ENGI 443 Line Integrals; Green s Theorem Page 8. 8. Line Integrals Two applications of line integrals are treated here: the evaluation of work done on a particle as it travels along a curve in the presence of a [vector field] force; and the evaluation of the location of the centre of mass of a wire. Work done: The work done by a force F in moving an elementary distance r along a curve is approximately the product of the component of the force in the direction of r and the distance r travelled: W F r Fcos r omponents of a force at right angles to the path do no work at all. Integrating along the curve yields the total work done by the force F in moving along the curve : W F dr dx dy dz f dx f dy f dz f f f 3 3 t t W F dr F dr
ENGI 443 Line Integrals; Green s Theorem Page 8. Example 8. Find the work done by F y x z T in moving once around the closed curve (defined in parametric form by x cos t, y sin t, z, t ). F y sin t x cost z x cos t sin t dr d d d y t t r t t z dr W F sin cos F sin cos dr F Note that Fv Fvˆ everywhere on the curve, so that dr W (the length of the path around the circle). Also note that y x z T curl ˆ F F k everywhere in The lesser curvature of the circular lines of force further away from the z axis is balanced exactly by the increased transverse force, so that curl F is the same in all of 3. We shall see later (Stokes theorem, hapter ) that the work done is also the normal component of the curl integrated over the area enclosed by the closed curve. In this case W F nˆ A k ˆ kˆ. 3.
ENGI 443 Line Integrals; Green s Theorem Page 8.3 Example 8. (continued) Example 8. Find the work done by x y z T F in moving around the curve (defined in parametric form by x cos t, y sin t, z, t ). F dr x cost y sin t z cost sin t d sin t cost dr F cos t sin t sin t cos t W In this case, the force is orthogonal to the direction of motion at all times and no work is done at any time.
ENGI 443 Line Integrals; Green s Theorem Page 8.4 If the initial and terminal points of a curve are identical and the curve meets itself nowhere else, then the curve is said to be a simple closed curve. Notation: When is a simple closed curve, write F dr as F dr. F is a conservative vector field if and only if F dr for all simple closed curves in the domain. Be careful of where the endpoints are and of the order in which they appear (the t t dr dr orientation of the curve). The identity F F leads to the result t t F dr F dr simple closed curves Another Application of Line Integrals: The Mass of a Wire Let be a segment t t t of wire of line density x, y, z. Then m x, y, z s First moments about the coordinate planes: ds ds m ds t t M r m r s t M r t ds The location r of the centre of mass of the wire is r M, where the moment m t t t t ds ds ds dr dx dy dz M r, m and.
ENGI 443 Line Integrals; Green s Theorem Page 8.5 Example 8.3 Find the mass and centre of mass of a wire (described in parametric form by x cos t, y sin t, z t, t ) of line density z. [The shape of the wire is one revolution of a helix, aligned along the z axis, centre the origin.] Let c = cos t, s = sin t. r c s d s r c t ds s c z t 3 ds t m ds t 3 m 3 3 t M r t ds 3 T t c t s t x component: Integration by parts. t c t s tc t c t s tc 4
ENGI 443 Line Integrals; Green s Theorem Page 8.6 Example 8.3 (continued) y component: For all integrable functions f t and for all constants a note that a a f t a if f t is an O function f t if f t is an EVEN function [The areas under odd graphs cancel out over [ a, +a] but the areas under even graphs double up.] t sin t is an odd function of t t s z component: 3 t is also an odd function of t 3 t Therefore M 4 ˆi M 3 6 r 4 ˆi ˆi 3 m 6 The centre of mass is therefore at,,
ENGI 443 Line Integrals; Green s Theorem Page 8.7 Green s Theorem Some definitions: r t x t ˆ i y t ˆ, j a t b ) is A curve on (defined in parametric form by closed iff xa, ya x b, y b. The curve is simple iff t t r r for all, t t such that a t t b ; (that is, the curve neither touches nor intersects itself, except possibly at the end points). Example 8.4 Two simple curves: open closed Two non-simple curves: open closed Orientation of closed curves: A closed curve has a positive orientation iff a point anticlockwise sense as the value of the parameter t increases. r t moves around in an
ENGI 443 Line Integrals; Green s Theorem Page 8.8 Example 8.5 Positive orientation Negative orientation Let be the finite region of bounded by. When a particle moves along a curve with positive orientation, is always to the left of the particle. For a simple closed curve enclosing a finite region of function T and for any vector F f f that is differentiable everywhere on and everywhere in, Green s theorem is valid: f f F dr x y da The region is entirely in the xy-plane, so that the unit normal vector everywhere on is ˆk. Let the differential vector da da k ˆ, then Green s theorem can also be written as ˆ F dr F k da curl F da f dx f f f dx f dy F dr da f dy x y and f f f x det x y f y det F z component of F Green s theorem is valid if there are no singularities in. A [non-examinable] proof is provided at the end of this chapter.
ENGI 443 Line Integrals; Green s Theorem Page 8.9 Example 8.6 F x r T : Green s theorem is valid for curve but not for curve. There is a singularity at the origin, which curve encloses. Example 8.7 For x y F x y and as shown, evaluate. F dr Q R P F dr F dr P F dr Q F dr R
ENGI 443 Line Integrals; Green s Theorem Page 8. Example 8.7 (continued) x y F x y Everywhere on the line segment from P to Q, y x (and the parameter t is just x) x dr r and x dx F x Q P F dr x dx 4 x dx 4x x 8 4 4 Everywhere on the line segment from Q to R, y x x dr x r and x dx F R F dr Q x dx x dx x = 4 = 4 Everywhere on the line segment from R to P, y = x dr x r and dx F x P x F dr x dx x dx R = = F dr 4 4
ENGI 443 Line Integrals; Green s Theorem Page 8. Example 8.7 (continued) OR use Green s theorem! f x det x y x y x y f y everywhere on f f da da x y By Green s theorem it then follows that F dr
ENGI 443 Line Integrals; Green s Theorem Page 8. Example 8.8 Find the work done by the force ˆ ˆ F xy i y j in one circuit of the unit square. By Green s theorem, W f f F dr x y da f f y xy x x y x y The region of integration is the square x, y W x da x dy dx x dy dx x y dx x dx x Therefore W The alternative method (using line integration instead of Green s theorem) involves four line integrals, each with different integrands! This alternative is Question 8 on Problem Set 7.
ENGI 443 Line Integrals; Green s Theorem Page 8.3 Path Independence Gradient Vector Fields: If F V, then T V V f f F x y Vyx Vxy x y (provided that the second partial derivatives are all continuous). It therefore follows, for any closed curve and twice differentiable potential function V that V dr Path Independence If F V or F V, then V is a potential function for F. Let the path travel from point P to point P : V V V V dr dx dy dz dv x y z F dr [chain rule] P P V V P V P which is independent of the path between the two points. Therefore work done difference in V by V between endpoints of V dr V P V P [again!] [work done = potential difference]
ENGI 443 Line Integrals; Green s Theorem Page 8.4 omain A region of is a domain if and only if ) For all points P in, there exists a circle, centre P, all of whose interior points are inside ; and ) For all points P and P in, there exists a piecewise smooth curve, entirely in, from P to P. Example 8.9 Are these domains? x, y y x, y x YES (but not simply connected) NO If a domain is not specified, then, by default, it is assumed to be all of.
ENGI 443 Line Integrals; Green s Theorem Page 8.5 When a vector field F is defined on a simply connected domain, these statements are all equivalent (that is, all of them are true or all of them are false): F V for some scalar field V that is differentiable everywhere in ; F is conservative; F dr is path-independent (has the same value no matter which path within is chosen between the two endpoints, for any two endpoints in ); Vend Vstart F dr (for any two endpoints in ); F dr for all closed curves lying entirely in ; f x f y everywhere in ; and F everywhere in (so that the vector field F is irrotational). There must be no singularities anywhere in the domain in order for the above set of equivalencies to be valid. Example 8. Evaluate 3, to x y dx x y dy where is any piecewise-smooth curve from,. x y F x 3y is continuous and differentiable everywhere in f f F is conservative and x y V V x y and x 3y x y F V A potential function that has the correct first partial derivatives is 3 (as does V x xy y c for any constant c), V 8 c, Therefore F dr c x y dx x 3y dy 3 V x xy y
ENGI 443 Line Integrals; Green s Theorem Page 8.6 Example 8. by direct evaluation of the line integral Let us pursue instead a particular path from (, ) to (, ). The straight line path is a segment of the line y x x y. 3 I x y dx x y dy 3 x x dx y y dy 3 4 x y y 8 An alternative evaluation of I F dr is to use x as the parameter in both integrals (that is, to express y in terms of x throughout). Then y x dy dx 3 d x 3 I x y dx x y y x dx x x dx x x dx x x 3 6 4 3 8 38 An alternative path involves going round the other two sides of the triangle, first from (, ) horizontally to (, ) then from there vertically to (, ). On the first leg y dy, so that the second part of the integral vanishes. On the second leg x dx, so that the first part of the integral vanishes. Therefore 3 3 I x y dx x y dy x dx y dy 3 x y y 8
ENGI 443 Line Integrals; Green s Theorem Page 8.7 Example 8. by direct evaluation of the line integral Yet another possibility is 3 an arc of the parabola y x dy 4x dx y x. 3 3 I x y dx x y dy x x dx x 3 x 4x dx x x x 5 dx x x 3 x 6 6 48 8 8, Note that the above suggests that I F dr might be path-independent, because, evaluations along three different paths have all produced the same answer. But this is not a proof of path independence. For a proof, one must establish that F is conservative, either by finding the potential function, or by showing that curl F.
ENGI 443 Line Integrals; Green s Theorem Page 8.8 Outline of a Proof of Green s Theorem [not examinable] Let Px, y ˆ Qx, y F i ˆj. onsider a convex region as shown. Left and right boundaries can be identified. Then Q da x d q y c p y Q dx dy x d c Q x, y x q y x p y dy d d c,,,, Qq y y Q p y y dy Qq y y dy Q p y y dy c c d x q y from y c to y d followed by the path along x p y But the path along from y d back to y c constitutes one complete circuit around the closed path. Q da x Q dy Lower and upper boundaries for the region can also be identified. b hx P P da dy dx y a gx y b a P x, y y h x y g x dx But the path along y g x from x a to x b b a a b,, P x h x P x g x dx,, P x h x dx P x g x dx followed by the path along y hx from x b back to x a constitutes one complete circuit around the closed path. P Q P da P dx y da P dx Q dy x y b a
ENGI 443 Line Integrals; Green s Theorem Page 8.9 Green s Theorem (continued) But F dr P dx P dx Q dy Q dy Therefore Q P da F dr x y This proof can be extended to non-convex regions. Simply divide them up into convex sub-regions and apply Green s theorem to each sub-region. The line integrals along common interior boundaries cancel out because they are travelled in opposite directions along the same line. The boundary of each convex sub-region is a simple closed curve theorem is valid:, for which Green s i i Q P F dr da x y i i Q P F dr da x y i i i i Therefore Green s theorem is also valid for any simply-connected region. [End of hapter 8]
ENGI 443 Line Integrals; Green s Theorem Page 8. [Space for Additional Notes]