize these curves. Let ′ C : x = x(t), y = y(t), t0 ≤ t ≤ t1 be a parametrization of the curve C ′ in the xy-plane; then C : x = x(t), y = y(t), z = f(x(t), y(t)), t0 ≤ t ≤ t1 gives a corresponding parametrization of the space curve C lying over it, since C lies on the surface z = f(x, y). Attacking the line integral first, we claim that (4) P (x, y, z) dz = P (x, y, f(x, y))(fxdx + fydy) . � C � C′ This looks reasonable purely formally, since we get the right side by substituting into the left side the expressions for z and dz in terms of x and y: z = f(x, y), dz = fxdx + fydy.
    [Show full text]
  • Generalized Stokes' Theorem
    Chapter 4 Generalized Stokes’ Theorem “It is very difficult for us, placed as we have been from earliest childhood in a condition of training, to say what would have been our feelings had such training never taken place.” Sir George Stokes, 1st Baronet 4.1. Manifolds with Boundary We have seen in the Chapter 3 that Green’s, Stokes’ and Divergence Theorem in Multivariable Calculus can be unified together using the language of differential forms. In this chapter, we will generalize Stokes’ Theorem to higher dimensional and abstract manifolds. These classic theorems and their generalizations concern about an integral over a manifold with an integral over its boundary. In this section, we will first rigorously define the notion of a boundary for abstract manifolds. Heuristically, an interior point of a manifold locally looks like a ball in Euclidean space, whereas a boundary point locally looks like an upper-half space. n 4.1.1. Smooth Functions on Upper-Half Spaces. From now on, we denote R+ := n n f(u1, ... , un) 2 R : un ≥ 0g which is the upper-half space of R . Under the subspace n n n topology, we say a subset V ⊂ R+ is open in R+ if there exists a set Ve ⊂ R open in n n n R such that V = Ve \ R+. It is intuitively clear that if V ⊂ R+ is disjoint from the n n n subspace fun = 0g of R , then V is open in R+ if and only if V is open in R . n n Now consider a set V ⊂ R+ which is open in R+ and that V \ fun = 0g 6= Æ.
    [Show full text]
  • Surface Normals and Tangent Planes
    Surface Normals and Tangent Planes Normal and Tangent Planes to Level Surfaces Because the equation of a plane requires a point and a normal vector to the plane, …nding the equation of a tangent plane to a surface at a given point requires the calculation of a surface normal vector. In this section, we explore the concept of a normal vector to a surface and its use in …nding equations of tangent planes. To begin with, a level surface U (x; y; z) = k is said to be smooth if the gradient U = Ux;Uy;Uz is continuous and non-zero at any point on the surface. Equivalently,r h we ofteni write U = Uxex + Uyey + Uzez r where ex = 1; 0; 0 ; ey = 0; 1; 0 ; and ez = 0; 0; 1 : Supposeh now thati r (t) =h x (ti) ; y (t) ; z (t) hlies oni a smooth surface U (x; y; z) = k: Applying the derivative withh respect to t toi both sides of the equation of the level surface yields dU d = k dt dt Since k is a constant, the chain rule implies that U v = 0 r where v = x0 (t) ; y0 (t) ; z0 (t) . However, v is tangent to the surface because it is tangenth to a curve on thei surface, which implies that U is orthogonal to each tangent vector v at a given point on the surface. r That is, U (p; q; r) at a given point (p; q; r) is normal to the tangent plane to r 1 the surface U(x; y; z) = k at the point (p; q; r).
    [Show full text]