Line integral and its application.

27. Line integrals Let I be an open interval and let �x: I −→ Rn , be a parametrised differentiable curve. If [a,b] ⊂ I then let C = �x([a,b]) be the image of [a,b] and let f : C −→ R be a function. Definition 27.1. The line integral of f along C is �� b f ds = f(�x(u))��x�(u)� du. Ca Let u: J −→ I be a diffeomorphism between two open intervals. Suppose that u is C1 . Definition 27.2. We say that u is orientation-preserving if u�(t) > 0 for every t ∈ J. We say that u is orientation-reversing if u�(t) < 0 for every t ∈ J. Notice that u is always either orientation-preserving or orientation-reversing (this is a consequence of the intermediate value theorem, appllie to the continuous function u�(t)). Define a function �y : J −→ Rn , by composition, �y(t)= �x(u(t)), so that �y = �x u.◦Now suppose that u([c,d]) = [a,b]. Then C = �y([c,d]), so that �y gives another parametrisation of C. Lemma 27.3. � b � d f(�x(u))��x�(u)� du = f(�y(t))��y�(t)� dt. ac Proof. We deal with the case that u is orientation-reversing. The case that u is orientation-preserving is similar and easier. 1 � As u is orientation-reversing, we have u(c)= b and u(d)= a and so, � d � d f(�y(t))��y�(t)� dt = f(�x(u(t)))�u(t)�x�(u(t))� dt cc � d = − f(�x(u(t)))��x�(u(t))�u�(t)dt �c a = − f(�x(u))��x�(u)� du b� b = f(�x(u))��x�(u)� du. � a Now suppose that we have a vector field on C, .F�: C −→ Rn Definition 27.4. The line integral of F�along C is �� b F�d�s = F�(�x(u)) �x�(u)du.·· Ca Note that now the orientation is very important: Lemma 27.5. � b F�(�x(u)) �x�(u)du = �� dF�(�y(t)) · �y�(t)dtu�(t) > 0 c · � cdF�(�y(t)) �y�(t)dtu�(t) < 0a −· Proof. We deal with the case that u is orientation-reversing. The case that u is orientation-preserving is similar and easier. As u is orientation-reversing, we have u(c)= b and u(d)= a and so, � d � d F�(�y(t)) �y�(t)dt = F�(�x(u(t))) �x�(u(t))u�(t)dt·· c �ca= F�(�x(u)) �x�(u)du· b � b = − F�(�x(u)) �x�(u)du. �· a Example 27.6. If C is a piece of wire and f(�x) is the mass density at �x ∈ C, then the line integral f ds, C is the total mass of the curve. Clearly this is always positive, whichever way you parametrise the curve. 2 � Example 27.7. If C is an oriented path and F�(�x) is a force field, then the line integral F�d�s,· C is the work done when moving along C. If we reverse the orientation, then the sign flips. For example, imagine C is a spiral staircase and F�is the force due to gravity. Going up the staircase costs energy and going down we gain energy. Definition 27.8. Let U ⊂ Rk and V ⊂ Rl be two open subsets. We say that f : U −→ V, is smooth if all higher order partials ∂nf (x1,x2,...,xk),∂xi1 ...∂xin exist and are continuous. Definition 27.9. Now suppose that X ⊂ Rk and Y ⊂ Rl are any two subsets. We say that a function f�: X −→ Y, is smooth, if given any point �a ∈ X we may find �a ∈ U ⊂ Rk open, and a smooth function ,F�: U −→ Rl such that f�(�x)= F�(�x), where �x ∈ X ∩U (equivalently f�|X∩U = F�|X∩U ), and we put Df�(�x)= DF�(�x). We say that f�is a (smooth) diffeomorphism if f�is bijective and both f�and f�−1 are smooth. Notice that in the definition of a diffeomorphism we are now requiriin more than we did (before we just required that f�and f�−1 were differentiable). Remark 27.10. Note that if X is not very “big” then Df(�x) might not be unique. For example, if X = {�x} is a single point, then there are very many different ways to extend f�to a function F�in an open neighbourhood of �x. In the examples we consider in this class, this will not be an issue (namely, manifolds with boundary). 3 Example 27.11. The map �x:[a,b] −→ Rn , is smooth if and only if there is a constant �> 0 and a smooth function �y :(a − �,b + �) −→ Rn , whose restriction to [a,b] is the function �x, �y(t)= �x(t) for all t ∈ [a,b]. Lemma 27.12. If �x:[a,b] −→ Rn , is injective for all t ∈ [a,b], then �x:[a,b] −→ C = �x([a,b]), is a diffeomorphism. 4 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.