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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 , for functions of one dimension, the definite is defined as the 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 between the 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 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 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 ˆ +ydy ˆ +zdz ˆ , with a magnitude of |dl| = 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 (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 √ 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 .) A path must still be specified between beginning point a and ending point b. The direction of dl as a vector is to the path but directed from a towards b along it.

2.1.1 Example: Let the force vector be F = 3( N/m)yxˆ − 4( N/m)xyˆ 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 = (3yxˆ−4xyˆ)·(ˆxdx+ˆydy+ˆzdz)( 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 = [ˆxdt +y ˆ(−2)dt +x ˆ(−1)dt]( m). F = 3( N/m)(−2t + 1)ˆx − 4( N/m)tyˆ. 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.