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Line Integrals 18.1 The Idea of a Line Integral 𝑏 As the integral ∫𝑎 𝑓(𝑥)dx integrates over the line y = 0, the line integral ∫𝑐 𝑓ds is a single integral that integrates over an oriented curve or path c instead than over a line, where ds is the element of arc length. 𝑏 The integral ∫𝑎 𝑓(𝑥)𝑑𝑥 represents, graphically, the area under the curve y = f (x) above the line y = 0. The line integral ∫𝑐 𝑓𝑑𝑠 represents, graphically, the area of a curtain formed by the projection of path c in the x-y plane on the surface z = f (x, y). A line integral can be expressed as ∫c f (x, y)dx; ∫c f (x, y)dy; ∫c f (x, y)ds in ℜ2 or ∫c f (x, y, z)dx; ∫c f (x, y, z)dy; ∫c f (x, y, z)dz; ∫c f (x, y, z)ds in ℜ3 ; with c a piecewise smooth oriented function. The line integral of the Vector Field 𝐹⃗ along an oriented curve c is given by ⃗⃗⃗⃗ ∙ 𝑑𝑟⃗ = lim ∑𝑛−1 ⃗⃗⃗⃗ ⃗𝑖 ) ∙ ∆𝑟⃗𝑖 , where the direction of ∆𝑟⃗ is in the direction of the orientation. ∫𝑐 𝐹 𝑖=0 𝐹 (𝑟 |∆𝑟⃗𝑖 |→0 The line integral measures the extent to which the path c is going with the field 𝐹⃗ or against the field. eg 1 Find the line integral of the vector field 𝐹⃗ = < – 2, 3 > along the path given by the linear segment from (1, 0) to (3, 0). 3 Since 𝐹⃗ = < – 2, 3 >, 𝑟⃗ = < 𝑥, 0 >, 1 ≤ 𝑥 ≤ 3, and 𝑑𝑟⃗ = < 𝑑𝑥, 0 >, ∫𝑐 𝐹⃗ ⋅ 𝑑𝑟⃗ = ∫1 − 2𝑑𝑥 = −4. eg 2 Find the line integral of the vector field 𝐹⃗ = < x, y > along the path given by the linear segments from (0,0) to (1,1) and from (1,1) to (1, – 1). Let c1 be the linear segment from (0,0) to (1, 1). Since 𝐹⃗ = < x, y >, 𝑟⃗ = < x, y > and d𝑟⃗ = < dx, dy >, 1 1 ∫𝑐1 𝐹⃗ . 𝑑𝑟⃗ = ∫𝑐1 < x, y > ⋅ < 𝑑𝑥, 𝑑𝑦 > = ∫0 𝑥𝑑𝑥 + ∫0 𝑦𝑑𝑦 = 1 . −1 Let c2 be the linear segment from (1,1) to (1, − 1). ∫ ⃗⃗⃗⃗ 𝐹 ⋅ 𝑑𝑟⃗ = ∫ < 𝑥, 𝑦 > ⋅ < 0, 𝑑𝑦 > = −1 ∫1 𝑦 𝑐2 𝑑𝑦 = 0. So ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫𝑐 𝐹⃗ ⋅ 𝑑𝑟⃗ + ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = 1. 1 2 1 1 Work: The work done on an object by a constant force while it moves on a straight line is given by W = 𝐹⃗ ⋅ 𝑑⃗ where 𝑑⃗ is the displacement vector in the direction of motion. If the force is not constant or the path is not a straight line, the element of work becomes ∆𝑊 = 𝐹⃗ ⋅ ⃗⃗⃗⃗⃗ 𝑑𝑟. The work along the curve 𝑐 will be the line integral ∫𝑐 𝐹⃗ ⋅ ⃗⃗⃗⃗⃗ 𝑑𝑟 . eg 3 A mass lying flat on a table is attached to a spring that is compressed 10 cm beyond its rest position. Find the work done by the force exerted by the spring to move the mass back to its equilibrium position. If we consider the equilibrium position to be zero, and the spring is compressed in the negative direction, by Hook’s law 𝐹⃗ = < −𝑘𝑥, 0 > where k is a positive constant. The line integral becomes 0 ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫−10 − 𝑘𝑠𝑑𝑠 = 50k. Since the force acts in the direction of motion, the work is positive. eg 4 Find the work done by the Earth’s gravity to lift a 4000N weight communications satellite to an altitude of 24,500km. Consider the radius of the earth to be r = 6500km. 𝑘 Since the force done by the earth gravity is F = − 𝑟 2 in the direction of the center of the earth, 11 31000 𝑘 1.69 𝑥 10 F = 4000 = 65002 , k = 1.69 X 1011. The work will be given by W = ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫6500 − 𝑟2 = − 20,548,387.1 N-km = −20,548,387,100J. Since the force acts in the direction opposite to the direction of motion, the work is negative. Homework 18.1 18.2 Computing Line Integrals Over Parameterized Curves We can describe integrals along a curve c parametrically by 𝑟⃗ = < x(t), y(t) > in ℜ2 or by 𝑏 𝑟⃗ = < x(t), y(t), z(t) > in ℜ3 . Let c be a smooth curve, ∫𝑐 𝑓(𝑥, 𝑦)𝑑𝑥 + 𝑔(𝑥, 𝑦)𝑑𝑦 = ∫𝑎 < 𝑓 (𝑥(𝑡), 𝑦(𝑡)), 𝑔(𝑥(𝑡), 𝑦(𝑡)) > ∙ < 𝑑𝑥 𝑑𝑦 , 𝑑𝑡 𝑑𝑡 > 𝑑𝑡 in t : [a, b]. eg 5 Find ∫𝑐 (𝑥 + 𝑦)𝑑𝑦 over c : y = x2 in x : [0, 1]. This integral can be seen as ∫𝑐 0𝑑𝑥 + (𝑥 + 𝑦)𝑑𝑦 𝑑𝑟⃗ Since c : 𝑟⃗ = < t, t2 > , t : [0, 1], 𝑑𝑡 = < ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ 𝑑𝑡 1 𝑑𝑥 𝑑𝑦 , 𝑑𝑡 𝑑𝑡 > = < 1, 2t >, and 𝐹⃗ = < 0, t + t2 > , 7 𝑑𝑡 = ∫0 < 0, t + t2 > ∙ < 1, 2t > dt = 6 eg 6 ∫𝑐 𝑒 𝑦−𝑥 dx + xdy over the line segment c : x = 2 + t, y = 3 – t in t : [0, 1]. We write this integral as ∫𝑐 < 𝑒 𝑦−𝑥 , x > ∙ < dx, dy >, so ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ 𝑑𝑡 1 1 𝑑𝑡 = ∫0 < 𝑒 (3−𝑡)−(2+𝑡) , (2 + t) > ⋅ < 1, – 1 > dt = ∫0 𝑒 (1−2𝑡) – (2 + t)dt = 2 𝑒−𝑒 −1 2 5 -2 𝑑𝑟 Vector form of Line Integrals 𝑏 𝑑𝑠 Let c be a smooth curve, ∫𝑐 𝑓 (x, y)ds = ∫𝑎 f (x(t), y(t)) 𝑑𝑡 dt in t : [a, b] where ds is the infinitesimal 𝑏 𝑑𝑠 element of arc length, = |𝑟⃗̇| = √𝑥̇ (𝑡) + 𝑦̇ (𝑡) in ℜ2 , and ∫ f (x, y, z)ds = ∫ 𝑓 (𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡)) 𝑐 𝑑𝑡 𝑎 𝑑𝑠 dt in t : [a, b] where 𝑑𝑡 = |𝑟⃗̇| = √𝑥̇ (𝑡) + 𝑦̇ (𝑡) + 𝑧̇ (𝑡) in ℜ3 . 2 𝑡 𝑑𝑟⃗(𝑢) Recall: Since s(t) = ∫𝑡 | Calculus 𝑑𝑠 𝑑𝑡 = 𝑑𝑡 2 𝑡 𝑑𝑥 𝑑𝑦 𝑑𝑧 | 𝑑𝑢 = ∫𝑡 √(𝑑𝑢) + (𝑑𝑢) + (𝑑𝑢) du, by the Fundamental Theorem of 𝑑𝑢 0 𝑡 𝑑𝑟⃗(𝑢) | 𝑑𝑢 ∫ | 𝑑𝑡 𝑡0 𝑑𝑢 𝑑 2 𝑑𝑠 0 = |𝑟⃗̇| . 𝑏 𝑏 𝑑𝑠 So ∫𝑐 f (x, y, z)ds = ∫𝑎 𝑓(𝑥(𝑡), 𝑦(𝑡), 𝑧 (𝑡)) 𝑑𝑡 𝑑𝑡 = ∫𝑎 𝑓(𝑥(𝑡), 𝑦(𝑡), 𝑧 (𝑡)) |𝑟⃗̇| 𝑑𝑡, in t: [a, b]. eg 7 Consider ∫𝑐 (2 + 𝑥 2 𝑦)𝑑𝑠 where c : x2 + y2 = 1, y > 0 in ℜ2 with a ccw orientation. Since this is the top of the unit semicircle, 𝑟⃗ = < cos(t), sin(t) > in t : [0, 𝜋]; 𝑟⃗̇ = < − sin(t), cos(t) > ; 𝜋 𝑑𝑠 𝑑𝑠 2 |𝑟⃗̇| = 𝑑𝑡 = 1, so ∫𝑐 (2 + 𝑥 2 𝑦) 𝑑𝑡 𝑑𝑡 = ∫0 (2 + cos2(t)sin(t))dt = 2𝜋 + 3 If C is the curve with pieces c1 and c2, ∫𝑐 𝑓 (𝑥, 𝑦, 𝑧)𝑑𝑠 = ∫𝑐1 𝑓 (𝑥, 𝑦, 𝑧)𝑑𝑠 + ∫𝑐 𝑓 (𝑥, 𝑦, 𝑧)𝑑𝑠 . 2 eg 8 Consider ∫𝐶 12 xds where C = { 𝑐1 : 𝑦 = 𝑥 2 in 𝑥 ∶ [0, 1] 𝑐2 : 𝑥 = 1 in 𝑦 ∶ [1, 2] Let c1 be represented by 𝑟⃗ = < t, t2 > t : [0, 1]; 𝑟⃗̇ = < 1, 2t > ; |𝑟⃗̇| = √1 + 4𝑡 2 Let c2 be represented by 𝑟⃗ = < 1, t > t : [1, 2]; 𝑟⃗̇ = < 0, 1 > ; |𝑟⃗̇| = 1 1 2 ∫𝑐 12xds = ∫𝑐1 12xds + ∫𝑐2 12xds = ∫0 12𝑡√1 + 4𝑡 2 𝑑𝑡 + ∫1 12𝑑𝑡 = 5√5 + 11 A change in orientation will change the sign of the line integral if it is along a coordinate axis, but will not change the sign along arc length. ∫−𝑐 f (x, y, z)dx = − ∫𝑐 f (x, y, z)dx; ∫𝑐 f (x, y, z)ds = ∫−𝑐 𝑓(x, y, z)ds. eg 9 Consider ∫𝑐 (x + 1)dx along both c1 : y = √1 − 𝑥 2 with CW orientation and c2 : y = √1 − 𝑥 2 with CCW orientation. For c1, 𝑟⃗ = < − cos(t), sin(t) > t : [0, 𝜋]; ∫𝑐 (x + 1) 1 𝑑𝑥 𝑑𝑡 𝜋 𝑑𝑥 𝑑𝑡 = sin(t) dt = ∫0 (− 𝑐𝑜𝑠(𝑡) + 1) 𝑠𝑖𝑛(𝑡)𝑑𝑡 = For c2, 𝑟⃗ = < cos(t), sin(t) > t : [0, 𝜋]; 𝑑𝑥 𝑑𝑡 𝑐𝑜𝑠2 (𝑡) 2 = − 𝑠𝑖𝑛(𝑡) 3 − 𝑐𝑜𝑠(𝑡) |𝜋0 = 2 ∫𝑐 (x + 1) 1 𝑑𝑥 𝑑𝑡 𝜋 dt = ∫0 (𝑐𝑜𝑠(𝑡) + 1)(− 𝑠𝑖𝑛(𝑡))𝑑𝑡 = 𝑐𝑜𝑠2 (𝑡) 2 + 𝑐𝑜𝑠(𝑡) |𝜋0 = −2 eg 10 Now consider the same integrals along the arc length of the curve ∫𝑐 (𝑥 + 1)ds along both c1 : y = √1 − 𝑥 2 with cw orientation and c2 : y = √1 − 𝑥 2 with ccw orientation. 𝑑𝑠 𝑑𝑠 For c1, 𝑟⃗ = < − cos(t), sin(t) > t : [0, 𝜋]; 𝑟⃗̇ = < sin(t), cos(t) > ; |𝑟⃗̇| = 𝑑𝑡 = 1 ∫𝑐1 (𝑥 − 1) 𝑑𝑡 𝑑𝑡 = 𝜋 𝜋 ∫0 ( −𝑐𝑜𝑠(𝑡) + 1)𝑑𝑡 = ( − sin(𝑡) + 𝑡) │ 0 = 𝜋 For c2, 𝑟⃗ = < 𝑐𝑜𝑠(𝑡) , 𝑠𝑖𝑛(𝑡) > 𝑡: [0, 𝜋]; 𝑟⃗̇ = < − 𝑠𝑖𝑛(𝑡) , 𝑐𝑜𝑠(𝑡) >; |𝑟⃗̇| = 𝑑𝑠 ∫𝑐 (𝑥 + 1) 𝑑𝑡 𝑑𝑡 = 2 𝜋 ∫0 (𝑐𝑜𝑠(𝑡) 𝑑𝑠 𝑑𝑡 =1 + 1)𝑑𝑡 = (𝑠𝑖𝑛(𝑡) + 𝑡) |𝜋0 = 𝜋 Average Value Along a Path C 1 The average value of f (x, y) on c will be fave = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝐶 ∫𝐶 𝑓(x, y)ds. eg 11 Find the average value of f (x, y) = x2y3 on c : y = x3 , x [0, 2] Let 𝑟⃗ = < t, t3 > ; 𝑟⃗̇ = < 1, 3𝑡 2 > ; |𝑟⃗̇| = √1 + 9𝑡 4 , t [0, 2] So fave = 2 ∫0 𝑡 8 √1+9𝑡 4 𝑑𝑡 2 ∫0 √1+9𝑡 4 𝑑𝑡 . This integral can be evaluated with a CAS. Line Integrals of Vector Fields Line integrals of vector fields will be given by ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ 𝑑𝑡 𝑑𝑡 eg 12 Find the line integral if 𝐹⃗ = < −𝑦, 𝑥 > along the semicircle or radius 1 oriented counterclockwise from the point (x, y) = (1, 0) to (x, y) = (− 1, 0). 𝑑𝑟⃗ Since the path is 𝑟⃗ = < cos(t), sin(t) >, 0 ≤ 𝑡 ≤ 𝜋, 𝑑𝑡 = 𝑟̇ = < −𝑠𝑖𝑛(𝑡), 𝑐𝑜𝑠(𝑡) > 𝜋 𝑑𝑟⃗ 𝑟̇ = ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫𝑐 𝐹⃗ ∙ 𝑑𝑡 = ∫ < − 𝑠𝑖𝑛(𝑡) , 𝑐𝑜𝑠𝑡(𝑡) > ∙ < − 𝑠𝑖𝑛(𝑡) , 𝑐𝑜𝑠(𝑡) > 𝑑𝑡 = 𝜋 𝑑𝑡 0 Since the field is perfectly aligned with the path, 𝜋 will be the highest value a line integral will obtain between those two points for any path C. eg 13 Find the line integral if 𝐹⃗ = < −𝑦, 𝑥 > along the line segments from (x, y) = (1, 0) to (0, 1), and from (0, 1) to (− 1, 0). Let C1 be the line segment from (1, 0) to (0, 1). If we parameterize the curve we obtain C1 = < 1 – t, t >, 1 0 ≤ 𝑡 ≤ 1 or ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫0 < −𝑡, 1 − 𝑡 > ∙ < −1,1 > 𝑑𝑡 = 1. 1 Let C2 be the line segment from (0, 1) to (−1, 0). If we parameterize the curve we obtain 4 1 C2 = < – t, 1 – t >, 0 ≤ 𝑡 ≤ 1, or ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫0 < − 𝑡 + 1, − 𝑡 > ∙ < −1, −1 > 𝑑𝑡 = 1. 2 The ∫ 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫ 𝐹⃗ ∙ 𝑑𝑟⃗ + ∫ 𝐹⃗ ∙ 𝑑𝑟⃗ = 2. 𝑐 𝑐1 𝑐2 eg 14 Find the line integral if 𝐹⃗ = < y, x > along the semicircle or radius 1 oriented clockwise from x = − 1 to x = 1. 𝑑𝑟⃗ Since the path is 𝑟⃗ = < − cos(t), sin(t) > 0 ≤ 𝑡 ≤ 𝜋, = 𝑟⃗̇ = < 𝑠𝑖𝑛(𝑡), 𝑐𝑜𝑠(𝑡) >, ∫ 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫ 𝐹⃗ ∙ 𝑑𝑟⃗ 𝑑𝑡 𝑑𝑡 = 𝜋 ∫0 𝑑𝑡 < 𝑠𝑖𝑛(𝑡) , − 𝑐𝑜𝑠(𝑡) > ∙ < 𝑠𝑖𝑛(𝑡), 𝑐𝑜𝑠(𝑡) > 𝑑𝑡 = 𝑐 𝜋 ∫0 − 𝑐𝑜𝑠(2𝑡) 𝑑𝑡 𝑐 = 0. Work: The work done on an object in a constant force field while it moves on a straight line is given by W = 𝐹⃗ ⋅ 𝑇̂ Δ𝑠 where 𝑇̂ is the unit tangent vector in the direction of motion and Δs is the length of the line segment. The work done by a variable force field 𝐹⃗ on an object that traverses the curve C is given by 𝑟⃗̇ 𝑑𝑠 W = ∫ 𝐹⃗ ∙ 𝑇̂ 𝑑𝑠 = ∫ 𝐹⃗ ∙ dt = ∫ 𝐹⃗ ∙ 𝑟⃗̇ dt 𝑐 𝑐 |𝑟⃗̇| 𝑑𝑡 𝑐 ∫𝑐 𝐹⃗ ⋅ < 𝑥̇ , 𝑦̇ , 𝑧̇ > 𝑑𝑡 = ∫𝑐 𝐹⃗ ∙ < 𝑑𝑥, 𝑑𝑦, 𝑑𝑧 > = ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗. eg 15 Find the work needed to move an object in a force field 𝐹⃗ = < 4, 7 > 𝑁 from P(5, 10) to Q (20, 30) with distance measured in meters. W = 𝐹⃗ ∙ 𝑇̂ ∆𝑠 = < 4, 7 > ∙ < 15, 20 > = 200N-m = 200J If the force is not constant or the path is not a straight line, the element of work becomes ∆𝑊 = 𝐹⃗ ∙ 𝑇̂ ∆𝑠 where the element of arc length of the curve is ∆𝑠 . The work will be the line integral ∫𝑐 𝐹⃗ ⋅ 𝑇̂ 𝑑𝑠. eg 16 Find the work done to move an object in the force field 𝐹⃗ = < − 𝑦, 𝑥𝑦 > along C : y = √1 − 𝑥 2 , x [0, 1] in a CCW direction with the force in N and the distance in m. 𝜋 Since 𝑟⃗ = < cos(t), sin(t) > ; d𝑟⃗ = < − sin(t), cos(t) > dt; t [0, 2 ] 𝜋/2 ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫0 < − 𝑠𝑖𝑛(𝑡) , 𝑐𝑜𝑠(𝑡) 𝑠𝑖𝑛(𝑡) > ∙ < − 𝑠𝑖𝑛(𝑡) , 𝑐𝑜𝑠(𝑡) > 𝑑𝑡 = 𝜋/2 ∫0 𝑠𝑖𝑛2 (𝑡) + 𝑐𝑜𝑠 2 (𝑡)𝑠𝑖𝑛(𝑡))𝑑𝑡 = 1 𝑡− 2 sin(2𝑡) 4 - 𝑐𝑜𝑠3 (𝑡) 𝜋/2 |0 3 𝜋 1 = ( 4 + 3) 𝐽. eg 17 Find the work done to move an object in the force field 𝐹⃗ = < 𝑥𝑦 − 𝑦𝑧, 𝑥 > 𝑙𝑏 along c: < t4, − t2, t3 >, t [0, 1] with distance in ft. ∫𝑐 < 𝑥𝑦 − 𝑦𝑧, 𝑥 > ∙ < 𝑑𝑥, 𝑑𝑦, 𝑑𝑧 > = ∫𝑐 < − 𝑡 6 , 𝑡 5 , 𝑡 4 > ∙ < 4𝑡 3 , −2𝑡, 3𝑡 2 > 𝑑𝑡 1 = ∫0 ( −4𝑡 9 − 2𝑡 6 + 3𝑡 6 )𝑑𝑡 = −9 35 𝑓𝑡 − 𝑙𝑏 Circulation: If 𝑣⃗ is a velocity vector field, the circulation of a fluid due to the velocity field 𝑟⃗ along the curve 𝑐 is given by ∫𝑐 𝑣⃗ ∙ 𝑑𝑟⃗ = ∫𝑐 𝑣⃗ ∙ 𝑇̂ Δ𝑠 where 𝑇̂ is the unit tangent vector in the direction of motion and Δs is the length of the line segment. 5 Flux of a Velocity Field in ℜ2 Flux: The flux (rate of flow) of a fluid across a line segment of a constant velocity field is given by ̂ ∆𝑠 where 𝛿 is the density of the fluid ∆𝑠 is length of the line segment and 𝑁 ̂ is the flux = ∅ = 𝛿𝑣⃗ ∙ 𝑁 unit normal vector to the line segment. eg 18 Find the flux of a fluid through a line segment of 20m in a constant velocity field of 50m/min at an angle of 𝜋/3 with the normal. If the velocity field is not constant or the path is not a straight line, the element of flux becomes ̂ ∆𝑠 where as is the element of arc length of the curve. The flux will be the line integral ∆∅ = 𝑣⃗ ∙ 𝑁 ̂ 𝑑𝑠. ∅ = ∫𝑐 𝐹⃗ ∙ 𝑑𝑟⃗ = ∫𝑐 𝛿𝑣⃗ ∙ 𝑁 Let 𝑣⃗ = < p, q > be a velocity vector field in ℜ2 . The flux ∅ across the curve c along the direction of the ̂ will be given by ∅ = ∫ 𝑣⃗ ∙ 𝑁 ̂ 𝑑𝑠. normal 𝑁 𝑐 Another form we can express this integral is as ∅ = ∫𝑐 𝑝𝑑𝑦 − 𝑞𝑑𝑥. ̂ = < 𝑢2 , −𝑢1 > Proof: Let 𝑇̂ = < 𝑢1 , 𝑢2 > be the unit tangent vector. The unit normal vector will be 𝑁 ̂ = 0 with 𝑁 ̂ to the right of 𝑇̂ (𝑇̂ x 𝑁 ̂ into the paper). (not < −𝑢2 , 𝑢1 > ) such that 𝑇̂ ∙ 𝑁 ̂ 𝑑𝑠 = < 𝑑𝑦, −𝑑𝑥 > and 𝑣⃗ = < 𝑝, 𝑞 >, ∅ = ∫ 𝑣⃗ ∙ 𝑁 ̂ 𝑑𝑠 = Since 𝑇̂𝑑𝑠 = < 𝑑𝑥, 𝑑𝑦 > , 𝑁 𝑐 ∫𝑐 < 𝑝, 𝑞 >∙ < 𝑑𝑦, −𝑑𝑥 > + ∫𝑐 𝑝𝑑𝑦 − 𝑞𝑑𝑥 . ̂ to the right of 𝑇̂ 𝑁 eg 19 Find the rate of flow of a fluid in the x-y plane, with distances measured in m, of a fluid with a velocity field 𝑣⃗ = < x3 , 1 > m/sec across the parabola x = y2 y : [−1, 1]. ̂ 𝑑𝑠 = ∅ = ∫ 𝑣⃗ ∙ 𝑁 ̂ Since 𝑟⃗ = < 𝑡 2 , 𝑡 > , 𝑡 ∶ [−1, 1], ∅ = ∫𝑐 𝑣⃗ ∙ 𝑁 𝑐 <1,−2𝑡> 1 𝑑𝑠 𝑑𝑡 1 𝑑𝑡 = ∫−1 < 𝑡 6 , 1 > ∙ √1 + 4𝑡 2 𝑑𝑡 = ∫−1(𝑡 6 − 2𝑡)𝑑𝑡 = 2/7m2/sec. This means that the net flow across the parabola is 2/7m2/sec. √1+4𝑡 2 1 If we use the other form ∅ = ∫𝑐 𝑝𝑑𝑦 − 𝑞𝑑𝑥 = ∫−1(𝑡 6 − 2𝑡)𝑑𝑡 = 2/7m2/sec. Homework 18.2 6 18.3 Gradient Fields and Path-Independent Fields If 𝐹⃗ (𝑥, 𝑦) =< 𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦) > can be represented as 𝐹⃗ (𝑥, 𝑦) = ⃗∇⃗𝑈(𝑥, 𝑦) the gradient of the potential U, then the line integral∫𝐶 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∫𝐶 ⃗∇⃗𝑈(𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟. The Fundamental theorem of Line Integrals states that if C is a smooth curve given by a vector function⃗⃗⃗(𝑡), 𝑟 𝑎 ≤ 𝑡 ≤ 𝑏, and if U (x,y) is a ⃗ ⃗ differentiable function whose gradient vector ∇𝑈(𝑥, 𝑦) is continuous in C, ∫𝑐 ⃗∇⃗𝑈 (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = 𝑈(𝑟⃗(𝑏)) − 𝑈(𝑟⃗(𝑎)). This means that the value of the integral depends on the end points and not on the path. If a force field can be represented as a gradient of a potential, we say that the field is independent of the path. Proof: Let 𝑟⃗ (𝑡) =< 𝑥 (𝑡), 𝑦 (𝑡) > describe the curve C, ⃗⃗⃗⃗⃗⃗ 𝑏 𝑏 𝜕𝑈 𝑑𝑥 𝑑𝑟 𝜕𝑈 𝑑𝑦 ∫𝐶 ⃗∇⃗𝑈 (𝑥, 𝑦). ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∫𝑎 ⃗∇⃗ 𝑈(𝑡) ∙ 𝑑𝑡 𝑑𝑡 = ∫𝑎 ( 𝜕𝑥 𝑑𝑡 + 𝜕𝑦 𝑑𝑡 ) 𝑑𝑡 = 𝑏 𝑑 ∫𝑎 𝑑𝑡 𝑈(𝑡)𝑑𝑡 = 𝑈(𝑏) − 𝑈(𝑎) . eg 20 Evaluate ∫𝐶 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 and ∫𝐶 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 where for 𝐹⃗ (𝑥, 𝑦) =< 𝑥, 𝑦 > 1 2 with C1 : y = x, 0 ≤ x ≤ 1 and C2 : x = 2y2 – y, 0 ≤ y ≤ 1. 1 ∫𝑐 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∫0 < 𝑡, 𝑡 > ∙ < 1, 1 > 𝑑𝑡 = 1 for 𝐶1 ∶< 𝑡, 𝑡 > 0 ≤ 𝑡 ≤ 1 1 1 ∫𝑐 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∫0 < 2𝑡 2 − 𝑡, 𝑡 > ∙ < 4𝑡 − 1, 1 > 𝑑𝑡 = 1 for 𝐶2 ∶ < 2𝑡 2 − 𝑡, 𝑡 > 1 0 ≤ 𝑡 ≤ 1. We see that this line integral is independent of the paths C1 and C2. Any vector field 𝐹⃗ (𝑥, 𝑦) = < 𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦) > can be represented as a gradient of a potential U (that is 𝜕𝑄 𝜕𝑃 𝐹⃗ (𝑥, 𝑦) = ⃗∇⃗𝑈 (𝑥, 𝑦)), if = . 𝜕𝑥 𝜕𝑦 Proof: If 𝐹⃗ (𝑥, 𝑦) = ⃗∇⃗𝑈 (𝑥, 𝑦) ⇒< 𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦) > = < So (𝑥, 𝑦) = 𝜕𝑈 𝜕𝑥 and 𝑄(𝑥, 𝑦) = 𝜕𝑈 𝜕𝑦 𝜕𝑈 𝜕𝑈 , . If we take the partials Since the second mixed partial are equal, 𝜕𝑄 𝜕𝑥 >. 𝜕𝑥 𝜕𝑦 𝜕𝑃(𝑥,𝑦) 𝜕𝑦 𝜕2 𝑈 = 𝜕𝑦𝜕𝑥 and 𝜕𝑄(𝑥,𝑦) 𝜕𝑥 𝜕2 𝑈 = 𝜕𝑥𝜕𝑦. 𝜕𝑃 = 𝜕𝑦 . For the previous example, since 𝐹⃗ (𝑥, 𝑦) =< 𝑥, 𝑦 > the path. Finding the Potential Function 7 𝜕𝑄 𝜕𝑥 𝜕𝑃 = 𝜕𝑦 = 0 so the line integral is independent of ⃗⃗⃗⃗⃗ 𝑓𝑜𝑟 𝐹⃗ (𝑥, 𝑦) = < 𝑒 𝑥 𝑐𝑜𝑠(𝑦) , −𝑒 𝑥 𝑠𝑖𝑛(𝑦) + 1 > for ⃗⃗⃗⃗ (𝑥, 𝑦) ∙ 𝑑𝑟 eg 21 Evaluate ∫𝐶 𝐹 C : y = x in x : [0, 1] (a) without finding the potential function, (b) by finding the potential function. 1 a. Since 𝑟⃗ =< 𝑡, 𝑡 > 0 ≤ 𝑡 ≤ 1, ∫𝐶 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∫0 (𝑒 𝑡 𝑐𝑜𝑠(𝑡) − 𝑒 𝑡 𝑠𝑖𝑛(𝑡) + 1)𝑑𝑡 (𝑠𝑖𝑛𝑡+𝑐𝑜𝑠𝑡) = 𝑒𝑡 ( 2 𝜕𝑄 − (𝑠𝑖𝑛𝑡−𝑐𝑜𝑠𝑡) 2 ) + 𝑡|10 = 𝑒 1 𝑐𝑜𝑠(1) 𝜕𝑃 b. Since 𝜕𝑥 = 𝜕𝑦 = −𝑒 𝑥 𝑠𝑖𝑛(𝑦), there is a potential function. Since 𝐹⃗ (𝑥, 𝑦) = ⃗∇⃗𝑈(𝑥, 𝑦) ⟹ < 𝑒 𝑥 𝑐𝑜𝑠(𝑦) , −𝑒 𝑥 𝑠𝑖𝑛(𝑦) + 1 > = < 𝜕𝑈 𝜕𝑈 , 𝜕𝑥 𝜕𝑦 > and by equating components, 𝜕𝑈 𝜕𝑥 = 𝑒 𝑥 𝑐𝑜𝑠 (𝑦) ↔ ↓ integrate in 𝑥 𝜕𝑈 𝜕𝑦 = −𝑒 𝑥 𝑠𝑖𝑛 (𝑦) + 1 diff wrt 𝑦 𝑈 = 𝑒 𝑥 𝑐𝑜𝑠(𝑦) + 𝑐(𝑦) ↓ compare 𝜕𝑈 ⟶ 𝜕𝑦 ⇒ 𝑐(𝑦) = 𝑦 + 𝑘, so = −𝑒 𝑥 𝑠𝑖𝑛(𝑦) + 𝑐 ′ (𝑦) ⃗⃗⃗⃗⃗ = 𝑒 𝑥 𝑐𝑜𝑠(𝑦) + 𝑦 + 𝑘|(1,1) = 𝑒 1 𝑐𝑜𝑠(1). The potential 𝑈 = 𝑒 𝑥 𝑐𝑜𝑠(𝑦) + 𝑦 + 𝑘, and ∫𝐶 𝐹⃗ (𝑥, 𝑦) ∙ 𝑑𝑟 (0,0) eg 22 Evaluate ∫𝐶 (𝑦𝑠𝑖𝑛(𝑥𝑦))𝑑𝑥 + (𝑥𝑠𝑖𝑛(𝑥𝑦) + 2𝑦)𝑑𝑦 between the points (1, 2) and (3, 4). 𝜕𝑄 𝜕𝑃 Since 𝐹⃗ (𝑥, 𝑦) =< 𝑦𝑠𝑖𝑛(𝑥𝑦), 𝑥𝑠𝑖𝑛(𝑥𝑦) + 2𝑦 >, 𝜕𝑥 = 𝜕𝑦 = 𝑥𝑦𝑐𝑜𝑠(𝑦) + 𝑠𝑖𝑛(𝑥𝑦), there is a potential function. Since < 𝜕𝑈 𝜕𝑥 𝜕𝑈 𝜕𝑈 , 𝜕𝑥 𝜕𝑦 = 𝑦𝑠𝑖𝑛(𝑥𝑦) > = < 𝑦𝑠𝑖𝑛(𝑥𝑦), 𝑥𝑠𝑖𝑛(𝑥𝑦) + 2𝑦 >, ⟷ ↓ integrate in 𝑥 𝑈 = − cos(𝑥𝑦) + 𝑐(𝑦) 𝜕𝑈 𝜕𝑦 = 𝑥𝑠𝑖𝑛(𝑥𝑦) + 2𝑦 diff wrt 𝑦 → ↓ compare 𝜕𝑈 = 𝑥𝑠𝑖𝑛(𝑥𝑦) + 𝑐 ′ (𝑦) 𝜕𝑦 ⇒ 𝑐(𝑦) = 𝑦 2 + 𝑘, so The potential 𝑈 = − 𝑐𝑜𝑠(𝑥𝑦) + 𝑦 2 + 𝑘 and (3,4) ⃗⃗⃗⃗⃗ = − 𝑐𝑜𝑠(𝑥𝑦) + 𝑦 2 + 𝑘| ⃗⃗⃗⃗(𝑥, 𝑦). 𝑑𝑟 ∫𝐶 𝐹 (1,2) = cos(2) − cos(12) + 12 . Conservative Vector Fields and Closed Curves If C is a smooth-closed curve in the x-y plane, then ∫𝑐 ⃗⃗⃗⃗ 𝐹 ∙ 𝑑𝑟⃗ = ∮𝐶 ⃗⃗⃗⃗ 𝐹 ∙ 𝑑𝑟⃗ . If 𝐹⃗ = < 𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦) > is a force field, ∮𝐶 𝐹⃗ ∙ 𝑑𝑟⃗ = ∮ 𝐹⃗ ∙ 𝑇̂ 𝑑𝑠 = ∮𝐶 (𝑃𝑑𝑥 + 𝑄𝑑𝑦) is the work 𝐶 done by the force field 𝐹⃗ . 8 ̂ 𝑑𝑠 = ∮ (𝑃𝑑𝑦 − 𝑄𝑑𝑥) will give the flow across If 𝑣⃗ =< 𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦) > is a velocity field, ∮𝐶 𝑣⃗ ∙ 𝑁 𝐶 the closed curve, and ∫𝑐 𝑣⃗ ∙ 𝑇̂ 𝑑𝑠 = ∮𝐶 𝑣⃗ ∙ 𝑑𝑟⃗ = ∮𝐶 (𝑃𝑑𝑥 + 𝑄𝑑𝑦) is the circulation to the fluid around C, and If 𝐹⃗ (𝑥, 𝑦) is a conservative vector field, the line integral along a closed curve C, ⃗⃗⃗⃗⃗⃗ 𝑎 𝑑𝑟 ∮ 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∫ ⃗∇⃗ 𝑈(𝑥(𝑡), 𝑦(𝑡)) ∙ 𝑑𝑡 = 𝑈(𝑟⃗(𝑎)) − 𝑈(𝑟⃗ (𝑎)) = 0, since its initial and final 𝐶 𝑎 𝑑𝑡 points are the same. If 𝐹⃗ (𝑥, 𝑦) is a conservative force field, the work along a closed curve is zero. If 𝑣⃗ (𝑥, 𝑦) is a conservative velocity field, the circulation along a closed curve is zero. A fluid like this is called irrotational (no vortices in the flow). NOTE: A line integral independent of the path can be zero with different initial and final points. (Open curve) (𝜋−1) eg 23 Evalute ∫(0,1) 𝑦𝑠𝑖𝑛(𝑥)𝑑𝑥 − 𝑐𝑜𝑠(𝑥) 𝑑𝑦 . 𝜕𝑄 𝜕𝑃 Since 𝜕𝑥 = 𝜕𝑦 = 𝑠𝑖𝑛(𝑥) , and < 𝜕𝑈 𝜕𝑥 = 𝑦𝑠𝑖𝑛(𝑥) ↓ integrate in 𝑥 𝜕𝑈 𝜕𝑈 𝜕𝑈 ⟷ 𝜕𝑦 diff wrt. 𝑦 𝑈 = −𝑦𝑐𝑜𝑠(𝑥) + 𝑐(𝑦) → , 𝜕𝑥 𝜕𝑦 > = < 𝑦𝑠𝑖𝑛(𝑥), − 𝑐𝑜𝑠(𝑥) >, = −𝑐𝑜𝑠(𝑥) ↓ compare 𝜕𝑈 𝜕𝑦 ⇒ 𝑐(𝑦) = 𝑘, so = − cos(𝑥) + 𝑐 ′ (𝑦) The potential 𝑈 = −𝑦𝑐𝑜𝑠(𝑥) + 𝑘 and (𝜋,−1) (𝜋,−1) ∫(0,1) 𝑦𝑠𝑖𝑛(𝑥)𝑑𝑥 − 𝑐𝑜𝑠(𝑥) 𝑑𝑦 = −𝑦𝑐𝑜𝑠(𝑥) + 𝑘|(0,1) = 0. Conservation of Energy 𝑚 𝑏 𝑑 (𝑟⃗̇ ∙ 𝑟⃗̇)𝑑𝑡 = 2 ∫𝑎 𝑑𝑡 |𝑟⃗̇|2 𝑑𝑡 𝑚 = (𝑣(𝑏) − 𝑣(𝑎)) = 𝑘(𝑏) − 𝑘(𝑎) = ∆𝑘 . The work done by the force field 𝐹⃗ along a path C on an 2 object is the change in its kinetic energy. ⃗⃗𝑃 with potential P (x, y, z), Since 𝐹⃗ is conservative, 𝐹⃗ = −∇ 𝑊 = ∫𝐶 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∫𝐶 − ⃗∇⃗𝑃 ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = −𝑃(𝑟(𝑏) − 𝑟(𝑎)) = −(𝑃(𝑏) − 𝑃(𝑎)) = −∆𝑃 the opposite of the change in potential energy Since ∆𝑘 + ∆𝑃 = 0, ∆𝑘 = −∆𝑃. Since the two integrals are the same, 𝑊 = ∫𝐶 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∆𝑘 = −∆𝑃 or ∆𝑘 = ∆𝑃. “Conservation of energy”. 𝑚 𝑊 = ∫𝐶 𝐹⃗ (𝑥, 𝑦) ∙ ⃗⃗⃗⃗⃗ 𝑑𝑟 = ∫𝐶 𝑚𝑟⃗̈ ∙ 𝑟⃗̇ 𝑑𝑡 = 2 ∫𝐶 𝑑 𝑑𝑡 Homework 18.3 9 Line Integrals Independent of the Path Worksheet 1) Evaluate the line integrals (a) by direct integration along any path, (b) by finding the potential. (1,3) a) b) y 2 dx 2 xydy ( 1,2) ( , 1) (0,1) Ans: 13 y sin xdx cos xdy Ans: 0 2) Integrate by any method. a) b) c) c y sin xdx cos xdy ( c any path) (3,2) (0,0) Ans: 9e2 4 2 xe y dx ( x 2e y 2 y )dy ( 1,0) (2, 2) Ans: 0 2 xy 3dx (3 y 2 x 2 1)dy Ans: 34 y x N on a particle that moves from the , 2 2 x y x y2 point P (0, 2) to the point Q (3,3) with distance measured in meters. Ans: / 4 J 3) Find the work done by the force F 2 4) Find the circulation of the velocity field v e y ye x , xe y e x of a fluid from the point (1, 0) to the point (-1, 0), without finding the potential function, a) along the x-axis. Ans: 2 b) along the semicircle 1 x 2 . Ans: 2 5) Find the flux of the velocity field v y, x of a fluid with density 1 from the point (1, 0) to the point (-1, 0), without finding the potential function, a) along the x-axis. Ans: 0 b) along the semicircle 1 x 2 . Ans: 0 18.4 Green’s Theorem 10 Green’s Theorem is a consequence of the First Fundamental Theorem of Calculus (Leibniz’s Rule) applied to double integrals. It give the relationship between a line integral of a simply connected closed curve C, with positive orientation (the region is at the left as the curve is traversed) and a double integral over a plane region R. Green’s Theorem: Let C be a positively oriented, piecewise smooth simply connected closed curve in ℜ2 and let R be the region bounded by C. If P and Q have continuous partial derivatives on an open region with no holes containing R, then the line integral of 𝐹⃗ around the closed curve C is given by 𝜕𝑄 𝜕𝑃 𝜕𝑄 𝜕𝑃 ∮ 𝑃 (𝑥, 𝑦)𝑑𝑥 + 𝑄(𝑥, 𝑦)𝑑𝑦 = ∫ ∫𝑅 (𝜕𝑥 (𝑥, 𝑦) − 𝜕𝑦 (𝑥, 𝑦)) 𝑑𝑥𝑑𝑦 , where 𝜕𝑥 (𝑥, 𝑦) − 𝜕𝑦 (𝑥, 𝑦) is called the two dimensional or scalar Curl of the vector field 𝐹⃗ =< 𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦) > 𝑐1 𝑐2 𝑐3 𝑐4 3 5 𝜕𝑃 eg 24 Given R, the square (1,1) → (5,1) → (5,3) → (1,3) → (1,1) find ∫1 ∫1 3 5 𝜕𝑃 5 3 𝜕𝑃 𝜕𝑦 (𝑥, 𝑦)𝑑𝑥𝑑𝑦. 5 ∫1 ∫1 𝜕𝑦 (𝑥, 𝑦)𝑑𝑥𝑑𝑦 = ∫1 ∫1 (𝜕𝑦 (𝑥, 𝑦)𝑑𝑦) 𝑑𝑥 = ∫1 (𝑃(𝑥, 3) − 𝑃(𝑥, 1))𝑑𝑥 by Leibniz’s Rule. Also, ∫𝑐 𝑃(𝑥, 𝑦)𝑑𝑥 = ∫𝑐 𝑃(𝑥, 𝑦)𝑑𝑥 + ∫𝑐 𝑃(𝑥, 𝑦)𝑑𝑥 + ∫𝑐 𝑃(𝑥, 𝑦)𝑑𝑥 + ∫𝑐 𝑃(𝑥, 𝑦)𝑑𝑥 = 5 ∫1 𝑃(𝑥, 1)𝑑𝑥 1 5 + ∫5 𝑃(𝑥, 𝑦)𝑑𝑥 2 + 1 ∫5 𝑃(𝑥, 3)𝑑𝑥 From this example, we see that ∫ ∫𝑅 𝜕𝑃 𝜕𝑦 𝑐1 3 + 1 ∫1 𝑃(𝑥, 𝑦)𝑑𝑥 4 =− 5 ∫1 (𝑃(𝑥, 3) (𝑥, 𝑦)𝑑𝑥𝑑𝑦 = −∫𝑐 𝑃(𝑥, 𝑦)𝑑𝑥. 𝑐2 𝑐3 𝑐4 3 5 𝜕𝑄 eg 25 Given R the square (1,1) → (5,1) → (5,3) → (1,3) → (1,1), find ∫1 ∫1 3 5 𝜕𝑄 − 𝑃(𝑥, 1)) 𝑑𝑥 . 3 𝜕𝑥 (𝑥, 𝑦)𝑑𝑥𝑑𝑦 ∫1 ∫1 𝜕𝑥 (𝑥, 𝑦)𝑑𝑥𝑑𝑦 = ∫1 (𝑄(5, 𝑦) − 𝑄(1, 𝑦))𝑑𝑦 by Leibniz’s Rule. Also ∫𝐶 𝑄(𝑥, 𝑦)𝑑𝑦 = ∫𝑐 𝑄(𝑥, 𝑦)𝑑𝑦 + ∫𝑐 𝑄(𝑥, 𝑦)𝑑𝑦 + ∫𝑐 𝑄(𝑥, 𝑦)𝑑𝑦 + ∫𝑐 𝑄(𝑥, 𝑦)𝑑𝑦 = 1 ∫1 𝑄(𝑥, 𝑦)𝑑𝑦 3 1 2 3 3 1 4 3 + ∫1 𝑄(5, 𝑥)𝑑𝑦 + ∫3 𝑄(𝑥, 𝑦)𝑑𝑦 + ∫3 𝑄(1, 𝑦)𝑑𝑦 = ∫1 (𝑄(5, 𝑦) − 𝑄(1, 𝑦))𝑑𝑦 . 𝜕𝑄 From this example, we see that ∫ ∫𝑅 𝜕𝑥 (𝑥, 𝑦)𝑑𝑥𝑑𝑦 = ∫𝑐 𝑄(𝑥, 𝑦)𝑑𝑦 . If we add the last two results we get Green’s Theorem. 𝜕𝑄 𝜕𝑃 ∮ 𝑃(𝑥, 𝑦)𝑑𝑥 + 𝑄(𝑥, 𝑦)𝑑𝑦 = ∫ ∫𝑅 ( 𝜕𝑥 (𝑥, 𝑦) − 𝜕𝑦 (𝑥, 𝑦)) 𝑑𝑥𝑑𝑦 . eg 26 Express ∫𝑐 𝑥𝑠𝑖𝑛(𝑦)𝑑𝑥 + 𝑥 2 𝑐𝑜𝑠 2 (𝑦)𝑑𝑦 as a double integral over R where c is a simply connected closed curve. ∫𝑐 𝑥𝑠𝑖𝑛(𝑦)𝑑𝑥 + 𝑥 2 𝑐𝑜𝑠 2 (𝑦)𝑑𝑦 = ∫ ∫𝑅 (2𝑥𝑐𝑜𝑠 2 (𝑦) − 𝑥𝑐𝑜𝑠(𝑦))𝑑𝑥𝑑𝑦. eg 27 Evaluate ∫𝑐 𝑥 4 𝑑𝑥 + 𝑥𝑦𝑑𝑦 where C is the triangle with vertices (0, 0), (1, 0), (0, 1). To evaluate without using Green’s Theorem, Let c1 : < t, 0 >, 0 < t < 1 the line segment from (0, 0) to (1, 0). Let c2 : < 1 – t, t >, 0 < t < 1 the line segment from (1, 0) to (0, 1). Let c3 : < 0, 1 – t >, 0 < t < 1 the line segment from (1, 0) to (0, 0). 1 1 1 1 1 1 1 [∫0 𝑡 4 𝑑𝑡 + ∫0 0𝑑𝑡] + [∫0 (1 − 𝑡)4 (−1)𝑑𝑡 + ∫0 (𝑡 − 𝑡 2 )𝑑𝑡] + [∫0 0𝑑𝑡 + ∫0 0𝑑𝑡] = 6 . 1 1−𝑥 If we apply Green’s Theorem, ∫𝑐 𝑥 4 𝑑𝑥 + 𝑥𝑦𝑑𝑦 = ∫0 ∫0 11 1 𝑦 𝑑𝑦𝑑𝑥 = 6. A consequence of Green’s Theorem is: ∫ 𝑥𝑑𝑦 𝐴𝑟𝑒𝑎 = ∫ ∫𝑅 𝑑𝑥𝑑𝑦 = { 𝐶 or if we combine both formulas, −∫𝑐 𝑦𝑑𝑥 1 𝐴𝑟𝑒𝑎 = 2 ∫𝑐 (𝑥𝑑𝑦 − 𝑦𝑑𝑥) . 𝑥2 𝑦2 eg 28 Find the area of the ellipse 𝑎2 + 𝑏2 = 1, 𝑎 > 𝑏 > 0 by a) double integration; b) using Green’s Theorem. 𝑎 𝑏 √𝑎2 −𝑥2 𝜋 𝑎𝑏 𝑏 a) 𝐴𝑟𝑒𝑎 = 4 ∫0 ∫0𝑎 𝑑𝑦𝑑𝑥 = 4 ∫0 𝑎 √𝑎2 − 𝑥 2 𝑑𝑥 = 4 ∫02 𝑎 𝑎2 𝑐𝑜𝑠 2 (𝜃)𝑑𝜃 = 𝑎𝑏𝜋. b) 𝑟⃗ =< 𝑎 𝑐𝑜𝑠(𝑡) , 𝑏 𝑠𝑖𝑛(𝑡) >, 𝑑 𝑟⃗ =< −𝑎 𝑠𝑖𝑛(𝑡) , 𝑏 𝑐𝑜𝑠(𝑡) >; 0 < 𝑡 < 2𝜋 1 2𝜋 𝐴𝑟𝑒𝑎 = 2 ∫0 (𝑎𝑏 𝑐𝑜𝑠 2 (𝑡) + 𝑎𝑏 𝑠𝑖𝑛2 (𝑡))𝑑𝑡 = 𝑎𝑏𝜋 . Green’s Theorem in Vector Form Let 𝐹⃗ (𝑥, 𝑦) =< 𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦) > 𝑎𝑛𝑑 𝑟⃗ =< 𝑥, 𝑦 >, 𝑑 𝑟⃗ =< 𝑑𝑥, 𝑑𝑦 >, Since ∫𝑐 𝑃(𝑥, 𝑦)𝑑𝑥 + 𝑄(𝑥, 𝑦)𝑑𝑦 = ∮ 𝐹⃗ (𝑥, 𝑦) ∙ 𝑑𝑟⃗, Green’s Theorem in vector form becomes 𝜕𝑄 𝜕𝑃 ∮ 𝐹⃗ (𝑥, 𝑦) ∙ 𝑑𝑟⃗ = ∫ ∫ ( (𝑥, 𝑦) − (𝑥, 𝑦)) 𝑑𝐴. 𝑅 𝜕𝑥 𝜕𝑦 eg 29 Evaluate ∫𝑐 < 𝑥 2 − 𝑦 2 , 𝑥 > ∙ 𝑑𝑟⃗ over the unit circle a) by using Green’s Theorem, b) by direct integration. a) ∫𝑐 < 𝑥 2 − 𝑦 2 , 𝑥 > ∙ 𝑑𝑟⃗ = ∫ ∫𝑅 (1 + 2𝑦)𝑑𝐴 = 2𝜋 1 2𝜋 1 2 ∫0 ∫0 (1 + 2𝑟 𝑠𝑖𝑛(𝜃) 𝑟𝑑𝑟𝑑𝜃 = ∫0 (2 + 3 sin(𝜃)) 𝑑𝜃 = 𝜋 b) Since 𝑟⃗⃗⃗ =< 𝑐𝑜𝑠(𝑡) , 𝑠𝑖𝑛(𝑡) >, 𝑑𝑟⃗ =< − 𝑠𝑖𝑛(𝑡) , 𝑐𝑜𝑠(𝑡) > 2𝜋 ∫𝑐 < 𝑥 2 − 𝑦 2 , 𝑥 > ∙ 𝑑𝑟⃗ = ∫0 (−𝑐𝑜𝑠 2 (𝑡) 𝑠𝑖𝑛(𝑡) + 𝑠𝑖𝑛3 (𝑡) + 𝑐𝑜𝑠 2 (𝑡))𝑑𝑡 = 𝜋 Curl Test for Vector Fields in 2-Space Let 𝐹⃗ (𝑥, 𝑦) =< 𝑃(𝑥, 𝑦), 𝑄(𝑥, 𝑦) > be a vector field with continuous partial derivatives such that the domain of 𝐹⃗ has the property that every closed curve that encircles it, encircles a region the lies entirely 𝜕𝑄 𝜕𝑃 in the domain (𝐹⃗ has no holes), and the curl of 𝐹⃗ is zero ( − = 0), then 𝐹⃗ is path independent, so 𝐹⃗ 𝜕𝑥 𝜕𝑦 is a gradient of some potential. 𝜕𝑄 𝜕𝑃 By Green’s Theorem ∮𝑐 𝐹⃗ (𝑥, 𝑦) ∙ 𝑑𝑟⃗ = ∫ ∫𝑅 ( (𝑥, 𝑦) − (𝑥, 𝑦)) 𝑑𝐴, if the curl of 𝐹⃗ is zero 𝜕𝑥 𝜕𝑄 𝜕𝑦 𝜕𝑃 (𝜕𝑥 − 𝜕𝑦 = 0), then ∮𝑐 𝐹⃗ (𝑥, 𝑦) ∙ 𝑑 𝑟⃗ = ∫ ∫𝑅 0 𝑑𝐴 = 0, so the vector field 𝐹⃗ (x, y) is path independent (conservative). eg 30 Is the vector field 𝐹⃗ (x, y) = < y, x > conservative over the unit circle? 𝜕𝑄 𝜕𝑃 Since 𝜕𝑥 = 𝜕𝑦 = 1, by the curl test for vector fields in 2-space is zero, so 𝐹⃗ ((x, y) = < y, x > is a conservative vector Field. By Green’s Theorem, ∫𝑐 𝑦𝑑𝑥 + 𝑥𝑑𝑦 = ∫ ∫𝑅 (1 − 1)𝑑𝐴. = 0. 12 eg 31 Is ∮𝑐 𝐹⃗ (𝑥, 𝑦) ∙ 𝑑𝑟⃗ independent of the path for 𝐹⃗ (𝑥, 𝑦) =< −𝑦 , 𝑥 𝑥 2 +𝑦 2 𝑥 2 +𝑦 2 > over the unit circle? 2𝜋 By Direct integration ∫0 < −𝑠𝑖𝑛𝑡, 𝑐𝑜𝑠𝑡 > ∙ < −𝑠𝑖𝑛𝑡, 𝑐𝑜𝑠𝑡 > 𝑑𝑡 = 2𝜋 ≠ 0 so the line integral is not independent of the path. 𝜕𝑄 𝑦 2 −𝑥 2 𝜕𝑃 Since 𝜕𝑥 = 𝜕𝑦 = (𝑥 2 +𝑦 2 )2, by the curl test for vector fields in 2-space is zero. This does not imply that the integral is independent of the path even though by Green’s Theorem, −𝑦 𝑦 2 −𝑥 2 𝑥 𝑦 2 −𝑥 2 ∮ 𝑥 2 +𝑦 2 𝑑𝑥 + 𝑥 2 +𝑦 2 𝑑𝑦 = ∫ ∫𝑅 ((𝑥 2 +𝑦 2 )2 − (𝑥 2 +𝑦2 )2 ) 𝑑𝐴. 0 . The discrepancy of the two results is that 𝐹⃗ (𝑥, 𝑦) has a hole at (0, 0) so Green’s Theorem can not be 𝜕𝑄 𝜕𝑃 used, or ∮𝑐 𝐹⃗ (𝑥, 𝑦) ∙ 𝑑 𝑟⃗ ≠ ∫ ∫𝑅 (𝜕𝑥 (𝑥, 𝑦) − 𝜕𝑦 (𝑥, 𝑦)) 𝑑𝐴 since the curl test does not apply. A vector field can be path independent even if the domain constrain of Green’s Theorem does not apply. 𝑥 𝑦 eg 32 Consider 𝐹⃗ (𝑥, 𝑦) =< 𝑥 2 +𝑦 2 , 𝑥 2 +𝑦 2 > 𝜕𝑄 𝜕𝑃 −2𝑥𝑦 Even though 𝜕𝑥 = 𝜕𝑥 = (𝑥 2 +𝑦2 )2 the curl test does not apply over any path containing the point (0, 0). Since 𝜕𝑈 𝑥 = 2 𝜕𝑥 𝑥 + 𝑦 2 ↔ ↓ integrate in x diff wrt y 1 𝑈 = 2 𝑙𝑛(𝑥 2 + 𝑦 2 ) + 𝑐(𝑦) 𝜕𝑈 𝑦 = 2 𝜕𝑦 𝑥 + 𝑦 2 ↓ compare → 𝜕𝑈 𝜕𝑦 ⇒ c(y) = k, so 𝑦 = 𝑥 2 +𝑦2 + 𝑐 ′ (𝑦) 1 The potential 𝑈 = 2 𝑙𝑛(𝑥 2 + 𝑦 2 ) so the vector field is conservative. To find a line integral for this vector field, we can use the potential function (since the integral is independent of the path), but we should not use Green’s Theorem for any closed path that includes the origin (the curl test will not apply). 13