Mth 32B iscussion ession ession 7 Notes August 28, 28 In tody s discussion we ll tlk bout surfce integrls both of sclr functions nd of vector fields nd we ll try to relte these to the mny other integrls we ve discussed so fr. As with the recent notes on chnge of vribles, mny of the detils presented here won t ctully show up in discussion, where we ll focus insted on computtion. urfce Integrls Recently we ve found ourselves needing to compute integrls over less-thn-idel objects, such s funny regions or curves in Eucliden spce. A technique we ve used to compute these integrls is to find some trnsformtion from nice region in Eucliden spce whose imge is the region or curve we cre bout, nd then to pull our integrl bck to the nice region using the trnsformtion. For instnce, when integrting over strnge regions in the plne, we found suitble trnsformtion T : R 2 R 2 which we used to rewrite our integrl s n integrl over rectngle. For line integrls we wnt to integrte over some curve, nd we use prmetriztion r of this curve to write our integrl s n integrl over n intervl in R. In both cses, the trnsformtion gives us some stretch fctor which determines the integrl over the nicer-looking spce. Figure : A ptch of the unit sphere. Now tht we wnt to integrte functions nd vector fields over surfces, we re going to follow this fmilir script. We hve prmetriztion T : R 2 of our surfce, nd we ll use this to pull integrls over (or regions within ) bck to integrls over regions in R 2. Along the wy, we ll pick up stretch fctor tht looks very similr to the Jcobin fctor we encountered erlier. Let s motivte this by wy of n exmple. Consider the prmetriztion of the unit sphere in R 3 given by G(u, v) = (cos u sin v, sin u sin v, cos v), for u 2π nd v π. A plot of the unit sphere cn be seen in Figure. In this plot the sphere hs few grid lines on it, where we ve held one vrible constnt nd llowed the other to vry. The blue curves on the sphere correspond to fixing v nd llowing u to vry, while the red
curves hve u fixed nd llow v to vry. Just s uv-coordinte system on R 2 subdivides the plne into smll ptches, these grid lines subdivide our sphere into smll regions, which we consider to be our most simple regions. Just s in the cse of trnsformtion T : R 2 R 2, we need to pproximte the re of these most simple regions, since they would be used in defining n integrl over the region. Recll tht in the previous cse, we pproximted these strnge regions in the plne by using tngent vectors to the grid lines. pecificlly, if we consider ptch in the plne where (u, v) strts t (u, v ) nd we llow u nd v to vry by u nd v, respectively, we pproximted this ptch by the prllelogrm in the plne spnned by the vectors T u nd T v. The sitution now is only slightly different: our ptch lies in the sphere, but the vectors G u nd G v lie in the plne tht s tngent to the sphere t (u, v ). ince the sphere is curving wy from the plne, this pproximtion of our ptch by prllelogrm isn t tht gret when u nd v re lrge. But when u nd v re very smll, this should be very good pproximtion indeed, the tngent plne is intended to represent the best first-order (flt) pproximtion we cn mke to the sphere t the point where the two meet. o for very smll u, v we should hve u u v, where is the (surfce) re of the smll ptch. Pssing to differentils puts this in the form d = u dudv. Here is intended to represent the surfce re of the surfce prmetrized by G, so tht d is the locl contribution to the surfce re. We cn use this, for exmple, to compute the surfce re of surfce : surfce re of = d = ˆ d ˆ b c u dudv, where we ssume G prmetrizes on the domin u [, b], v [c, d]. Exmple. Use the provided prmetriztion to compute the surfce re of the unit sphere. (olution) ince G(u, v) = (cos u sin v, sin u sin v, cos v), we hve (u, v) = sin u sin v, cos u sin v, u nd (u, v) = cos u cos v, sin u cos v, sin v. Then the cross product is given by u = cos u sin2 v, sin u sin 2 v, cos 2 u cos v sin v cos v sin 2 u sin v. 2
o we hve d = u cos dudv = 2 v sin 2 v + sin 4 vdudv = Finlly, the surfce re is given by A = ˆ π ˆ 2π sin vdudv = 2π ˆ π As you might expect, the fudge fctor for surfce integrls. sin 2 vdudv = sin vdudv. sin vdv = 2π [ cos v] π = 2π ( ( )) = 4π. u finds its wy into the more generl formul Theorem. Let G(u, v) be prmetriztion of surfce with prmeter domin. Assume tht G is continuously differentible, one-to-one, nd regulr, except possibly t the boundry of. Then f(x, y, z)d = f(g(u, v)) N(u, v) dudv, where N(u, v) is the norml vector obtined from the prmetriztion. N(u, v) = u Exmple. We cn use surfce integrls to compute the mss of thin lmin. If our lmin is represented by the surfce nd this lmin hs density δ(x, y, z), then the mss of the lmin is given by integrting δ over. Find the mss of the lmin tht is the portion of the surfce y 2 = 4 z between the plnes x =, x = 3, y =, nd y = 3 if the density is δ(x, y, z) = y. (olution) First we need prmetriztion of our surfce, which cn be seen in Figure 2. ince both x nd y re llowed to vry from to 3 without ny other condition, it mkes sense to tke x = u nd y = v. Then we hve z = 4 y 2 = 4 v 2, so our prmetriztion is G(u, v) = (u, v, 4 v 2 ), u 3, v 3. We compute the relevnt tngent vectors: u =, =. 2v Then the norml vector ssocited to our prmetriztion is N(u, v) = u = i j k 2v = 2v. This exmple is tken from the eighth edition of Anton, Bivens, nd vis s Clculus. 3
Figure 2: A thin lmin. Our fudge fctor is the mgnitude of this norml vector: N(u, v) = 4v 2 +. Finlly, we notice tht δ(g(u, v)) = δ(u, v, 4 v 2 ) = v, so the mss is given by M = = ˆ 3 ˆ 3 δ(x, y, z)d = ˆ 3 ˆ 3 v 4v 2 + dudv = δ(g(u, v)) N(u, v) dudv ˆ 3 3v 4v 2 + dv. Here we cn mke substitution w = 4v 2 +, so tht dw = 8vdv. As v vries from to 3, w will vry from to 37, so we hve M = ˆ 37 [ ] 3 3 2 37 wdw = 8 8 3 w3/2 = 4 (373/2 ) for the mss of our lmin. 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: 4
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 ˆ 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: 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 (signed) 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, 5
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. 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 7.5, Exercise 26) upose we hve net given by y = x 2 z 2, where y, 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. (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) = r 2, 6
for r nd 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,, 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 r cos θ = 2r2 cos θ, r, 2r 2 sin θ. Notice tht since r <, 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 o we hve N(r, θ) = 2r 2 cos θ, r, 2r 2 sin θ. v(g(r, θ)) = r cos θ ( r 2 ), r sin θ + ( r 2 ) + 4, r 2 sin 2 θ. v(g(r, θ)) N(r, θ) = 2r 3 cos 2 θ 2r 2 cos θ( r 2 ) + r 2 sin θ + 5r r 3 + 2r 4 sin 3 θ Finlly, our flow rte cross is given by ˆ 2π ˆ v d = v(g(r, θ)) N(r, θ)drdθ = ˆ 2π ˆ = 2r 2 cos θ(r 2 + r cos θ) + 2r 4 sin 3 θ + r(5 r 2 + r sin θ). 2r 2 cos θ(r 2 + 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, C 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 C, nd we write n for unit norml vector to curve C 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 = ± u. The ± is needed here becuse we wnt N(u, v) to be outwrd-pointing, nd this my require negting G u G v. 7
rc length/surfce re integrl of sclr function f integrl of vector field F fluid flowing through C or ˆ C Line Integrls urfce Integrls ˆ b r (t) dt u dudv ˆ b f(r(t)) r (t) dt f(g(u, v)) u dudv ˆ F dr := C (F T)ds = ˆ b F(r(t)) r (t)dt F d := (F n)d = F(G(u, v)) N(u, v)dudv ˆ C F nds = ˆ b F(r(t)) n(t) r (t) dt F d = F(G(u, v)) N(u, v)dudv 8
Group Work. how tht G(u, v) = (2u +, u v, 3u + v) prmetrizes the plne 2x y z = 2. Then () Clculte G u, G v, nd n(u, v). (b) Find the re of = G(), where = {(u, v) : u 2, v }. (c) Express f(x, y, z) = yz in terms of u nd v nd evlute f(x, y, z)d. 2. Clculte f(x, y, z)d, where is the prt of the surfce z = x 3 for which x nd y nd f(x, y, z) = z. 3. Let be the disk x 2 +y 2 in the xy-plne, oriented with norml in the positive z-direction. etermine F d for ech of the following constnt vector fields: () F =,, (b) F =,, (c) F =,, (Hint: rw picture.) 4. Let T be the tringulr region in R 3 with vertices (,, ), (,, ), nd (,, ), oriented so tht its norml vector hs positive y-component. Assume tht the units of R 3 re mesured in meters. A fluid flows in R 3 with constnt velocity field v = j m/s. Answers () Clculte the flow rte through T. (b) Clculte the flow rte through the projection of T onto the xz-plne (i.e., through the tringle with vertices (,, ), (,, ), nd (,, )).. () G u = 2,, 3, G v =,,, n(u, v) = 4, 2, 2. (b) Are() = 4 6. (c) f(x, y, z)d = 32 6/3. 2. 5 /27 3. () (b) π (c) π 4. () /2 m 3 /s (b) /2 m 3 /s 9