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Line Integrals Dr. E. Jacobs Introduction Applications of Integration to Physics and Engineering Require an Extension of the Integral Called a Line Integral

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Line Integrals Dr. E. Jacobs Introduction Applications of Integration to Physics and Engineering Require an Extension of the Integral Called a Line Integral Line Integrals Dr. E. Jacobs Introduction Applications of integration to physics and engineering require an extension of the integral called a line integral. Line integrals are necessary to express the work done along a path by a force. Line integrals are needed to describe circulation of fluids. They are also important in describing the relationship between electric and magnetic fields. The name line integral is somewhat of a misnomer because it refers to adding up quantities along curved paths as well as along straight lines. You have already encountered the notion of a line integral when you learned about arclength in Calculus II and III. Review of Arclength If a curve C is described by the parametric equations x = x(t) and y = y(t) for a · t · b then the length along curve C is given by: s Z µ ¶ µ ¶ b dx 2 dy 2 Length(C) = + dt a dt dt The equation is simple enough to use. For example, to find the length ¼ along the segment of the circle x = 2 cos t, y = 2 sin t for 0 · t · 8 we just calculate: Z q Z ¼=8 ¼=8 p ¼ Length = (x0(t))2 + (y0(t))2 dt = 4 sin2 t + 4 cos2 t dt = 0 0 4 R q b 0 2 0 2 The expression a (x (t)) + (y (t)) dt can be understood with either a geometric or physical interpretation. The usual geometric interpretation is to split up the curve C into small segments of length ds. The length of a typical segment is approximated as a hypotenuse of a right triangle of base dx and height dy and therefore, by the Pythagorean Theorem: sµ ¶ µ ¶ p dx 2 dy 2 ds = (dx)2 + (dy)2 = + dt dt dt R If we use the notation C ds to symbolize the limit of the sum of all the segments of length ds as t ranges from a to b thenr it becomes clearer that R R ¡ ¢ ³ ´2 b dx 2 dy the length along C is represented by C ds = a dt + dt dt. For a more physical interpretation, we consider a particle moving along C with it’s position after t seconds given by ~r = hx(t); y(t)i. The velocity of d~r 0 0 the particle is ~v = dt = hx (t); y (t)i and so the speed of the particle is: p j~vj = (x0(t))2 + (y0(t))2 We can look at the formula : p ds = j~vj dt = (x0(t))2 + (y0(t))2 dt as: change in distance = (speed)(change in time) R therefore, if we look at the length of curve C, C ds as the sum of all the (speed)(time) expressions, so Z b Z b p Total distance traveled = j~vj dt = (x0(t))2 + (y0(t))2 dt a a In three dimensions, the velocity vector has three coordinates ~v = hx0(t); y0(t); z0(t)i p and the speed is j~vj = (x0(t))2 + (y0(t))2 + (z0(t))2 so the total distance along C is: Z Z b Z b p ds = j~vj dt = (x0(t))2 + (y0(t))2 + (z0(t))2 dt C a a The notion of adding quantities up over a curve C applies not only to distance but to other physical quantities as well. Suppose, for example, we had electric charge distributed along this path C with a charge density at point ~r = hx; y; zi of f(x; y; z) coulombs per meter. Then, a small segment of length ds meters would hold a charge of f(x; y; z) ds coulombs. The total charge along this path would be the limit of the sum of such quantities, which would give us: Z Total Charge along C = f(x; y; z) ds C The computation of such an integral is similar to an arc length calculation. For example, suppose C is a helical segment, given by ~r = hx(t); y(t); z(t)i = h¼ cos t; ¼ sin t; ti 0 · t · ¼ Let us suppose that the linear charge density at any point (x; y; z) on this 1 curve is given by f(x; y; z) = x2+y2+z2 coulomb per meter. This would mean that as we move along the helix from (¼; 0; 0) at t = 0 to (¡¼; 0; ¼) at t = ¼, the charge density gets smaller and smaller. The density at ~r = h¼ cos t; ¼ sin t; ti is: 1 1 1 f(x; y; z) = = = x2 + y2 + z2 ¼2 cos2 t + ¼2 sin2 t + t2 ¼2 + t2 The charge (in coulombs) along a segment of length ds is: f(x; y; z) ds = f(x; y; z)j~vj dt 1 p = (x0(t))2 + (y0(t))2 + (z0(t))2 dt ¼2 + t2 1 p = (¡¼ sin t)2 + (¼ cos t)2 + 1 dt ¼2 + t2 p ¼2 + 1 = dt ¼2 + t2 Therefore, the total charge is: p p p Z Z ¼ 2 2 2 ¼ + 1 ¼ + 1 ¡1 ¼ + 1 f(x; y; z) ds = 2 2 dt = tan 1 = C 0 ¼ + t ¼ 4 Work Work is defined as the integral of force over distance. Z b W = F (x) dx a This definition is only applicable if a force F is pushing something along a straight line. You might expect that if a force is acting along a curve C then the definition of work needs to be extended to be: Z W = F (x; y; z) ds C since ds represents the change in distance along the curve. However, there is a complication. Suppose the force is directed at an angle θ to the path C. It is only the tangential component of the force that is doing work along the path. Therefore, we should be using jF~ j cos θ in our integral. If T~ is the unit tangent vector, then jF~ j cos θ = F~ ² T~ and the expression for the work done along path C is: Z W = F~ ² T~ ds C The unit tangent vector T~ can be found by dividing the tangent vector d~r ~v = dt by its length. ~v T~ = j~vj We already know from before that ds = j~vj dt so: ~v T~ ds = j~vj dt = ~v dt j~vj Therefore, if the initial point on the curve occurs at t = a and the final point is at t = b, the integral for work is given by: Z b W = F~ ² ~v dt a d~r Note that ~v dt = dt dt = d~r, so the integral for work can be written more succinctly as: Z W = F~ ² d~r C This is called a line integral whether the path C is a straight line or a curve. Example: Let’s consider the same helical segment C as before: ~r = hx(t); y(t); z(t)i = h¼ cos t; ¼ sin t; ti 0 · t · ¼ and let’s consider the work along this path by the force: F~ = (z ¡ ¼)~i = hz ¡ ¼; 0; 0i The force is parallel to the x-axis at all points and it is shrinking to 0 as we approach the last point (¡¼; 0; ¼) on the helical segment. At any point (x; y; z) = (¼ cos t; ¼ sin t; t), the force is F~ = ht ¡ ¼; 0; 0i and d~r is given by: d~r d~r = dt = h¡¼ sin t; ¼ cos t; 1i dt dt Therefore, F~ ² d~r = ht ¡ ¼; 0; 0i ² h¡¼ sin t; ¼ cos t; 1i dt = ¼(¼ ¡ t) sin t dt and the work can be easily calculated using integration by parts: Z Z ¼ W = F~ ² d~r = ¼ (¼ ¡ t) sin t dt = ¼2 C 0 There are two interesting things to be observed about this integral. As you learned in Calculus I, reversing the limits of integration changes the answer by a negative sign: Z 0 ¼ (¼ ¡ t) sin t dt = ¡¼2 ¼ We interpret this as reversing direction along the path. If we start with t = ¼ then we are beginning at the point (¡¼; 0; ¼) and if we stop with t = 0 then we are ending up at (¼; 0; 0). We use the notation ¡C to denote the path C traversed in the opposite direction. Therefore, Z Z F~ ² d~r = ¡ F~ ² d~r ¡C C The second interesting thing to point out is the effect of changing the path between (¼; 0; 0) and (¡¼; 0; ¼). Consider the following path : hx; y; zi = h¼ ¡ 2¼t; 0; 2¼t ¡ ¼t2i for 0 · t · 1 Let’s call this path Ω. Ω is a parabolic segment connecting (¼; 0; 0) to (¡¼; 0; ¼). At any point along this path: d~r F~ ² d~r = hz ¡ ¼; 0; 0i ² dt dt = h2¼t ¡ ¼t2 ¡ ¼; 0; 0i ² h¡2¼; 0; 2¼ ¡ 2¼ti dt = 2¼(¼t2 + ¼ ¡ 2¼t) dt Therefore, the work done by F~ along Ω is: Z Z 1 2 F~ ² d~r = 2¼ (¼t2 + ¼ ¡ 2¼t) dt = ¼3 Ω 0 3 R R ~ ~ Note that C F ² d~r is not the same as Ω F ² d~r even though C and Ω have the same starting point and the same ending point. We say that the integral is path dependent. An important result we will encounter later is that, for certain vector fields, called conservative vector fields, the path connecting a point P to a point Q will not affect the answer. Since the path does makes a difference for F~ = (z ¡¼)~i, this would be an example of a nonconservative vector field. Example: Let F~ = hy ¡ x; z ¡ y; x ¡ zi. Integrate this vector field over the straight line segment L connecting (0; 0; 1) to (1; 2; 2).
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