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Line, Surface and Volume Integrals

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Line, Surface and Volume Integrals Line, Surface and Volume Integrals 1 Line integrals Z Z Z Á dr; a ¢ dr; a £ dr (1) C C C (Á is a scalar ¯eld and a is a vector ¯eld) We divide the path C joining the points A and B into N small line elements ¢rp, p = 1;:::;N. If (xp; yp; zp) is any point on the line element ¢rp, then the second type of line integral in Eq. (1) is de¯ned as Z XN a ¢ dr = lim a(xp; yp; zp) ¢ rp N!1 C p=1 where it is assumed that all j¢rpj ! 0 as N ! 1. 2 Evaluating line integrals The ¯rst type of line integral in Eq. (1) can be written as Z Z Z Á dr = i Á(x; y; z) dx + j Á(x; y; z) dy C CZ C +k Á(x; y; z) dz C The three integrals on the RHS are ordinary scalar integrals. The second and third line integrals in Eq. (1) can also be reduced to a set of scalar integrals by writing the vector ¯eld a in terms of its Cartesian components as a = axi + ayj + azk. Thus, Z Z a ¢ dr = (axi + ayj + azk) ¢ (dxi + dyj + dzk) C ZC = (axdx + aydy + azdz) ZC Z Z = axdx + aydy + azdz C C C 3 Some useful properties about line integrals: 1. Reversing the path of integration changes the sign of the integral. That is, Z B Z A a ¢ dr = ¡ a ¢ dr A B 2. If the path of integration is subdivided into smaller segments, then the sum of the separate line integrals along each segment is equal to the line integral along the whole path. That is, Z B Z P Z B a ¢ dr = a ¢ dr + a ¢ dr A A P 4 Example R Evaluate the line integral I = C a ¢ dr, where a = (x + y)i + (y ¡ x)j, along each of the paths in the xy-plane shown in the ¯gure below, namely, 1. the parabola y2 = x from (1; 1) to (4; 2), 2. the curve x = 2u2 + u + 1, y = 1 + u2 from (1; 1) to (4; 2), 3. the line y = 1 from (1; 1) to (4; 1), followed by the line y = x from (4; 1) to (4; 2). FIG. 1: Di®erent possible paths between points (1; 1) and (4; 2). 5 Answer Since each of the path lies entirely in the xy-plane, we have dr = dx i + dy j. Therefore, Z Z I = a ¢ dr = [(x + y)dx + (y ¡ x)dy]: (2) C C We now evaluate the line integral along each path. Case (i). Along the parabola y2 = x we have 2y dy = dx. Substituting for x in Eq. (2) and using just the limits on y, we obtain Z (4;2) I = [(x + y)dx + (y ¡ x)dy] (1;1) Z 2 1 = [(y2 + y)2y + (y ¡ y2)]dy = 11 1 3 6 Case (ii). The second path is given in terms of parameter u. We could eliminate u between two equations to obtain a relationship between x and y directly, and proceed as above, but it us usually quicker to write the line integral in terms of parameter u. Along the curve x = 2u2 + u + 1, y = 1 + u2, we have dx = (4u + 1) du and dy = 2u du. Substituting for x and y in Eq. (2) and writing the correct limits on u, we obtain Z (4;2) I = [(x + y)dx + (y ¡ x)dy] (1;1) Z 1 = [(3u2 + u + 2)(4u + 1) ¡ (u2 + u)2u]du 0 2 = 10 3 7 Case (iii). For the third path the line integral must be evaluated along the two line segments separately and the results added together. First, along the line y = 1, we have dy = 0. Substituting this into Eq. (2) and using just the limits on x for this segment, we obtain Z (4;1) Z 4 [(x + y)dx + (y ¡ x)dy] = (x + 1)dx (1;1) 1 1 = 10 2 Along the line x = 4, we have dx = 0. Substituting this into Eq. (2) and using just the limits on y, we obtain Z (4;2) Z 2 [(x + y)dx + (y ¡ x)dy] = (y ¡ 4)dy (4;1) 1 1 = ¡2 2 Therefore, the value of the line integral along the 1 1 whole path is 10 2 ¡ 2 2 = 8. 8 Connectivity of regions FIG. 2: (a) A simply connected region; (b) a doubly connected region; (c) a triply connected region. A plane region R is simply connected if any closed curve within R can be continuously shrunk to a point without leaving the region. If, however, the region R contains a hole then there exits simple closed curves that cannot be shrunk to a point without leaving R. Such a region is doubly connected. Similarly, a region with n ¡ 1 holes is said to be n-fold connected, or multiply connected. 9 Green's theorem in a plane Suppose the functions P (x; y), Q(x; y) and their partial derivatives are single-valued, ¯nite and continuous inside and on the boundary C of some simply connected region R in the xy-plane. Green's theorem in a plane then states that I ZZ µ ¶ @Q @P (P dx + Q dy) = ¡ dx dy (3) C R @x @y To prove this, let us consider the simply connected region R below. FIG. 3: A simply connected region R bounded by the curve C. 10 Let y = y1(x) and y = y2(x) be the equations of the curves ST U and SV U respectively. We then write ZZ Z Z @P b y2(x) @P = dx dy R @y a y1(x) @y Z b = dx [P (x; y)]y=y2(x) y=y1(x) a Z b = [P (x; y2(x)) ¡ P (x; y1(x))] dx a Z b Z b = ¡ P (x; y1(x))dx ¡ P (x; y2(x))dx Ia a = ¡ P dx C 11 If we now let x = x1(y) and x = x2(y) be the equations of the curves T SV and TUV respectively, we can similarly show that ZZ Z Z @Q d x2(y) @Q dx dy = dy dx R @x c x1(y) @x Z d = dy [Q(x; y)]x=x2(y) x=x1(y) c Z d = [Q(x2(y); y) ¡ Q(x1(y); y)] dy c Z d Z d = Q(x1; y)dy + Q(x2; y)dy Ic c = Qdy C Subtracting these two results gives Green's theorem in a plane. 12 Example Show that the area of a region R enclosed by a simple closed curve C is given by H H H 1 A = 2 C (x dy ¡ y dx) = C x dy = ¡ C y dx. Hence, calculate the area of the ellipse x = a cos Á, y = b sin Á. Answer In Green's theorem, put P = ¡y and Q = x. Then I ZZ ZZ (x dy¡y dx) = (1+1)dx dy = 2 dx dy = 2A C R R Therefore, the area of the region is H 1 A = 2 C (xdy ¡ ydx). Alternatively, we could put P = 0 and Q = x and H obtain A = xdy, or put P = ¡y and Q = 0, C H which gives A = ¡ C ydx. 13 The area of the ellipse x = a cos Á, y = b sin Á is given by I 1 A = (xdy ¡ ydx) 2 C Z 1 2¼ = ab(cos2 Á + sin2 Á) dÁ 2 0 Z ab 2¼ = dÁ = ¼ab: 2 0 14 Conservative ¯elds and potentials R For line integrals of the form C a ¢ dr, there exists a class of vector ¯elds for which the line integral between two points is independent of the path taken. Such vector ¯elds are called conservative. A vector ¯eld a that has continuous partial derivatives in a simply connected region R is conservative if, and only if, any of the following is true. R B 1. The integral A a ¢ dr, where A and B lie in the region R, is independent of the path from A to H B. Hence the integral C a ¢ dr around any closed loop in R is zero. 2. There exits a single-valued function Á of position such that a = rÁ. 3. r £ a = 0 4. a ¢ dr is an exact di®erential. 15 If the line integral from A to B is independent of the path taken between the points, then its value must be a function only of the positions of A and B. We write Z B a ¢ dr = Á(B) ¡ Á(A) (4) A which de¯nes a single-valued scalar function of position Á. If the points A and B are separated by an in¯nitesimal displacement dr, then Eq. (4) becomes a ¢ dr = dÁ which requires a ¢ dr to be an exact di®erential. But dÁ = rÁ ¢ dr, so (a ¡ rÁ) ¢ dr = 0: Since dr is arbitrary, a = rÁ, which implies r £ a = 0. 16 Example R B Evaluate the line integral I = A a ¢ dr, where a = (xy2 + z)i + (x2y + 2)j + xk, A is the point (c; c; h) and B is the point (2c; c=2; h), along the di®erent paths 1. C1, given by x = cu, y = c=u, z = h, and 2. C2, given by 2y = 3c ¡ x, z = h. Show that the vector ¯eld a is in fact conservative, and ¯nd Á such that a = rÁ. 17 Answer Expanding out the integrand, we have Z (2c;c=2;h) £ ¤ I = (xy2 + z)dx + (x2y + 2)dy + xdz (c;c;h) (5) 2 (i).
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