Surface Integrals of Vector Fields

Transcription

1 Mth 32B iscussion ession Week 7 Notes Februry 21 nd 23, 2017 In lst week s notes we introduced surfce integrls, integrting sclr-vlued functions over prmetrized surfces. As with our previous integrls, we used trnsformtion (nmely, the prmetriztion) to rewrite our integrl over more fmilir domin, nd picked up fudge fctor long the wy. This week we wnt to integrte vector fields over surfces. urfce Integrls of Vector Fields uppose we hve surfce R 3 nd vector field F defined on R 3, such s those seen in the following figure: We wnt to mke sense of wht it mens to integrte the vector field over the surfce. Tht is, we wnt to define the symbol F d. When defining integrtion of vector fields over curves we set things up so tht our integrl would mesure the work done by the vector field long the curve. This doesn t lift very nicely to surfces which direction in the surfce should be considered positive? Insted of mesuring the extent to which F is pushing long the surfce, we ll mesure the extent to which F is pushing through. o insted of work, we wnt the surfce integrl of vector field to mesure the flux of the vector field through the surfce. We cn mesure the pointwise contribution to flux in much the sme wy we mesured the pointwise contribution to work for line integrls. We focus on fixed point p in our surfce. At this point, the flux through is the sme s the flux through the tngent plne to t p, s indicted here: 1

2 The blue vector in this figure is the vector ssocited to p by the vector field F. Only tht prt of this vector which is perpendiculr to the tngent plne is mking ny contribution to the flux. Tht is, the blue vector cn be decomposed s sum of two vectors, one of which is perpendiculr to the tngent plne, nd one of which is in the tngent plne. The vector tht is in the tngent plne is not pushing through the surfce, nd thus mkes no contribution to the flux. All of the flux dt of our blue vector is encoded in its projection onto the line norml to the tngent plne. The pointwise contribution to flux is then the mgnitude of this vector, so we see tht pointwise contribution to flux = F n, where n is the unit norml vector to t p pointing in the positive direction. For this reson we mke the following definition. efinition. Let R 3 be surfce nd suppose F is vector field whose domin contins. We define the vector surfce integrl of F long to be F d := (F n)d, where n(p ) is the unit norml vector to the tngent plne of t P, for ech point P in. The sitution so fr is very similr to tht of line integrls. When integrting sclr vlued functions we pick up strnge fudge fctor, nd when integrting vector fields we compute the dot product of our vector field with some distinguished unit vector field. Just s in the line integrl cse, the fudge fctor nd the distinguished vector field re relted in wy tht gretly simplifies the computtionl difficulty of integrting vector fields. Theorem 1. Let G(u, v) be n oriented prmetriztion of n oriented surfce with prmeter domin. Assume tht G is one-to-one nd regulr, except possibly t the boundry of. Then F d = F(G(u, v)) N(u, v)dudv. Our lone ppliction of this theorem will be fluid flow exmple. If fluid is flowing with velocity vector field v, its flow rte cross, mesured in units of volume per unit time, is given by flow rte cross = v d. Exmple. (ection 17.5, Exercise 26) upose we hve net given by y = 1 x 2 z 2, where y 0, nd this net is dipped in river. If the river hs velocity vector field given by v = x y, z + y + 4, z 2, determine the flow rte of wter through the net in the positive y-direction. 2

3 (olution) Here s plot of the net in question, long with severl vectors depicting the flow of the river. Note tht these velocity vectors hve been given unit length for purposes of the figure, but in fct hve vrying mgnitudes. Now we need prmetriztion of this surfce. Becuse the projection of this surfce onto the xz-plne is unit circle, it s very nturl to define G(r, θ) = (x(r, θ), y(r, θ), z(r, θ)), where x(r, θ) = r cos θ, z(r, θ) = r sin θ, nd then y(r, θ) = 1 r 2, for 0 r 1 nd 0 θ 2π. We cn now compute our tngent vectors t ech point (r, θ): T r (r, θ) = cos θ, 2r, sin θ nd T θ (r, θ) = r sin θ, 0, r cos θ. o we hve norml vector is given by i j k T r (r, θ) T θ (r, θ) = cos θ 2r sin θ r sin θ 0 r cos θ = 2r2 cos θ, r, 2r 2 sin θ. Notice tht since r < 0, this vector points in the negtive y-direction, opposite our orienttion. o we choose the norml vector pointing in the opposite direction nd hve Next, notice tht N(r, θ) = 2r 2 cos θ, r, 2r 2 sin θ. v(g(r, θ)) = r cos θ (1 r 2 ), r sin θ + (1 r 2 ) + 4, r 2 sin 2 θ. 3

4 o we hve v(g(r, θ)) N(r, θ) = 2r 3 cos 2 θ 2r 2 cos θ(1 r 2 ) + r 2 sin θ + 5r r 3 + 2r 4 sin 3 θ Finlly, our flow rte cross is given by v d = = 2π π = 2r 2 cos θ(r r cos θ) + 2r 4 sin 3 θ + r(5 r 2 + r sin θ). v(g(r, θ)) N(r, θ)drdθ 2r 2 cos θ(r r cos θ) + 2r 4 sin 3 θ + r(5 r 2 + r sin θ)drdθ = 5π. This lst integrl isn t hrd, it s just relly ugly. ince this integrl is not very nice, one might expect sphericl prmetriztion such s G(θ, φ) = (cos θ sin φ, cos 2 φ, sin θ sin φ) would work out better. I didn t find the resulting integrl to be ny nicer. Line Integrls nd urfce Integrls We ll finish by summrizing the vrious integrls we ve considered in the lst few sections. Throughout, is curve prmetrized by r(t) with t [, b] nd is surfce prmetrized by G(u, v) with (u, v). We write T for unit tngent vector to curve, nd we write n for unit norml vector to curve or surfce. In both cses we ssume tht the vectors re positively oriented. Given prmetriztion G of, we lso write N(u, v) = ±G u G v = ± G v. The ± is needed here becuse we wnt N(u, v) to be outwrd-pointing, nd this my require negting G u G v.

5 rc length/surfce re integrl of sclr function f integrl of vector field F fluid flowing through or F dr := F nds = Line Integrls urfce Integrls r (t) dt f(r(t)) r (t) dt (F T)ds = F(r(t)) r (t)dt F(r(t)) n(t) r (t) dt F d := F d = f(g(u, v)) G v (F n)d = dudv G v dudv F(G(u, v)) N(u, v)dudv F(G(u, v)) N(u, v)dudv For midterm 2 we re primrily concerned with the first three rows of the left column nd the first two rows of the right column.