16.2 Line Integrals. Lukas Geyer. M273, Fall Montana State University. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
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1 16.2 Line Integrals Lukas Geyer Montana State University M273, Fall 211 Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
2 Scalar Line Integrals Definition f (x) ds = lim { s i } N f (P i ) s i i=1 Riemann sum illustration. Subdivision into arcs by red crosses, sample points P i as green stars, the s i are the length of the red line segments. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
3 Applications of Scalar Line Integrals Length length() = 1 ds Mass of a wire of linear density ρ(x) mass = ρ(x)ds Surface area of (one side of) a fence of height h(x) area = h(x)ds Electric Potential of a charged wire with linear charge density ρ(x) ρ(x) V (y) = k ds, where k is oulomb s constant. x y Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
4 alculation of Scalar Line Integrals Theorem Let c(t) be a parametrization of a curve for a t b. If f (x) and c (t) are continuous, then b f (x) ds = f (c(t)) c (t) dt. Differential version a ds = c (t) dt. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
5 Scalar Line Integral I alculate xy + z ds over the helix parametrized by c(t) = cos 3t, sin 3t, 4t for t π. ompute ds c (t) = 3 sin 3t, 3 cos 3t, 4 c (t) = 9 sin 2 3t + 9 cos 2 3t + 16 = 25 = 5 ds = c (t) dt = 5dt. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
6 Scalar Line Integral II alculate xy + z ds over the helix parametrized by c(t) = cos 3t, sin 3t, 4t for t π. ompute ds Write out integrand and evaluate xy + z ds = ds = c (t) dt = 5dt. = 5 = 5 π π (cos 3t sin 3t + 4t)5 dt cos 3t sin 3t dt + 2 ] π [ t2 [ 1 6 sin2 3t π ] π t dt = 1π 2 Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
7 Scalar Line Integral Applied I Find the mass of a wire in the shape of a circle of radius 2 centered at (3,4) with linear mass density ρ(x, y) = y 2. Parametrize c(t) = 3, cos t, sin t = cos t, sin t, where t 2π. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
8 Scalar Line Integral Applied II Find the mass of a wire in the shape of a circle of radius 2 centered at (3,4) with linear mass density ρ(x, y) = y 2. Parametrize ompute ds c(t) = cos t, sin t, t 2π. c (t) = 2 sin t, 2 cos t c (t) = 4 sin 2 t + 4 cos 2 t = 4 = 2 ds = c (t) dt = 2dt. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
9 Scalar Line Integral Applied III Find the mass of a wire in the shape of a circle of radius 2 centered at (3,4) with linear mass density ρ(x, y) = y 2. Parametrize, compute ds c(t) = cos t, sin t, t 2π, ds = 2dt. Write out integral and evaluate m = = 2π y 2 ds = 2π (4 + 2 sin t) 2 2 dt sin t + 8 sin 2 t dt = [32t 32 cos t + 4(t sin t cos t)] 2π = 64π + 8π = 72π Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
10 Vector Line Integrals Definition The line integral of a vector field F along an oriented curve is the scalar line integral of the tangential component of F. F ds = (F T)ds. Vector field F(x, y) =, 1 with an oriented curve (in green) and one unit tangent vector (in red). Question Is F ds positive, negative, or zero? Positive Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
11 Applications of Vector Line Integrals Work done by a force field F on a particle moving along W = F ds. Work done against a force field F, moving along W = F ds. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
12 omputing a Vector Line Integral Theorem If c(t) is a regular parametrization of an oriented curve for a t b, then b F ds = F(c(t)) c (t) dt. Differential version ds = c (t)dt c(t) is regular if it is smooth and c (t). a Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
13 Scalar vs. Vector Line Integrals Orientation matters for vector line integrals: If is the curve in negative orientation, then F ds = F ds Orientation does not matter for scalar line integrals: F ds = Fds F ds F ds Why? The tangential component F T always satisfies F T F. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
14 Vector Line Integral I Evaluate F ds, where F(x, y) =, 1, and is the part of the graph of y = 1 2 x 3 x from (2, 2) to ( 2, 2). Parametrize c(t) = t, 1 2 t3 t, 2 t 2. Problem The orientation is wrong. What to do about this? The easiest solution is to calculate with the wrong orientation and reverse the sign of the result. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
15 Vector Line Integral II Evaluate F ds, where F(x, y) =, 1, and is the part of the graph of y = 1 2 x 3 x from (2, 2) to ( 2, 2). Parametrize c(t) = t, 1 2 t3 t, 2 t 2. Write out integral and evaluate F ds = = F(c(t)) c (t) dt = 3 2 t2 + 1 dt = = = [ 1 2 t3 + t, 1 1, 3 2 t2 1 dt Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21 ] 2 2
16 Vector Line Integral III Evaluate F ds, where F(x, y) =, 1, and is the part of the graph of y = 1 2 x 3 x from (2, 2) to ( 2, 2). Write out integral and evaluate F ds = 4 Fix the orientation F ds = F ds = ( 4) = 4. Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
17 Alternative Notation for Vector Line Integrals 2-D Version For F(x, y) = F 1 (x, y), F 2 (x, y) we write ds = dx, dy and F ds = F 1 dx + F 2 dy 3-D Version For F(x, y, z) = F 1 (x, y, z), F 2 (x, y, z), F 3 (x, y, z) we write ds = dx, dy, dz and F ds = F 1 dx + F 2 dy + F 3 dz Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
18 Alternative Notation II 3-D Parametrized Version For F(x, y, z) = F 1 (x, y, z), F 2 (x, y, z), F 3 (x, y, z) and a regular parametrization c(t) = x(t), y(t), z(t), a t b of, F ds = F 1 dx + F 2 dy + F 3 dz = b a F 1 (c(t))x (t) + F 2 (c(t))y (t) + F 3 (c(t))z (t) dt Differential version dx = x (t) dt, dy = y (t) dt, dz = z (t) dt Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
19 3-D Line Integral I alculate yz dx + xz dy + xy dz over the line segment from (1, 1, 1) to (3, 2, ). Parametrize c(t) = 1, 1, 1 + t( 3, 2, 1, 1, 1 ) = 1 + 2t, 1 + t, 1 t, t 1. Write out differentials dx = x (t) dt = 2 dt dy = y (t) dt = dt dz = z (t) dt = dt Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
20 3-D Line Integral II alculate yz dx + xz dy + xy dz over the line segment from (1, 1, 1) to (3, 2, ). Parametrize, write out differentials c(t) = 1 + 2t, 1 + t, 1 t, t 1, dx = 2 dt, dy = dt, dz = dt Write out integral and evaluate yz dx + xz dy + xy dz = 1 (1 + t)(1 t) 2 + (1 + 2t)(1 t) + (1 + 2t)(1 + t)( 1) dt Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
21 3-D Line Integral III alculate yz dx + xz dy + xy dz over the line segment from (1, 1, 1) to (3, 2, ). Write out integral and evaluate yz dx + xz dy + xy dz = = 1 1 (1 + t)(1 t) 2 + (1 + 2t)(1 t) + (1 + 2t)(1 + t)( 1) dt 2 2t t 2t 2 1 3t 2t 2 dt = = [ 2t t 2 2t 3] 1 = = t 6t 2 dt Lukas Geyer (MSU) 16.2 Line Integrals M273, Fall / 21
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