Line Integrals

Total Page:16

File Type:pdf, Size:1020Kb

Line Integrals Line Integrals Kenneth H. Carpenter Department of Electrical and Computer Engineering Kansas State University September 12, 2008 Line integrals arise often in calculations in electromagnetics. The evaluation of such integrals can be done without error if certain simple rules are followed. This rules will be illustrated in the following. 1 Definitions 1.1 Definite integrals In calculus, for functions of one dimension, the definite integral is defined as the limit of a sum (the “Riemann” integral – other kinds are possible – see an advanced calculus text if you are curious about this :). Z b X f(x)dx = lim f(ξi)∆xi (1) ∆xi!0 a i where the ξi is a value of x within the incremental ∆xi, and all the incremental ∆xi add up to the length along the x axis from a to b. This is often pictured graphically as the area between the curve drawn as f(x) and the x axis. 1 EECE557 Line integrals – supplement to text - Fall 2008 2 1.2 Evaluation of definite integrals in terms of indefinite inte- grals The fundamental theorem of integral calculus relates the indefinite integral to the definite integral (with certain restrictions as to when these exist) as Z b dF f(x)dx = F (b) − F (a); when = f(x): (2) a dx The proof of eq.(2) does not depend on having a < b. In fact, R b f(x)dx = R a a − b f(x)dx follows from the values of ∆xi being signed so that ∆xi equals the value of x on the side of the increment next to b minus the value of x on the side of the increment next to a. 1.3 Line integrals The line integral extends the concept of the Riemann integral in one dimension to integration along a curved path in three dimensional space. Z b X f(l)dl = lim f(ξi)∆li (3) ∆li!0 a i where the line integral is along a path in space from beginning point a to ending point b, and where ∆li is an increment of length along this path. Since the path has a beginning point and an ending point, it is a directed path. If the direction is reversed, the value of the integral is negated, just as for the R b R a ordinary integral: a f(l)dl = − b f(l)dl. In order to apply the fundamental theorem of eq.(2) to the line integral, one reduces the form of eq.(3) to that of an ordinary integral by one of the following methods. The integrand of eq.(3) is a function of the point in space along the path: f(l) = g(x; y; z) in Cartesian coordinates, e.g.. In the same way, the increment, dl can be expressed in terms of the increments of the coordinates. For Cartesian co- ordiantes, dl =xdx ^ +ydy ^ +zdz ^ , with a magnitude of jdlj = pdx2 + dy2 + dz2. There is an ambiguity, however, in choice of the sign to be given to the scalar dl in terms of the magnitude of the vector dl. To keep the fundamental theorem of eq.(2) applicable, one must have an ordered length along the path. That is, the value l for which ∆l is an increment, must be defined as either decreasing or in- creasing as one goes from point a to point b. Once this definition is made, then the EECE557 Line integrals – supplement to text - Fall 2008 3 value of the line integral is fixed in terms of the sum, and the value of the lower and upper limits to be subsituted in the integrand are the values of l at the point of the lower limit and at the point of the upper limit, respectively. This is carried out, in fact, by using a parametric representation for the path along which the in- tegration is carried out, and by expressing the integrand in terms of the parameter (t in the following): Z b Z tb dl f(l)dl = f[l(t)] dt (4) a ta dt This latter integral is just an ordinary definite integral in one dimension. 1.3.1 Example: Let the path lie in the x-y plane along the first one-fourth of a period of a sine wave, y = sin(x), beginning at the origin for point a and ending at point (π=2; 1; 0) for point b. Let the function to be integrated be f(l) = g(x; y; z) = 2x2 − y + z2. The positive length begins at the origin, and increase towards point b. Then one may even use x itself as the parameter, but to keep the notation as it will be for more complicated cases, let the parameter be t. Then the path is given by x = t, y = sin(t), and z = 0. The length along the path is the integrated value of dl. In this case, dl = pdx2 + dy2 = dxp1 + (dy=dx)2 = dtp1 + cos2(t). The integral becomes, in terms of t, Z b Z π=2 p f(l)dl = [2t2 − sin(t) + 0] 1 + cos2 tdt: (5) a 0 Note that the square root factor in eq.(5) is essential to this being a line integral. 1.3.2 Example: One could integrate the function of the previous example along the path following the x axis from x = 0 to x = π=2 and then following the line x = π=2 from y = 0 to y = 1. Using x as the parameter along the first part of the path, rather the t, and noting that the increment of length is just dx this time, the integration is much simpler. Along the second part of the path one could use y itself as the parameter, and note that the increment of length is dy, so dl=dy = 1. Thus on this new path the total line integral is Z b Z π=2 Z 1 f(l)dl = [2x2 − 0 + 0]dx + [2(π=2)2 − sin(y) + 0]dy: (6) a 0 0 EECE557 Line integrals – supplement to text - Fall 2008 4 One does not expect the numerical result for eq.(5) to be the same as for eq.(5) since the paths are different. 2 Line integrals in vector analysis The concept of line integrals may be extended in vector analysis by having the integrand expressed in terms of the vector line element, dl and a vector or scalar field or fields operating on it. The result can be reduced to one, in the case of a scalar result, or three, in the case of a vector result, ordinary line integrals. But sometimes the vector formulation of the problem is even easier than the ordinary line integral, since it eliminates the need to take the square root to get a magnitude for dl. 2.1 Scalar product line integrals When the integrand is a vector field, and the operation is a scalar product, one obtains the symbolic line integral Z b W = F · dl: (7) a (If F is force, then w is work.) A path must still be specified between beginning point a and ending point b. The direction of dl as a vector is tangent to the path but directed from a towards b along it. 2.1.1 Example: Let the force vector be F = 3( N/m)yx^ − 4( N/m)xy^ and let the path be along the straight line from Cartesian point a(0; 2 m; 1 m) to b(1 m; 0; 0). Then we may write Z b Z 1;0;0 Z 1 Z 0 Z 0 F ·dl = (3yx^−4xy^)·(^xdx+^ydy+^zdz)( J) = [ 3ydx+ (−4x)dy+ 0dz]( J): a (0;2;1) 0 2 1 (8) Here we see that the single line integral has become three ordinary integrals along the three coordinates. The upper and lower limits for each of these integrals must be kept in the same order as in the original points. This takes care of the direction of the vector increment dl. One must use the equation defining the path to express EECE557 Line integrals – supplement to text - Fall 2008 5 each integrand in terms of the variable of integration. For this example the path is given by the equations y = −2x + 2 and z = −x + 1. An alternate method of solution would be obtained by using t as the parameter on the path. Then the equations of the path would be x = t, y = −2t + 1, and z = −t + 1. Now dl = [^xdt +y ^(−2)dt +x ^(−1)dt]( m). F = 3( N/m)(−2t + 1)^x − 4( N/m)ty^. The result is Z b Z 1 F · dl = [3(−2t + 1) − 4t(−2)]dt( J): (9) a 0 The numerical results for eq.(8) and for eq.(9) are the same..
Recommended publications
  • Lecture 11 Line Integrals of Vector-Valued Functions (Cont'd)
    Lecture 11 Line integrals of vector-valued functions (cont’d) In the previous lecture, we considered the following physical situation: A force, F(x), which is not necessarily constant in space, is acting on a mass m, as the mass moves along a curve C from point P to point Q as shown in the diagram below. z Q P m F(x(t)) x(t) C y x The goal is to compute the total amount of work W done by the force. Clearly the constant force/straight line displacement formula, W = F · d , (1) where F is force and d is the displacement vector, does not apply here. But the fundamental idea, in the “Spirit of Calculus,” is to break up the motion into tiny pieces over which we can use Eq. (1 as an approximation over small pieces of the curve. We then “sum up,” i.e., integrate, over all contributions to obtain W . The total work is the line integral of the vector field F over the curve C: W = F · dx . (2) ZC Here, we summarize the main steps involved in the computation of this line integral: 1. Step 1: Parametrize the curve C We assume that the curve C can be parametrized, i.e., x(t)= g(t) = (x(t),y(t), z(t)), t ∈ [a, b], (3) so that g(a) is point P and g(b) is point Q. From this parametrization we can compute the velocity vector, ′ ′ ′ ′ v(t)= g (t) = (x (t),y (t), z (t)) . (4) 1 2. Step 2: Compute field vector F(g(t)) over curve C F(g(t)) = F(x(t),y(t), z(t)) (5) = (F1(x(t),y(t), z(t), F2(x(t),y(t), z(t), F3(x(t),y(t), z(t))) , t ∈ [a, b] .
    [Show full text]
  • A Brief Tour of Vector Calculus
    A BRIEF TOUR OF VECTOR CALCULUS A. HAVENS Contents 0 Prelude ii 1 Directional Derivatives, the Gradient and the Del Operator 1 1.1 Conceptual Review: Directional Derivatives and the Gradient........... 1 1.2 The Gradient as a Vector Field............................ 5 1.3 The Gradient Flow and Critical Points ....................... 10 1.4 The Del Operator and the Gradient in Other Coordinates*............ 17 1.5 Problems........................................ 21 2 Vector Fields in Low Dimensions 26 2 3 2.1 General Vector Fields in Domains of R and R . 26 2.2 Flows and Integral Curves .............................. 31 2.3 Conservative Vector Fields and Potentials...................... 32 2.4 Vector Fields from Frames*.............................. 37 2.5 Divergence, Curl, Jacobians, and the Laplacian................... 41 2.6 Parametrized Surfaces and Coordinate Vector Fields*............... 48 2.7 Tangent Vectors, Normal Vectors, and Orientations*................ 52 2.8 Problems........................................ 58 3 Line Integrals 66 3.1 Defining Scalar Line Integrals............................. 66 3.2 Line Integrals in Vector Fields ............................ 75 3.3 Work in a Force Field................................. 78 3.4 The Fundamental Theorem of Line Integrals .................... 79 3.5 Motion in Conservative Force Fields Conserves Energy .............. 81 3.6 Path Independence and Corollaries of the Fundamental Theorem......... 82 3.7 Green's Theorem.................................... 84 3.8 Problems........................................ 89 4 Surface Integrals, Flux, and Fundamental Theorems 93 4.1 Surface Integrals of Scalar Fields........................... 93 4.2 Flux........................................... 96 4.3 The Gradient, Divergence, and Curl Operators Via Limits* . 103 4.4 The Stokes-Kelvin Theorem..............................108 4.5 The Divergence Theorem ...............................112 4.6 Problems........................................114 List of Figures 117 i 11/14/19 Multivariate Calculus: Vector Calculus Havens 0.
    [Show full text]
  • Appendix a Short Course in Taylor Series
    Appendix A Short Course in Taylor Series The Taylor series is mainly used for approximating functions when one can identify a small parameter. Expansion techniques are useful for many applications in physics, sometimes in unexpected ways. A.1 Taylor Series Expansions and Approximations In mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the English mathematician Brook Taylor. If the series is centered at zero, the series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin. It is common practice to use a finite number of terms of the series to approximate a function. The Taylor series may be regarded as the limit of the Taylor polynomials. A.2 Definition A Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function f(x) about a point x ¼ a is given by; f 00ðÞa f 3ðÞa fxðÞ¼faðÞþf 0ðÞa ðÞþx À a ðÞx À a 2 þ ðÞx À a 3 þÁÁÁ 2! 3! f ðÞn ðÞa þ ðÞx À a n þÁÁÁ ðA:1Þ n! © Springer International Publishing Switzerland 2016 415 B. Zohuri, Directed Energy Weapons, DOI 10.1007/978-3-319-31289-7 416 Appendix A: Short Course in Taylor Series If a ¼ 0, the expansion is known as a Maclaurin Series. Equation A.1 can be written in the more compact sigma notation as follows: X1 f ðÞn ðÞa ðÞx À a n ðA:2Þ n! n¼0 where n ! is mathematical notation for factorial n and f(n)(a) denotes the n th derivation of function f evaluated at the point a.
    [Show full text]
  • How to Compute Line Integrals Scalar Line Integral Vector Line Integral
    How to Compute Line Integrals Northwestern, Spring 2013 Computing line integrals can be a tricky business. First, there's two types of line integrals: scalar line integrals and vector line integrals. Although these are related, we compute them in different ways. Also, there's two theorems flying around, Green's Theorem and the Fundamental Theorem of Line Integrals (as I've called it), which give various other ways of computing line integrals. Knowing what to use when and where and why can be confusing, so hopefully this will help to keep things straight. I won't say much about what line integrals mean geometrically or otherwise. Hopefully this is something you have some idea about from class or the book, but I'm always happy to talk about this more in office hours. Here the focus is on computation. Key Point: When you are trying to apply one of these techniques, make sure the technique you are using is actually applicable! (This was a major source of confusion on Quiz 5.) Scalar Line Integral • Arises when integrating a function f over a curve C. • Notationally, is always denoted by something like Z f ds; C where the s in ds is just a normal s, as opposed to ds or d~s. • To compute, find parametric equations x(t); a ≤ t ≤ b for C and use Z Z b f ds = f(x(t)) kx0(t)k dt: C a • In the special case where f is the constant function 1, Z ds = arclength of C: C • For scalar line integrals, the orientation (i.e.
    [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]
  • Multivariable Calculus Math 21A
    Multivariable Calculus Math 21a Harvard University Spring 2004 Oliver Knill These are some class notes distributed in a multivariable calculus course tought in Spring 2004. This was a physics flavored section. Some of the pages were developed as complements to the text and lectures in the years 2000-2004. While some of the pages are proofread pretty well over the years, others were written just the night before class. The last lecture was "calculus beyond calculus". Glued with it after that are some notes from "last hours" from previous semesters. Oliver Knill. 5/7/2004 Lecture 1: VECTORS DOT PRODUCT O. Knill, Math21a VECTOR OPERATIONS: The ad- ~u + ~v = ~v + ~u commutativity dition and scalar multiplication of ~u + (~v + w~) = (~u + ~v) + w~ additive associativity HOMEWORK: Section 10.1: 42,60: Section 10.2: 4,16 vectors satisfy "obvious" properties. ~u + ~0 = ~0 + ~u = ~0 null vector There is no need to memorize them. r (s ~v) = (r s) ~v scalar associativity ∗ ∗ ∗ ∗ VECTORS. Two points P1 = (x1; y1), Q = P2 = (x2; y2) in the plane determine a vector ~v = x2 x1; y2 y1 . We write here for multiplication (r + s)~v = ~v(r + s) distributivity in scalar h − − i ∗ It points from P1 to P2 and we can write P1 + ~v = P2. with a scalar but usually, the multi- r(~v + w~) = r~v + rw~ distributivity in vector COORDINATES. Points P in space are in one to one correspondence to vectors pointing from 0 to P . The plication sign is left out. 1 ~v = ~v the one element numbers ~vi in a vector ~v = (v1; v2) are also called components or of the vector.
    [Show full text]
  • Line Integral Along a Piecewise C1 Curve: 1 Let C Be a Continuous Curve Which Is Piecewise of Class C I.E
    Properties of line integrals: i) Linearity: (F + G) · dx = F·dx + G·dx, (2.17) Z Z Z C C C (λF) · dx = λ F·dx, (2.18) Z Z C C where λ is a constant scalar. These properties follow immediately from the definition (2.14) and the corresponding properties of the Riemann integral. ii) Additivity: If C is a C1 curve that is the union of two curves C1 and C2 joined end-to-end and consis- tently oriented ( C = C1 ∪ C 2), then F·dx = F·dx + F·dx (2.19) Z Z Z C C1 C2 Line integral along a piecewise C1 curve: 1 Let C be a continuous curve which is piecewise of class C i.e. C = C1 ∪ ··· ∪ C n, where 1 the individual pieces Ci, i = 1 ,...,n are of class C . Motivated by equation (2.19), we define the line integral of a vector field F along C by F·dx = F·dx + · · · + F·dx. (2.20) Z Z Z C C1 Cn Each line integral on the right is, of course, defined as a Riemann integral by equation (2.14). Example 2.3: Compute the line integral of the vector field F = ( y, −2x) along the piecewise C1 curve consisting of the two straight line segments joining ( −b, b ) to (0 , 0) and (0 , 0) to (2 b, b ), where b is a positive constant. Solution: For C1, x = g1(t) = ( t, −t), with −b ≤ t ≤ 0, giving ′ g1(t) = (1 , −1) , and F(g1(t)) = ( −t, −2t). By the definition (2.14), 0 F·dx = (−t, −2t) · (1 , −1) dt Z Z−b C1 0 1 2 = t dt = − 2 b .
    [Show full text]
  • Multivariable Calculus 1 / 130 Section 16.2 Line Integrals Types of Integrals
    Section 16.2 Line Integrals Section 16.2 Line Integrals Goals: Compute line integrals of multi variable functions. Compute line integrals of vector functions. Interpret the physical implications of a line integral. Multivariable Calculus 1 / 130 Section 16.2 Line Integrals Types of Integrals We have integrated a function over The real number line The plane R b f (x)dx RR Three space a D f (x; y)dA RRR R f (x; y; z)dV Multivariable Calculus 2 / 130 Section 16.2 Line Integrals Line Integrals In this chapter we will integrate a function over a curve (in either two or three dimensions, though more are possible). A two-variable function f (x; y) over a plane curve r(t). Multivariable Calculus 3 / 130 Section 16.2 Line Integrals Parameterizations and the Line Integral The naive approach to integrating a function f (x; y) over a curve r(t) = x(t)i + y(t)j would be to plug in x(t) and y(t). Now we are integrating a function of t. But this is dependent on our choice of parameterizations. x(2t)i + y(2t)j defines the same curve but moves twice as fast. Integrating this composition would give half the ∆t and half the area: Multivariable Calculus 4 / 130 Section 16.2 Line Integrals Integrating Independent of Parameterization Instead, our variable of integration should measure something geometrically inherent to the curve: Multivariable Calculus 5 / 130 Section 16.2 Line Integrals Arc Length A more geometrically relevant measure is to integrate with respect to distance traveled.
    [Show full text]
  • Line Integrals
    Line integrals What to know: 1. Be able to parametrize curves 2. Be able to set up and compute line integrals with respect to arc length (ds) and of vector fields (·d~r). 3. Be able to compute line integrals dx and dy (interpreted as special cases of integrals of vector fields). 4. Know which line integrals depend on the parametrization of the curve and which ones don't. In this section, we'll see how to integrate over curves. The curves we'll talk about are called piecewise smooth curves, which means that they are finite unions of smooth curves, parametrized as c(t) = (x(t); y(t)), t 2 [a; b], like in figures 1 and 21 . Figure 2: Another piecewise smooth curve Figure 1: A piecewise smooth curve Line integrals of scalar functions Suppose that f(x; y) is a continuous non-negative function. Our goal is to be able to evaluate the area of a fence that lies above the curve c and under the graph of f. To do this, we approximate the area in the following way: We first approximate the curve 3 c by line segments [aj; aj+1] and form rectangles (living in R ) with base [aj; aj+1] and height ∗ ∗ f(aj ), where aj is the midpoint of each line segment. Then, we evaluate the area of each of those rectangles and sum the areas. As we make the approximation of the curve more and more accurate, this procedure gives rise to a special type of integral, called line integral with respect to to arc length.
    [Show full text]
  • A Calculus Oasis on the Sands of Trigonometry
    φ θ b 3 A Calculus Oasis on the sands of trigonometry A h Conal Boyce A Calculus Oasis on the sands of trigonometry with 86+ illustrations by the author Conal Boyce rev. 130612 Copyright 2013 by Conal Boyce All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the written prior permission of the author. Cover design and layout by CB By the same author: The Chemistry Redemption and the Next Copernican Shift Chinese As It Is: A 3D Sound Atlas with First 1000 Characters Kindred German, Exotic German: The Lexicon Split in Two ISBN 978-1-304-13029-7 www.lulu.com Printed in the United States of America In memory of Dr. Lorraine (Rani) Schwartz, Assistant Professor in the Department of Mathematics from 1960-1965, University of British Columbia Table of Contents v List of Figures . vii Prologue . 1 – Vintage Calculus versus Wonk Calculus. 3 – Assumed Audience . 6 – Calculus III (3D vector calculus) . 9 – A Quick Backward Glance at Precalculus and Related Topics . 11 – Keeping the Goal in Sight: the FTC . 14 I Slopes and Functions . 19 – The Concept of Slope . 19 – The Function Defined. 22 – The Difference Quotient (alias ‘Limit Definition of the Derivative’). 29 – Tangent Line Equations . 33 – The Slope of e . 33 II Limits . 35 – The Little-δ Little-ε Picture. 35 – Properties of Limits . 38 – Limits, Continuity, and Differentiability. 38 III The Fundamental Theorem of Calculus (FTC) . 41 – A Pictorial Approach (Mainly) to the FTC.
    [Show full text]
  • BEYOND CALCULUS Math21a, O
    5/7/2004 CALCULUS BEYOND CALCULUS Math21a, O. Knill DISCRETE SPACE CALCULUS. Many ideas in calculus make sense in a discrete setup, where space is a graph, curves are curves in the graph and surfaces are collections of "plaquettes", polygons formed by edges of the graph. One can look at functions on this graph. Scalar functions are functions defined on the vertices of the INTRODUCTION. Topics beyond multi-variable calculus are usually labeled with special names like "linear graphs. Vector fields are functions defined on the edges, other vector fields are defined as functions defined on algebra", "ordinary differential equations", "numerical analysis", "partial differential equations", "functional plaquettes. The gradient is a function defined on an edge as the difference between the values of f at the end analysis" or "complex analysis". Where one would draw the line between calculus and non-calculus topics is points. not clear but if calculus is about learning the basics of limits, differentiation, integration and summation, then Consider a network modeled by a planar graph which forms triangles. A scalar function assigns a value f multi-variable calculus is the "black belt" of calculus. Are there other ways to play this sport? n to each node n. An area function assigns values fT to each triangle T . A vector field assign values Fnm to each edge connecting node n with node m. The gradient of a scalar function is the vector field Fnm = fn fm. HOW WOULD ALIENS COMPUTE? On an other planet, calculus might be taught in a completely different − The curl of a vector field F is attaches to each triangle (k; m; n) the value curl(F )kmn = Fkm + Fmn + Fnk.
    [Show full text]