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Unformatted text preview: Section 17.2 Line Integrals “Integrating Vector Fields and Functions along a Curve” In this section we consider the problem of integrating functions, both scalar and vector (vector fields) along a curve C in the plane. We want the definition to generalize the ideas of integration we have already developed in Calculus I and II. We start with the scalar problem. 1. Line Integrals of a Scalar Function with respect to Distance, x and y We want to define the integral of a continuous function f ( x, y ) of two variables over a general smooth curve C in the plane. We do this as follows: ( i ) Suppose C is some curve in space parameterized by the func tions ( x ( t ) , y ( t )) with a lessorequalslant t lessorequalslant b , or equivalently with vector equation vector r ( t ) = x ( t ) vector i + y ( t ) vector j . ( ii ) Suppose f ( x, y ) is a function which is continuous on C . ( iii ) Break up the interval [ a, b ] into n equal sized intervals of lengths Δ s = ( b a ) /n . ( iv ) Fix a point (¯ x i , ¯ y i ) in each interval. ( v ) We can construct the Riemann sum n summationdisplay i =1 f (¯ x i , ¯ y j )Δ s ( vi ) Observe that this Riemann sum approximates the area bounded between the function f ( x, y ) and the curve C weighted by the xyplane (see illustration below). In particular, this really does generalize the idea of the single variable integral. 1 2 We are now ready to formally define a line integral with respect to distance. Definition 1.1. If f ( x, y ) is defined on a smooth curve C given by the equation vector r ( t ) = x ( t ) vector i + y ( t ) vector j , then the integral of f along C is defined by integraldisplay C f ( x, y ) ds = lim n →∞ n summationdisplay i =1 f (¯ x i , ¯ y j )Δ s provided this limit exists. Observe that the integral is with respect to s and the function itself is with respect to x and y . This means we cannot directly integrate it. However, we can integrate using the following method: ( i ) Rewrite f ( x, y ) in terms of t by substituting x = x ( t ) and y = y ( t ). ( ii ) Replace ds in the integral by  vector r ′ ( t )  = radicalBigg parenleftbigg dx dt parenrightbigg 2 + parenleftbigg dy dt parenrightbigg 2 dt (substitution). ( iii ) We can now integrate with respect to t with limits a lessorequalslant t lessorequalslant b . Summarizing, we evaluate line integrals using the following: Result 1.2. integraldisplay C f ( x, y ) ds = integraldisplay b a f ( x ( t ) , y ( t )) radicalBigg parenleftbigg dx dt parenrightbigg 2 + parenleftbigg dy dt parenrightbigg 2 dt = integraldisplay b a f ( x ( t ) , y ( t ))  vector r ′ ( t )  dt By the definition of the line integral, we are integrating with respect to the arc length s . There are two other line integrals we shall be interested  line integrals with respect to x and y . Specifically, we can modify the formula, so instead of integrating with respect to s , we integrate with respect to x or y . In this case, since x = x ( t ) and y =...
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 Fall '08
 Stefanov
 Calculus, Derivative, Integrals, Scalar, Vector Space, Manifold

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