12/3/2002, LINE INTEGRALS Math 21a, O. KnillHOMEWORK:Due 12/5:Section 13.1: 6, 29-32Section 13.2: 10,14Due 12/10:Section 13.3: 16, 26Section 13.4: 8, 18, 20PLANAR VECTOR FIELD EXAMPLES.1) F (x, y) = (y, −x) is a planar vector field which you see in a pictureon the right.2) F (x, y) = (x − 1, y)/((x − 1)2+ y2)3/2− (x + 1, y)/((x + 1)2+ y2)3/2is the electric field of positive and negative point charge. It is calleddipole field. See picture.3) Gradient field. If f (x, y, z) is a function of three variables, then∇f(x, y, z) is called the gradient field or conservative. For example,(2x, 2y, −2z) is the vector field which is orthogonal to hyperboloids.4) If H(x, y) is a function of two variables, then (Hy(x, y), −Hx(x, y))is called a Hamiltonian vector field. Example: Harmonic OscillatorH(x, y) = x2+ y2. (Hy(x, y), Hx(x, y)) = (y, −x) is the vector field in1).LINE INTEGRALS. If F (x, y, z) is a vector field and γ : t 7→ r(t) is a curve, thenRbaF (r(t)) · r0(t) dtis calledthe line integral of F along the curve γ. The short-hand notationRγF · ds is also used. In the literature,where curves are sometimes written as r(t) = (x(t), y(t), z(t)) or r(t), the notationRγF ·dr orRγF ·dr appears.EXAMPLE: Work. If F (x, y, z) is a force field, then the line integralRbaF (r(t)).r0(t) dt is called work.EXAMPLE: Electric potential. If E(x, y, z) is an electric field, then the line integralRbaE(r(t)) · r0(t) dt iscalled electric potential.EXAMPLE: Gradient field. If F (x, y, z) = ∇U(x, y, z) is a gradient field, thenRbaF (r(t))·r0(t) dt = U(r(b))−U(r(a)). The gradient field has physical relevance. For example, if U (x, y, z) is the pressure distribution in theatmosphere, then ∇U (x, y, z) is the pressure gradient. It coincides roughly with the wind velocity field.EXAMPLE 1. Let γ : t 7→ r(t) = (cos(t), sin(t)) be a circle parametrized byt ∈ [0, 2π] and let F(x, y) = (−y, x). Calculate the line integral I =RγF (r)·dr.ANSWER: We have I =R2π0F (r(t)) · r0(t) dt =R2π0(− cos(t), sin(t)) ·(− cos(t), sin(t)) dt =R2π0cos2(t) + sin2(t) dt = 2πEXAMPLE 2. Let r(t) be a curve given in polar coordinates as r(t) = cos(t), φ(t) = t defined on [0, π]. Let Fbe the vector field F (x, y) = (−xy, 0). Calculate the line integralRγF · dr.SOLUTION. In Cartesian coordinates, the curve is r(t) = (cos2(t), cos(t) sin(t)). The velocity vector is thenr0(t) = (−2 sin(t) cos(t), − sin2(t) + cos2(t)) = (x(t), y(t)). The line integral isZπ0F (r(t)) · r0(t) dt =Zπ0(cos4(t) sin(t), 0) · (−2 sin(t) cos(t), − sin2(t) + cos2(t)) dt= −2Zπ0sin2(t) cos4(t) dt = −2(t/16 + sin(2t)/64 − sin(4t)/64 − sin(6t)/192)|π0= −π/8FUNDAMENTAL THEOREM OF LINE INTEGRALS. If F = ∇U, thenRbaF (r(t)) · r0(t) dt = U(r(b)) − U (r(a))The line integral gives the potential difference between the points r(b) and r(a).EXAMPLE. Let U (x, y, z) be the temperature distribution in a room and let r(t) the path of a fly in the room,then U (r(t)) is the temperature, the fly experiences at the point r(t) at time t. The change of temperature forthe fly isddtU(r(t)). The line-integral of the temperature gradient ∇U along the path of the fly coincides withthe temperature difference.SPECIAL CASES.r(t) parallel to level curve means d/dtU(r(t)) = 0 and r0(t) orthogonal to ∇U (r(t))r(t) orthogonal to level curve means |d/dtU(r(t))| = |∇U ||r0(t)| and r0(t) parallel to ∇U (r(t)).PROOF OF THE FUNDAMENTAL THEOREM. Use the chain rule in the second equality and the fundamentaltheorem of calculus in the third equality:ZbaF (r(t)) · r0(t) dt =Zba∇U(r(t)) · r0(t) dt =ZbaddtU(r(t)) dt = U (r(b)) − U (r(a)) .ADDING AND SUBTRACTING CURVES.If γ1, γ2are curves, then γ1+γ2denotes the curve obtainedby traveling first along γ1, then along γ2. One writes −γfor the curve γ traveled backwards and γ1−γ2= γ1+(−γ2).EXAMPLES. If γ1(t) = (t, 0) for t ∈ [0, 1], γ2(t) = (1, (t −1)) for t ∈ [1, 2], γ3(t) = (1 − (t − 2), 1) for t ∈ [2, 3],γ4(t) = (0, 1−(t−3)) for t ∈ [3, 4], then γ = γ1+γ2+γ3+γ4for t ∈ [0, 4] is the path which goes around a the unit square.The path −γ travels around in the clockwise direction.xyCALCULATING WITH LINE-INTEGRALS.•RγF · dr +RγG · dr =Rγ(F + G) · dr.•RγcF · dr = cRγF · dr.•Rγ1+γ2F · dr =Rγ1F · dr +Rγ2F · dr•R−γF · dr = −RγF · dr.CLOSED CURVES. We see from the last example that the line integral along a closed curve is zero if the vectorfield is a gradient field. The work done along a closed path is zero. This is a form of energy conservation.PERPETUUM MOBILES. A machine which imple-ments a force field which is not a gradient fieldis called a perpetuum mobile. Mathematically,it realizes a force field for which there exist someclosed loops along which the energy gain is nonnega-tive. (By possibly changing the direction, the energychange can be made positive). The first law of ther-modynamics forbids the existence of such a machine.It is informative to stare at some of the ideas peoplehave come up with and to see why they don’t work.The drawings of Escher appear also to produce situ-ations, where a force field can be used to gain energy.Escher uses genius graphical tricks however.WHEN IS A VECTOR FIELD A GRADIENT FIELD (2D)?F (x, y) = ∇U (x, y) implies Fy(x, y) = Fx(x, y). If this does not hold at some point, F is no gradient field. Wewill see in the next hour, how the condition curl(F ) = Fy− Fx= 0 can assure that F is
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