(Vector) Line IntegralsDr Aiden PriceEGB241Week 81/15(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week2/15Last Week• Covered the processes for finding surfaces and volumes.• It is also possible to find integrals along a line, known asline integrals.• To make things more complex, the focus of this lecture ison line integrals of vector fields.(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week3/15Scalar Vs Vector Line IntegralsTo get a better idea of the difference between the two, let’scheck out some quick animations:(1) Check out this Wikipedia .gif on line integrals (scalar).• Magnitude at a point on the line is the distance to the axis.(2) Check out this Wikipedia .gif on line integrals (scalar).• Magnitude at a point on the line is how much the vectorat that point in aligned with the direction of the line.(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week4/15Line Integrals of Vector Fields I• To begin, consider a vector field in Cartesian coordinates,with potentially changing direction at every point.• This vector field, F, can be written asF(x, y, z) = Fx(x, y, z)ˆ x + Fy(x, y, z)ˆ y + Fz(x, y, z)ˆ z.• Now consider some line, C, travelling through F frompoint a to point b.(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week5/15Line Integrals of Vector Fields II• To find how much of F is aligned with the direction of Cat every point along C, we must integrate!• The change along our line? The differential length!• Remember, in Cartesian coordinates:• dl = dx xˆ + dy yˆ + dz zˆ.• Working with vectors – use dot products!(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week6/15Line Integrals of Vector Fields III• By using dot products, some familiar ideas resurface:ZC F · dl = ZC Fxxˆ · dl + ZC Fyyˆ · dl + ZC Fzzˆ · dl= ZC Fxxˆ · (dx xˆ) + ZC Fyyˆ · (dy yˆ) + ZC Fzzˆ · (dz zˆ)= ZC Fxdx + ZC Fydy + ZC Fzdz• This equation will only work, however, if C changes inonly one direction at a time.• What about curved, diagonal or otherwise “squiggly” lines?(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week7/15Parameterisation• When the line traverses more than one dimension at once,parameterisation is used to express F as a series ofimplicit equations, with independent variable t (recallMXB125).• Mathematically, we replace parts of F as follows:F(x(t), y(t), z(t)) = F(r(t)).• Which, when integrated (derivation excluded) givesZC F · dr = Zab F(r(t)) · r0(t) dt.(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week8/15Parameterisation II• There are infinitely many lines with infinitely many shapes.• In EGB241, consider straight lines and perfectly roundlines.• Mainly for ease; there are plenty more plausible curves butnone so easy to consider as a straight or circular line.(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week9/15Parameterisation – Straight• Think of a straight line as one having an intercept and aslope.• If our line is represented solely as a function of t, then itessentially becomes a linear line (albeit with x, y and zcomponents).• Treat one point as the starting point and another as theend point. Then the starting point, c1, acts as ourintercept (when t = 0) and the difference from the start tothe end point, c2 – c1, is our slope.r(t) = (c2 – c1)t + c1, 0 ≤ t ≤ 1.(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week10/15Parameterisation – Example• Calculate the integral of a line on a vector field, where• F(x, y, z) = xz xˆ – yz zˆ, and• C is the line segment from (-1, 2, 0) to (3, 0, 1).(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week11/15Parameterisation – Circular• Consider a circular line in Cartesian coordinates.• There is no change in height, assuming the circle is flat, sothe z component can be ignored.• What if we consider cylindrical coordinates?• There is no height, so z is not a factor. There is nochange in radius, as our circle is centered correctly.• Therefore, F in cylindrical coordinates will have only onedifference length, dlφ.l(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week12/15Line Integral – Example I• Ampere’s law: The line integral of the magnetic field, H,along a closed path, C, is equal to the current, I, passingthrough the surface enclosed by the path. Writtenmathematically:IC H · dl = I,where H represents a closed line integral.• Consider an infinite long conductor aligned along thez-axis, carrying a steady current, I.(1) Draw a simple diagram illustrating this line integral.(2) Determine the magnetic field, H, due to the current, I.(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week13/15Line Integral – Example II• Consider now an electric field, E, which has a potentialdifference between two points of Vp1,p2 = V1 – V2, orVp1,p2 = – Zp2p1 E · dl.• Consider a parallel plate capacitor with an electric fieldE = -1000V/m2 zˆ and a plate separation of d = 1mm.(1) Draw a simple diagram illustrating a line integral from apoint p1 in the first plate to a point p2 in the second plate.(2) Calculate the potential difference between the plates.(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week14/15Line Integral – Example III• Use this image to help answer the following questions:• Given a vector function F = xyxˆ + (3x – y2)2yˆ, evaluatethe line integral R F · dl from point P1 = (5, 6) toP2 = (3, 3) along(1) The direct path, P1P2,(2) and along the path P1AP2, where A = (5, 3).(Vector) LineIntegralsEGB241Line IntegralsLine Integralsof VectorFieldsParametricEquationsNext Week15/15Next Week• Gradients.• Divergence.
