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Lecture 33 Classical Integration Theorems in the Plane

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Lecture 33 Classical Integration Theorems in the Plane Lecture 33 Classical Integration Theorems in the Plane In this section, we present two very important results on integration over closed curves in the plane, namely, Green’s Theorem and the Divergence Theorem, as a prelude to their important counterparts in R3 involving surface integrals. Green’s Theorem in the Plane (Relevant section from Stewart, Calculus, Early Transcendentals, Sixth Edition: 16.4) 2 1 Let F(x,y) = F1(x,y) i + F2(x,y) j be a vector field in R . Let C be a simple closed (piecewise C ) curve in R2 that encloses a simply connected (i.e., “no holes”) region D ⊂ R. Also assume that the ∂F ∂F partial derivatives 2 and 1 exist at all points (x,y) ∈ D. Then, ∂x ∂y ∂F ∂F F · dr = 2 − 1 dA, (1) IC Z ZD ∂x ∂y where the line integration is performed over C in a counterclockwise direction, with D lying to the left of the path. Note: 1. The integral on the left is a line integral – the circulation of the vector field F over the closed curve C. 2. The integral on the right is a double integral over the region D enclosed by C. Special case: If ∂F ∂F 2 = 1 , (2) ∂x ∂y for all points (x,y) ∈ D, then the integrand in the double integral of Eq. (1) is zero, implying that the circulation integral F · dr is zero. IC We’ve seen this situation before: Recall that the above equality condition for the partial derivatives is the condition for F to be conservative. From this, we would suspect that the circulation integral over a closed curve C would be zero, because of the Generalized Fundamental Theorem of Calculus 247 (the endpoints are the same point). But also recall that we have to be a bit careful – we must be concerned about the possibility of “interference” from singularities that might make the circulation integral nonzero. In this theorem, however, we have assumed that there are no such singularities – the derivatives are assumed to exist and F is therefore defined over the region. Therefore, we may finally conclude that, yes, the line integral is zero. In fact, the integrand in Green’s Theorem, Eq. (1) is the k component of the curl of F. To see this, let us compute it: i j k ~ curl F = ∇× F = ∂/∂x ∂/∂y ∂/∂z F1(x,y) F2(x,y) 0 ∂F ∂F = 0 i + 0j + 2 − 1 k. ∂x ∂y It is important to keep in mind that that the curl of a vector field in R2 is a vector that points in the z-direction. This is related to the convention of assigning a velocity vector that points along an axis of rotation using the right-hand rule. With the above result, we can rewrite Green’s Theorem as F · dr = [curl F]z dA, (3) IC Z ZD where [v]z denotes the z-component of vector v. In this case, the z-component of curl F is its only component. For this reason, we shall omit the subscript z in this section. Examples: 1. The vector field F = −ωyi + ωxj. You will recall that this is the velocity vector field of a thin dθ plate on the xy-plane that is rotating about the z-axis with angular speed = ω. Now let C dt R denote the circle of radius R> 0 centered at the origin. Earlier in the course, we computed the circulation integral of F over CR to be F · dr = 2πωR2. (4) I This was done by a direct calculation of the line integral using the parametrization of CR as r(t) = (R cos t, R sin t). Let us now compute this result using Green’s Theorem. Here F1 = −ωy and F2 = ωx so that ∂F ∂F 2 − 1 = ω − (−ω) = 2ω. (5) ∂x ∂y 248 Then the double integral in Green’s Theorem is ∂F ∂F 2 − 1 dA = 2ω dA = 2ωA(D) = 2ωπR2, (6) Z ZD ∂x ∂y Z ZD where A(D) denotes the area of region D (area of a circular region, radius R). It is sometimes misleading to present this example because people may associate the curl of F, which is 2ω, with the origin, about which the rotation is taking place. But the curl of this vector field is 2ω everywhere. This means that for any simple closed curve in the plane, F · dr = 2ωA(D), (7) IC where A(D) denotes the area of the region D enclosed by C. 1. Compute the circulation of the vector field F = −y2i + xj around the circle of radius 1 centered 2 at the origin. Here, F1 = −y and F2 = x. Then ∂F ∂F 2 − 1 dA = (1 + 2y) dA (8) Z ZD ∂x ∂y Z ZD = dA + 2 y dA Z ZD Z ZD = π. The first integral is the area of the region enclosed by the unit circle. The second integral is zero. One may confirm this result by computing the integral explicitly, either with Cartesian or planar polar coordinates. One may also conclude that it is zero because the function y is an odd function – y > 0 over the region above the x-axis and y < 0 over the region below the x-axis which is a mirror image of the region above. Another way of deducing that the integral is zero is to note that it is related to the center of mass of region D when the density is constant, i.e., the centroid: 1 y¯ = y dA. (9) A(D) Z ZD But by symmetry,y ¯ = 0. Therefore the integral is zero. You can also confirm the above result by explicitly computing the circulation integral using the parametrization r(t) = (cos t, sin t). 2. The vector field y x F = − i + j. (10) x2 + y2 x2 + y2 249 You may recall that this vector field described, up to a constant, the magnetic field around a thin current-carrying wire of infinite length lying on the z-axis. Also recall that curl F = ∇×~ F = 0, (x,y) 6= 0. (11) Because the curl is not defined at (0,0), we can use Green’s Theorem only for simple closed curves C that do not enclose the origin (0,0). (Recall that one of the assumptions in Green’s Theorem was that the partial derivatives existed at all points (x,y) ∈ D, the region enclosed by C.) In this case, all circulation integrals are zero: F · dr = 0. (12) IC We cannot use Green’s Theorem to conclude anything about circulation integrals over closed curves C that enclose the origin. In this case, we computed earlier in this course that the circulation integral over a circle CR of radius R centered at the origin is 2π. From this result, one can actually conclude that the circulation integral over any simple curve C enclosing the origin is 2π. 3. Some additional special cases: The vector fields 1 F = −yi, F = xj, F = (−yi + xj). (13) 2 In all of these cases, ∂F ∂F 2 − 1 = 1. (14) ∂x ∂y For these vector fields F · dr = dA = A(D), the area of D. (15) IC Z ZD In fact, there are many, many other vector fields F = F1i + F2j for which Eq. (14) is satisfied. Can you find any? Can you find a general family of such fields? 250 Physical interpretation of the curl in terms of the circulation integral You may recall that when the curl and divergence operators were introduced some time ago, a physical interpretation of the divergence operator could be provided in terms of the total outward flow of a vector field from a tiny box of dimensions ∆x and ∆y centered at a point (x,y). A “box-based” interpretation of the curl operator could not be made at that time because it would have required the notion of the circulation integral which, of course, was not developed until the previous lecture. Now that we know about the circulation integral, we can obtain an interpretation of the curl operator. To start, we consider the rectangular region ABCD centered at (x,y) with dimensions ∆x and ∆y, as sketched below. We shall estimate the circulation of a planar vector field F(x,y)= F1(x,y) i + F2(x,y) j (16) along the closed curve C composed of the four line segments C1, C2, C3 and C4, identified in the figure. D C3 C y + ∆y/2 C4 P (x, y) y C2 y − ∆y/2 A B C1 x − ∆x/2 x x + ∆x/2 From the additivity property of integrals, the circulation integral of F will be given by F · dr = F · dr + F · dr + F · dr + F · dr. (17) IC ZC1 ZC2 ZC3 ZC4 Recall that each of these integrals is computed by means of a parametrization of the form b ′ F · dr = F(r(t)) · r (t) dt, (18) ZCi Za ′ ′ ′ where the curve Ci is parametrized as r(t) = (x(t),y(t)), a ≤ t ≤ b. The vector r (t) = (x (t),y (t)) is the tangent vector to the curve Ci. We now consider the line integral over each of these curves. 251 1. Curve C1: We may parametrize this curve as ∆x ∆y x(t)= x − + t, y(t)= y − , 0 ≤ t ≤ ∆x, (19) 2 2 implying that the tangent vector is ′ ′ ′ r (t) = (x (t),y (t)) = (1, 0). (20) As such, the integrand in (18) will be given by ′ ∆y ∆y F(r(t)) · r (t)= F x(t),y − =∼ F x,y − .
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