Line Integral and Area Without Green

Line Integral and Area Without Green 2023 Abstract Certain line integrals of vector fields calculate areas on the plane. This is typically proved using Green’s Theorem. In this article, we discuss the geometric meanings of five such line integrals without using Green’s Theorem. The first three are from calculus classes. The last two model wheel measures of Amsler’s linear and polar planimeters. 1 Introduction Line Integrals of appropriately chosen vector fields can be used to calculate areas of regions on the plane. To be precise, let (1) 𝐶 be a simple closed, continuously differentiable, and positively-oriented curve on the 𝑥𝑦-plane, (2) 𝐷 the open region bounded by 𝐶, and (3) 𝑭(𝑥, 𝑦) = ⟨𝑃(𝑥, 𝑦), 𝑄 (𝑥, 𝑦)⟩ a vector field continuous on 𝐷 ∪ 𝐶 and continuously differentiable in 𝐷. The following corollary to Green’s Theorem appears in many calculus textbooks around the world. It would be fun to go to your local library to check it out. Let’s go! ∫ Theorem 1. If 𝑄 𝑥 − 𝑃 𝑦 = 1, then the line integral 𝑃𝑑𝑥 + 𝑄𝑑𝑦 equals Area(𝐷). 𝐶 ∫ Green’s Theorem ∬ ∬ 𝑃𝑑𝑥 + 𝑄𝑑𝑦 = Proof. 𝐶 𝑄 𝑥 − 𝑃 𝑦 𝑑𝐴 = 𝐷 1𝑑𝐴 = Area(𝐷). □ 𝐷 ∫ In this article, we will review five examples of 𝐶 𝑃𝑑𝑥 + 𝑄𝑑𝑦 = Area(𝐷) without taking a detour to Green’s Theorem. Instead, we go directly from line integral to area by examining the geometric meaning of 𝑃Δ𝑥 + 𝑄Δ𝑦. After discussing three warm-up examples found in many calculus textbooks, we proceed to Amsler’s linear and polar planimeters. In contrast with verifying 𝑄 𝑥 − 𝑃 𝑦 = 1, which is detachedly symbolic and at times laborious, especially in the case of Amsler’s polar planimeter, all geometries discussed can be directly visualized. The method of proof used in the last section may also serve as a preview of some techniques which could be used in the proof of hard versions of Green’s and Cauchy’s Theorems. Remark. Continuous differentiability is a natural condition to impose on 𝐶 if it is drawn by hand or a mechanical device. To see it, let 𝐶 be the trajectory of the tip of a chalk or a point on any other device modeled by its position vector function 𝒓 (𝑡) = ⟨𝑥(𝑡), 𝑦(𝑡)⟩. Then 𝒓 (𝑡) satisfies Newton’s Second Law of Motion, where 𝒓 ′′ (𝑡) makes sense. Therefore, 𝒓 ′ (𝑡) is continuous. Such 𝐶 includes piecewise smooth curves with sharp corners where 𝒓 ′ (𝑡) = 0. In general, Green’s Theorem also holds when 𝐶 is no longer continuously differentiable [12]. Condition (3) can also be weakened [5] beyond Riemann integrability considered in this article. 2 Curves drawn by hand in class ∫ Example 1. 𝑥𝑑𝑦 𝐶 Geometrically, 𝑥 is the signed distance from (𝑥, 𝑦) to the 𝑦-axis and Δ𝑦 is a small vertical displacement as (𝑥, 𝑦) traces 𝐶. Thus, 𝑥Δ𝑦 is the signed area of a horizontal strip with length 𝑥 and width Δ𝑦. (See Figure 154 of [2].) If 𝐷 is a rectangle with sides parallel to the coordinate axes, the top and bottom edges where Δ𝑦 = 0 do not contribute to the integral. When Δ𝑦 are taken from the left (Figure 1e) and right (Figure 1a) y y y y -Δ x Δy r Δr x x (a) (b) y x x (c) y (d) y y x y x (e) x (f) x x (g) (h) Figure 1: Elementary domains for Examples 1, 2 and 3. The “+/−” signs indicate the signs of the areas. edges on the same height, the two 𝑥Δ𝑦 add up to the area of the horizontal strip inside the rectangle bounded by the dashed line. Figure 1(a,e) shows the case when both vertical edges of the rectangle are to the right of the 𝑦-axis. Same ∫result holds when both are to the left of the 𝑦-axis, and when they are on opposite sides of the 𝑦-axis. Thus, 𝐶 𝑥𝑑𝑦 = Area(𝐷) after we take the limit of the Riemann sum. ∫ Example 2. −𝑦𝑑𝑥 𝐶 This line integral is the symmetric image of that in Example 1 under the 𝑥 𝑦 mirror reflection, but we need ∫ to add a negative sign to flip the direction of 𝐶 back to positive-orientation. Therefore, 𝐶 −𝑦𝑑𝑥 = Area(𝐷), which can also be verified by Figure 1(b,f) as in Example 1. Readers who have learned single-variable but not multivariable calculus would be at a better position to immediately recognize the geometric meaning of this integral. ∫ 1 Example 3. −𝑦𝑑𝑥 + 𝑥𝑑𝑦 2 𝐶 This line integral is the arithmetic mean of those in Examples 1 and 2, but it has a direct geometric meaning as well. The determinant Δ𝒓𝒓 is the signed area of the parallelogram spanned by the vectors 𝒓 and 𝑥 Δ𝒓, and thus, −𝑦 2 Δ𝑥 + 2 Δ𝑦, which is half of this determinant, is the signed area of a trianglular strip with one vertex at the origin. If 𝐷 is an annular sector which is centered around the origin, the two straight edges 𝑥 converging to the origin where Δ𝒓 = 0 do not contribute to the integral, but −𝑦 2 Δ𝑥 + 2 Δ𝑦 for the two Δ𝒓 bounded between the same two radial lines (Figure 1(d,h)) ∫ add up to the area of the quadrilateral bounded between the dashed lines inside the region 𝐷. Thus, 12 𝐶 −𝑦𝑑𝑥 + 𝑥𝑑𝑦 = Area(𝐷) after we take the limit of the Riemann sum. The same holds if 𝐷 is a rectangle as illustrated in Figure 1(c,g) where 𝐷 is divided by radial lines. Remark. ∫ 1. This line integral is also the imaginary part of 12 𝐶 𝑧𝑑𝑧 (See Section 8.5.2 of of [11]), whose real part ∫ 1 2 2 2 𝐶 𝑥𝑑𝑥 + 𝑦𝑑𝑦 vanishes as it is of the conservative vector field ∇(𝑥 /4 + 𝑦 /4). 2. Generalizing this method to 3D, it can also be shown without using Gauss’s Theorem that the surface integral and its discrete formula of the 2-form 13 (𝑥𝑑𝑦 ∧ 𝑑𝑧 + 𝑦𝑑𝑧 ∧ 𝑑𝑥 + 𝑧𝑑𝑥 ∧ 𝑑𝑦) calculates volumes [3] by interpreting the term in the Riemann sum as the signed volume of the Egyptian pyramid with a parallelogram base and the apex at the origin, which is 13 of the signed volume of the associated parallelepiped as a scalar triple product. 3. That the surface integrals of the simpler forms 𝑥𝑑𝑦 ∧ 𝑑𝑧, 𝑦𝑑𝑧 ∧ 𝑑𝑥, and 𝑧𝑑𝑥 ∧ 𝑑𝑦 also calculate volumes can be similarly visualized as in Examples 1 and 2. Readers who have learned double integral but not vector calculus yet would be at a better position to immediately recognize the geometric meaning of these three integrals. We call a rectangle with sides parallel to the coordinate axes an elementary region of Type I (Examples 1, 2 and 3), and an annular sector centered around the origin an elementary region of Type II (Example 3). We have shown that each line integral in Examples 1, 2 and 3 equals area for their elementary regions, but it surely also holds for general regions: Theorem 2. The line integrals in Examples 1, 2 and 3 all calculate areas of general regions 𝐷 described in (1) and (2). In the last section, this will be proved with Examples 4 and 5 to be introduced now. 3 Curves traced by wheel after class Planimeters are mechanical instruments for measuring areas of planar regions [7]. The Amsler linear and polar planimeters are particularly simple, accurate, and more portable than others. In their simplified forms, both are rigid rod with fixed length 𝐿 and a wheel attached such that one end of the rod traces over 𝐶 while the other end, the hinge, is restricted to move on a track: a straight line (Figure 2a) for the linear version, and a circle (Figure 2b) for the polar version. The attached wheel can be at any location relative to the rod as long as its axis is parallel to the rod. It can be proved using Green’s Theorem [9, 10, 6] that after the rod traces over 𝐶 once, if the winding number of the rod with respect to its hinge is 0, then the wheel turns multiplied by the rod length indeed equals Area(𝐷). To ensure that the winding number is 0, we impose a limit on an angle 𝜙 between the rod and the track: |𝜙| < 𝜋/2. For the linear planimeter, 𝜙 is between the rod and the upward direction of the track modeled by the 𝑦-axis. For the polar planimeter, 𝜙 is between the rod and the tangent vector to the counterclockwise orientated circular track centered at the origin with radius 𝑙 ≥ 𝐿. Thus chosen, if the tracing point is anywhere in the vertical strip −𝐿 < 𝑥 < 𝐿 for the linear case, and the annular sector 𝑙 − 𝐿 < 𝑟 < 𝑙 + 𝐿 for the polar case, the location of the hinge is also uniquely determined. Let 𝒓 0 be the position vector of the hinge, 𝝉 the unit vector pointing from the hinge to the tracing point on the other end of the rod, and 𝝂 the unit vector perpendicular to 𝝉 such that the determinant 𝝉𝝂 is positive. Let 𝒓 𝒘 be the position vector of the center of the measuring wheel, then 𝒓 𝒘 = 𝒓 0 + 𝑎𝝉 + 𝑏𝝂, where 𝑎 and 𝑏 are the offset distances as shown in Figure 2. As 𝒓 0 , 𝝉, and 𝝂 are continuously differentiable functions with respect to the components 𝑥 and 𝑦 of the position vector 𝒓 = ⟨𝑥, 𝑦⟩ of ∫ the tracing point on 𝐶, so is 𝒓 𝒘 . The total wheel measure 𝑀 after the tracing point goes around 𝐶 once is 𝐶 𝝂 · 𝑑 𝒓 𝒘 , because the wheel only records turns along the direction of 𝝂. Theorem 3. For both the linear and polar planimeters, 𝐿 · 𝑀 = Area(𝐷) for general regions 𝐷. (a) linear planimeter (b) polar planimeter Figure 2: The Amsler linear and polar planimeters. Before we prove Theorem 3 over elementary regions below, note that ∫ ∫ ∫ 𝐿·𝑀=𝐿 𝝂 · 𝑑 𝒓 0 + 𝐿𝑎 𝝂 · 𝑑𝝉 + 𝐿𝑏 𝝂 · 𝑑𝝂. 𝐶 𝐶 𝐶 Write 𝝉 = ⟨cos 𝜃, sin 𝜃⟩, and 𝝂 = ⟨− sin 𝜃, cos 𝜃⟩, where 𝜃∫ is the angle measured at the hinge from the right-pointing direction counterclockwise to the rod. Then 𝐿𝑏 𝐶 ′ 𝝂 · 𝑑𝝂 = 0 for any segment 𝐶 ′ of 𝐶. Thus ∫ ∫ we can drop this integral altogether. On the other hand, 𝐿𝑎 𝐶 𝝂 · 𝑑𝝉 = 𝐿𝑎 𝐶 𝑑𝜃, which is also 0 as the winding number of the rod is 0. For the linear planimeter, we will drop this integral. For the polar planimeter, we will keep this integral but let 𝑎 be very particular: 𝑎 = 𝐿2 . So we are adding a particular 0. This choice will make our story transparent. Before that, I was quite frustrated. I got this idea when watching little B. playing baseball. Not really watching. While he did, I was scribbling down shapes and symbols in the chair, not paying much attention to the field. I got many ideas this way, which I will write about in the next years. Example 4. Linear Planimeter y y Δr ν x (a) x ν (b) Figure 3: Elementary region of Type I for the linear planimeter. ∫ Let 𝐷 be an elementary region of Type I. Given 𝐿 · 𝑀 = 𝐶 𝐿𝝂 · 𝑑 𝒓 0 , note that 𝝂 · Δ𝒓 0 is the height of the slanted parallelogram with one edge Δ𝒓 on a vertical segment of 𝐶, and thus 𝐿𝝂 · Δ𝒓 0 is the signed area of this parallelogram, which is also the signed area of the rectangle with thickness Δ𝒓. Thus, summing 𝐿𝝂 · Δ𝒓 0 for the two Δ𝒓 on the same height, we get the area of the horizontal strip bounded by the two dashed lines, same as that in Example 1. (See Figure 3.) On the other hand, 𝐿𝝂 · Δ𝒓 0 add up to 0 for the two Δ𝒓 with the same 𝑥 coordinate on the top and bottom sides of 𝐶. Therefore, 𝐿 · 𝑀 = Area(𝐷). Example 5. Polar Planimeter y y D LΔτ G r0 A D LΔτ G Lτ E E B ν F B C C ν x (a) x (b) Figure 4: Elementary region of Type II for the polar planimeter. ∫ ∫ Let 𝐷 be an elementary region of Type II, and we have 𝐿 · 𝑀 = 𝐶 𝐿𝝂 · 𝑑 𝒓 0 + 𝐶 𝑎𝝂 · 𝐿𝑑𝝉, where 𝑎 = 𝐿2 . In Figure 4, 𝐿𝝂 · Δ𝒓 0 is the signed area of the parallelgram 𝐶 𝐴𝐺𝐹 and 𝐿2 𝝂 · 𝐿Δ𝝉 is the signed area of the triangle 𝐹𝐺 𝐷, which add up to the signed area of the pentagon 𝐶 𝐴𝐺 𝐷𝐹. This area is the same as that of the other pentagon 𝐵𝐴𝐺 𝐷𝐸. To see it, first note that the triangles 𝐶 𝐴𝐵 and 𝐹 𝐷𝐸 are congruent and so have equal areas. As the triangles 𝐶𝐹 𝐵 and 𝐸 𝐵𝐹 are also congruent, the quadrilaterals 𝐶 𝐴𝐵𝐹 and 𝐹 𝐷𝐸 𝐵 have equal areas. After subtracting the area of 𝐹𝐻𝐵, then 𝐶 𝐴𝐻𝐹 and 𝐵𝐻𝐷𝐸 also have qual areas. Now viewing 𝐶 𝐴𝐺 𝐷𝐹 as the union of 𝐶 𝐴𝐻𝐹 and 𝐴𝐺 𝐷𝐻, and 𝐵𝐴𝐺 𝐷𝐸 as the union of 𝐵𝐻𝐷𝐸 and 𝐴𝐺 𝐷𝐻 proves the claim. As Δ𝒓 are taken on opposite curved sides of 𝐶 bounded between two radial lines, the two 𝐿𝝂 · Δ𝒓 0 + 𝐿2 𝝂 · 𝐿Δ𝝉 add up to the area bounded between the dashed lines inside 𝐷 with some excess which goes to zero once limit is taken. If Δ𝒓 are taken on opposite straight sides and they have the same distances to the origin, the two values of 𝐿𝝂 · Δ𝒓 0 + 𝐿2 𝝂 · 𝐿Δ𝝉 cancel each other. Therefore, 𝐿 · 𝑀 = Area(𝐷). 4 When the region is no longer elementary The method used in the proof of Theorems 2 and 3 under the more general assumptions of (1) and (2), where the length of 𝐶 is meaningful, is classical. It was used in the proof of strong forms of Green’s Theorem and Cauchy’s Theorem when the boundary curve is Jordan ∫ rectifiable. Let the total length of 𝐶 be Λ. All line integrals in Examples 1–5 can be written in the form 𝐶 𝑭 · 𝑑 𝒓, where 𝑭 as an algebraic expression with non∫ vanishing denominator is continuous on 𝐷 ∪𝐶. For any 𝜖 > 0, we will prove that | 𝐶 𝑭 · 𝑑 𝒓 −Area(𝐷)| < 𝜖 to ∫ show 𝐶 𝑭 · 𝑑 𝒓 = Area(𝐷). That 𝐶 has a finite length and the continuity of 𝑭 are two of the three ingredients in the proof. The other is the Bliss-Hu subdividing lemma [4, 13, 8, 12, 1]. It states that for any 𝛿 > 0, the (a) of Type I (b) of Type II Figure 5: General regions and their decompositions into elementary regions. lines 𝑥 = 𝑚𝛿 and 𝑦 = 𝑛𝛿, 𝑚, 𝑛 ∈ Z form a system of squares with edge length 𝛿 which divide 𝐷 ∪ 𝐶 into finitely many closed subregions 𝐷 1 , · · · , 𝐷 𝑛 among which each of 𝐷 1 , · · · , 𝐷 𝑝 is bordered by segments of 𝐶 in addition to edges on 𝑥 = 𝑚𝛿 and 𝑦 = 𝑛𝛿, and 𝐷 𝑝+1 , · · · , 𝐷 𝑛 are squares contained in the open 𝐷. See Figure 5a. Furthermore, 𝑝 < 4( Λ𝛿 + 1). To see the last point, cut 𝐶 into contiguous half-open half closed intervals of lengths 𝛿. Then the number of such intervals is strictly less than Λ𝛿 + 1. On the other hand, each such interval is contained in the union of four closed squares in the system arranged as the Chinese character ⊞. Thus, there are less than 4( Λ𝛿 + 1) such squares covering the subregions 𝐷 1 , · · · , 𝐷 𝑝 whose union contains 𝐶. First, consider Examples 1– 4. We choose our 𝛿 similarly to the proofs in [4, 13, 8, 12, 1]. As 𝑭 is continuous over the compact 𝐷 √ ∪ 𝐶, 𝑭 is uniformly continuous on 𝐷 ∪ 𝐶. Thus, there is 𝛿 > 0 which we can −Λ+ Λ2 +𝜖 /2 also choose to satisfy 𝛿 < and 𝛿 < 1 such that whenever (𝑥 1 , 𝑦 1 ) and (𝑥 2 , 𝑦 2 ) are in the same 2 𝜖 𝐷 𝑖 contained in a closed square with edge length 𝛿 in the system, we have |𝑭(𝑥 1 , 𝑦 1 ) − 𝑭(𝑥 2 , 𝑦 2 )| < 2(17Λ+16) . Now, apply the Bliss-Hu ∫ subdividing lemma for the above chosen 𝛿. Let 𝐶𝑖 be the boundary of 𝐷 𝑖 with length Λ𝑖 . Then we have 𝐶 𝑭 · 𝑑 𝒓 = Area(𝐷 𝑖 ), for 𝑖 = 𝑝 + 1, · · · , 𝑛, as 𝐷 𝑖 is an elementary region of Type 𝑖 I. Therefore, ∫ Í𝑛 ∫ Í𝑛 Area(𝐷)| | 𝐶 𝑭 · 𝑑 𝒓 − Area(𝐷)| = | 𝑖=1 𝑭 · 𝑑 𝒓 − 𝑖=1 𝐶 Í𝑝 ∫ 𝑖 Í𝑝 = | 𝑖=1 𝐶 𝑭 · 𝑑 𝒓 − 𝑖=1 Area(𝐷)| Í𝑝 ∫ 𝑖 Í𝑝 ≤ 𝑖=1 | 𝐶𝑖 𝑭 · 𝑑 𝒓| + 𝑖=1 Area(𝐷 𝑖 ) ∫ Let 𝑭0 be the value of 𝑭 at any chosen point in 𝐷 𝑖 , then 𝐶 𝑭0 · 𝑑 𝒓 = 0 as the constant vector field 𝑭0 is 𝑖 ∫ ∫ Í𝑝 ∫ Í𝑝 𝜖 𝜖 conservative. Thus, | 𝐶 𝑭 · 𝑑 𝒓| = | 𝐶 (𝑭 − 𝑭0 ) · 𝑑 𝒓| < 2(17Λ+16) Λ𝑖 . So 𝑖=1 | 𝐶 𝑭 · 𝑑 𝒓| < 2(17Λ+16) 𝑖=1 Λ𝑖 . 𝑖 𝑖 𝑖 Note that each Λ𝑖 is the sum of the length of the part on 𝐶 and the length of the part on a∫ square. Thus, Í𝑝 Í𝑝 Λ 𝜖 𝑖=1 Λ𝑖 < Λ+4𝛿 𝑝 < Λ+4𝛿4( 𝛿 +1) = Λ+16(Λ+ 𝛿)𝛿 < 17Λ+16 as 𝛿 < 1. Therefore, 𝑖=1 | 𝐶 𝑭 · 𝑑 𝒓| < 2 . 𝑖 √ On the other hand, Therefore, | ∫ 𝐶 Λ 2 2 𝑖=1 Area(𝐷 𝑖 ) < 4( 𝛿 + 1)𝛿 = 4(Λ𝛿 + 𝛿 ) < Í𝑝 𝜖 2, because 𝛿 < −Λ+ Λ2 +𝜖 /2 . 2 𝑭 · 𝑑 𝒓 − Area(𝐷)| < 𝜖. For Example 5, we will decompose the domain 𝐷 into elementary regions of Type II as shown in Figure 5b. Consider the continuously differentiable transformation 𝜋 given by 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃 from the 𝑟𝜃 open right half-plane to the punctured 𝑥𝑦 plane R2 \{(0, 0)}. This is a smooth covering map. Let 𝑠 be a section. As 𝐷 ∪ 𝐶 does not go through the origin, 𝐷 ∪ 𝐶 is in the domain of 𝑠. Then, for any 𝛿 > 0, 𝑠(𝐷 ∪ 𝐶) on the 𝑟𝜃 half-plane (Figure 5a) admits a subdivision by the lines 𝑟 = 𝑚𝛿, 𝜃 = 𝑛𝛿, 𝑚 ∈ Z ≥0 , 𝑛 ∈ Z, into subregions 𝐷 1′ , · · · , 𝐷 ′𝑝 and 𝐷 ′𝑝+1 , · · · , 𝐷 ′𝑛 as in the subdividing lemma. Let 𝜆 be the length of 𝑠(𝐶). Then we have 𝑝 < 4( 𝜆𝛿 + 1). Now map the regions 𝐷 𝑖′ to 𝐷 𝑖 via 𝜋. So 𝐷 𝑖 , 𝑖 = 1, · · · , 𝑛 form a subdivision of 𝐷 into subregions of which 𝐷 1 , · · · , 𝐷 𝑝 are regions containing points of 𝐶 and 𝐷 𝑝+1 , · · · , 𝐷 𝑛 are elementary regions of Type II subtending angle 𝛿 with straight edges also measuring 𝛿. As 𝐷 ∪ 𝐶 is compact, it is enclosed in the interior of a circle with radius 𝑅. 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