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Unformatted text preview: Section 17.3 The Fundamental Theorem for Line Integrals FTC Number 2 Recall that in single variable calculus, in order to evaluate a definite integral, we use the fundamental theorem of calculus. Specifically, we have integraldisplay b a F ( x ) dx = F ( b ) F ( a ) . Provided F ( x ) is a continuous function, an antiderivative always exists, so FTC can always be used to evaluate single variable integrals of continuous functions. In this section, we consider a generalization of FTC to line integrals. The difference in this case however is that we shall not always be able to apply it. 1. The Fundamental Theorem In multivariable calculus, it doesnt make sense to talk about the derivative since a function of more than one variable can be differ entiated with respect to different variables. In our earlier discussions on differentiation, we concluded that though we could not define the derivative, we could define a generalization of the derivative  specif ically, we said that the gradient vector of a function f was the closest generalization to the derivative we could get. With this in mind, we have the following generalization of FTC in single variable. Result 1.1. Let C be a smooth curve parameterized by vector r ( t ) with a lessorequalslant t lessorequalslant b and suppose f is a differentiable function of two or three variables whose gradient vector is continuous on C . Then integraldisplay C f dvector r = f ( vector r ( b )) f ( vector r ( a )) . This results tells us that if vector F is a conservative vector field (meaning it is the gradient vector field of some function f ( x, y )), then rather than evaluating the integral by hand, we can simply find the potential function and take its difference at the end points. We illustrate. Example 1.2. ( i ) Suppose vector F = 2 x vector i +2 y vector j +2 z vector k . Evaluate integraltext C vector F dvector r where vector r = 3 t 2 vector i + 3 t 2 vector j + 2 t vector k with 0 lessorequalslant t lessorequalslant 1. Observe that vector F is a gradient vector field with potential func tion f ( x, y, z ) = x 2 + y 2 + z 2 , so we can use FTC to evaluate the integral. Since the end points are (0 , , 0) and (3 , 3 , 2), we have integraldisplay C vector F dvector r = f (3 , 3 , 2) f (0 , , 0) = 9 + 9 + 4 = 22 1 2 ( ii ) Suppose vector F = 2 x vector i + 2 y vector j + 2 z vector k . Evaluate integraltext C vector F dvector r where vector r is drawn below. (0,0,0) (3,3,2) Since this is a conservative vector field, we can use FTC. Since the end points are (0 , , 0) and (3 , 3 , 2), we have integraldisplay C vector F dvector r = f (3 , 3 , 2) f (0 , , 0) = 9 + 9 + 4 = 22 In the last two examples, observe that the answer was the same. This is simply because the vector field vector F is conservative, so the integral can be evaluated using FTC. This observation motivates the following definition....
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This note was uploaded on 10/28/2010 for the course MA 261 taught by Professor Stefanov during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 Stefanov
 Calculus, Fundamental Theorem Of Calculus, Integrals

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