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Line Integral Along a Piecewise C1 Curve: 1 Let C Be a Continuous Curve Which Is Piecewise of Class C I.E

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Line Integral Along a Piecewise C1 Curve: 1 Let C Be a Continuous Curve Which Is Piecewise of Class C I.E Properties of line integrals: i) Linearity: (F + G) · dx = F·dx + G·dx, (2.17) Z Z Z C C C (λF) · dx = λ F·dx, (2.18) Z Z C C where λ is a constant scalar. These properties follow immediately from the definition (2.14) and the corresponding properties of the Riemann integral. ii) Additivity: If C is a C1 curve that is the union of two curves C1 and C2 joined end-to-end and consis- tently oriented ( C = C1 ∪ C 2), then F·dx = F·dx + F·dx (2.19) Z Z Z C C1 C2 Line integral along a piecewise C1 curve: 1 Let C be a continuous curve which is piecewise of class C i.e. C = C1 ∪ ··· ∪ C n, where 1 the individual pieces Ci, i = 1 ,...,n are of class C . Motivated by equation (2.19), we define the line integral of a vector field F along C by F·dx = F·dx + · · · + F·dx. (2.20) Z Z Z C C1 Cn Each line integral on the right is, of course, defined as a Riemann integral by equation (2.14). Example 2.3: Compute the line integral of the vector field F = ( y, −2x) along the piecewise C1 curve consisting of the two straight line segments joining ( −b, b ) to (0 , 0) and (0 , 0) to (2 b, b ), where b is a positive constant. Solution: For C1, x = g1(t) = ( t, −t), with −b ≤ t ≤ 0, giving ′ g1(t) = (1 , −1) , and F(g1(t)) = ( −t, −2t). By the definition (2.14), 0 F·dx = (−t, −2t) · (1 , −1) dt Z Z−b C1 0 1 2 = t dt = − 2 b . Z−b 39 y (−b,b) (2b,b) C 1 C 2 (0,0) x For C2, x = g2(t) = (2 t, t ), 0 ≤ t ≤ b, and a similar calculation yields F·dx = −b2. Z C2 Finally, by (2.20), F·dx = F·dx + F·dx Z Z Z C C1 C2 1 2 2 3 2 = − 2 b − b = − 2 b . Exercise 2.8: Calculate the line integral of the vector field F = ( y, −2x) along the piecewise smooth curve consisting of two parabolic segments as drawn. Answer: 2b2. y (b,b) x (0,0) (2b,0) Reversal of orientation: If C is a curve in Rn with a specific orientation, we denote by −C the curve that is obtained by reversing the orientation. Specifically if C is given by x = g(t), a ≤ t ≤ b, then −C is given by x = gˆ(τ), a ≤ τ ≤ b, where gˆ(τ) = g(a + b − τ). 40 Observe that gˆ(a) = g(b) and gˆ(b) = g(a), which reverses the orientation. It is an important result that reversing the orientation of a curve changes the sign of the line integral of a vector field along the curve . Proposition 2.2: If C is a piecewise-smooth curve and F is a vector field continuous on C, then F·dx = − F·dx. Z Z −C C Proof: By definition of line integral b F·dx = F(ˆg(τ)) · gˆ′(τ)dτ Z Zτ=a −C b ′ = − F(g(a + b − τ)) · g (a + b − τ)dτ (by the equation for gˆ(τ)) Zτ=a a ′ = − F(g(t)) · g (t)( −1) dt (by the change of variable t = a + b − τ) Zt=b b ′ = − F(g(t)) · g (t)dt (reverse the limits of integration) Zt=a = − F·dx. (by definition of line integral) Z C Comment: The change in sign is physically reasonable if one thinks in terms of work done by a force field F; for example, if the height of an object above the earth’s surface is increased, the work done by the gravitational field is negative , whereas if an object falls, the work done is positive . 2.3 Path-independent line integrals Let F : U → Rn be a continuous vector field on the connected open set U ∈ Rn. Consider two points x1, x2 in U, and imagine all possible piecewise smooth curves in U joining x1 to x2. In general, the value of the line integral F·dx will depend on the particular curve C joining Z C x1 to x2. In physical terms, thinking of F as a force field, the work done on the particle as it moves from x1 to x2 in general depends on the path followed by the particle. There are, however, certain special vector fields (force fields) with the property that the line integral (the work done) depends only on the endpoints of the curve and not on the particular curve joining the two points. 41 As an example, consider the vector field (see Problem Set 1, #14): x y E(x, y ) = −kq , . (2.21) (x2 + y2)3/2 (x2 + y2)3/2 As endpoints, consider x1 = (1 , 0) and x2 = (0 , 1). You will find that for any piecewise smooth curve C joining x1 to x2, E·dx has the same value, namely zero. Z C Exercise 2.9: Show that the line integral of the vector field (2.21) equals zero for each of the curves C1 and C2 joining (1 , 0) to (0 , 1), where C1 is the quarter circle. (0,1) C 1 C 2 (0,0) (1,0) Definition: Let F be a continuous vector field on a connected open set in Rn. The line integral F·dx Z C is path-independent in U mean that given any two points x1, x2 in U, the line integral has the same value for all piecewise smooth curves in U that join x1 to x2. Exercise 2.9 gives a hint that the line integral of the vector field (2.21) is path-independent in U = R2 − { (0 , 0) }. Of course we cannot prove that a line integral is path-independent by calculating its value all different curves joining different pairs of points because there are infinitely many possibilities! So an important question is: how can we tell whether a given line integral is path-independent or not ? To find out, let us be guided by one of the most important results in elementary calculus, the first Fundamental Theorem. 2.3.1 First Fundamental Theorem for Line Integrals Recall that if f is continuous on an interval [ a, b ], then the new function g defined by x g(x) = f(t) dt, a ≤ x ≤ b, (2.22) Za is such that g′(x) = f(x). (2.23) This result is the first Fundamental Theorem of Calculus (FTC). 42 Q: Can we extent this theorem to line integrals? A: Yes, provided the line-integral is path-independent . If the line integral F·dx is path-independent and C joins x0 to x, we denote the line- Z C integral by x F·dx. Zx0 In this case we can define a new function – a scalar field φ – by x φ(x) = F·dx, Zx0 in analogy with (2.22). We can now state the first Fundamental Theorem for line integrals . Theorem 2.1: Let U be a connected open subset of Rn, and let F : U → Rn be a continuous vector field whose line integral is path independent in U. If x φ(x) = F·dx, (2.24) Zx0 where x0 is a specified point, then, ∇φ(x) = F(x) (2.25) for all x ∈ U . Proof: For simplicity we give the proof in R2. With φ defined by (2.24), we have to prove that ∂φ ∂φ = F1, = F2, ∂x ∂y where F1 and F2 are the components of F. The key idea is this: since the line integral is path-independent, we are free to make a special choice of the curve joining x0 = ( x0, y 0) to x = ( x, y ), i.e. to choose a “custom- designed” curve. Figure 2.6 shows the curve we need. Suitable parametrizations for C1 and C2 are x = g1(t) = ( x0, t ), y 0 ≤ t ≤ y, x = g2(t) = ( t, y ), x 0 ≤ t ≤ x. 43.
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