Chapter Fifiteen. Surfaces Revisited

Transcription

1 Chapte Ffteen ufaces Revsted 15.1 Vecto Descpton of ufaces We look now at the vey specal case of functons : D R 3, whee D R s a nce subset of the plane. We suppose s a nce functon. As the pont ( s, t) D moves aound n D, f we place the tal of the vecto ( s, t) at the ogn, the nose of ths vecto wll tace out a suface n thee-space. Look, fo example at the functon : D R 3, whee ( s, t) = s + tj + ( s + t ) k, and D = {( s, t) R : 1 s, t 1 }. It shouldn't be dffcult to convnce youself that f the tal of ( s, t ) s at the ogn, then the nose wll be on the paabolod z = x + y, and fo all ( s, t) D, we get the pat of the paabolod above the squae 1 x, y 1. It s sometmes helpful to thnk of the functon as povdng a map fom the egon D to the suface. The vecto functon s called a vecto descpton of the suface. Ths s, of couse, exactly the two dmensonal analogue of the vecto descpton of a cuve. 15.1

2 Fo a cuve, s a functon fom a nce pece of the eal lne nto thee space; and fo a suface, s a functon fom a nce pece of the plane nto thee space. Let's look at anothe example. ee, let ( s, t) = coss snt + sn s sn t j + cost k, fo t π and s π. What have we hee? Fst, notce that ( s, t) = (coss sn t) + (sn s sn t) + (cos t ) = sn t (cos s + sn s) + cos t = sn t + cos t = 1 Thus the nose of s always on the sphee of adus one and centeed at the ogn. Notce next, that the vaable, o paamete, s s the longtude of ( s, t ) ; and the vaable t s the lattude of ( s, t ). (Moe pecsely, t s co-lattude.) A moment's eflecton on ths wll convnce you that as s a descpton of the ente sphee. We have a map of the sphee on the ectangle 15.

3 Obseve that the ente lowe edge of the ectangle (the lne fom (, ) to ( π, ) ) s mapped by onto the Noth Pole, whle the uppe edge s mapped onto the outh Pole. Let ( s, t), ( s, t ) D be a vecto descpton of a suface, and let p = ( s, t ) be a pont on. Now, c( s) = ( s, t ) s a cuve on the suface that passes though he pont p. Thus the vecto d c ds = ( s s, t ) s tangent to ths cuve at the pont p. We see n the same way that the vecto ( s, t ) s tangent to the cuve ( s, t) at p. t 15.3

4 At the pont p = ( s, t ) on the suface, the vectos ence the vecto s nomal to. s t s and t ae thus tangent to. Example Let's fnd a vecto nomal to the suface gven by the vecto descpton ( s, t) = s + tj + ( s + t ) k s : at a pont. We need to fnd the patal devatves s and = + sk, and = j + tk. s t The nomal N s N j k = = 1 s = s tj + k. s t 1 t Medtate on the geomety hee and convnce youself that ths esult s at least easonable. Execses 1. Gve a vecto descpton fo the suface z = x + y, x, y.. Gve a vecto descpton fo the ellpsod 4x + y + 8z = Gve a vecto descpton fo the cylnde x + y =

5 4. Descbe the suface gven by ( s, t) = scost + ssn tj + sk, t π, 1 s Descbe the suface gven by ( s, t) = scost + ssn tj + s k, t π, 1 s. 6. Gve a vecto descpton fo the sphee havng adus 3 and centeed at the pont (1,,3). 7. Fnd an equaton (I.e., a vecto descpton) of the lne nomal to the sphee a a a x + y + z = a at the pont (,, ) Fnd a scala equaton (I.e., of the fom f ( x, y, z) = ) of the plane tangent to the a a a sphee x + y + z = a at the pont (,, ) Fnd all ponts on the suface ( s, t ) = ( s + t ) + ( s + 3 t) j stk at whch the tangent plane s paallel to the plane 5x 6y + z = 7, o show thee ae no such ponts. 1. Fnd an equaton of the plane that contans the pont (1,-,3) and s paallel to the plane tangent to the suface ( s, t ) = ( s + t ) + s j t k at the pont (1, 4,-18). 15. Integaton uppose we have a nce suface and a functon f : R defned on the suface. We want to defne an ntegal of f on as the lmt of some sot of Remann sum n the way n whch we have aleady defned vaous ntegals. ee we have a slght poblem n that we eally ae not sue at ths pont exactly what we mght mean by the aea of a 15.5

6 small pece of suface. We assume the suface s suffcently smooth to allow us to appoxmate the aea of a small pece of t by a small plana egon, and then add up these appoxmatons to get a Remann sum, etc., etc. Let's be specfc. We subdvde nto a numbe of small peces,, K, each havng aea A, 1 select ponts * = ( x *, y *, z * ), and fom the Remann sum n n R = f ( * ) A = 1. Then, of couse, we take fne and fne subdvsons, and f the coespondng Remann. sums have a lmt, ths lmt s the thng we call the ntegal of f on : f ( ) d Now, how do fnd such a thng. We need a vecto descpton of, say : D ( D) =. The suface s subdvded by subdvdng the egon D R nto ectangles n the usual way: 15.6

7 The mages of the vetcal lnes, s = constant, fom a famly of "paallel" cuves on the suface, and the mages of the hozontal lnes t = constant, also fom a famly of such cuves: Let's look closely at one of the subdvsons: 15.7

8 We paste a paallelogam tangent to the suface at the pont ( s, t ) as shown. The lengths of the sdes of ths paallelogam ae ( s s t s, ) and ( s, t ) t. The aea t s then s s t s (, ) ( s, t ) t, and we use the appoxmaton t A s s t s t s t (, ) (, ) t n the Remann sums: n R f s t s s t = s t s t ( (, )) (, ) t (, ). = 1 These ae just the Remann sums fo the usual old tme double ntegal of the functon n F s t f s t s s t (, ) = ( (, )) (, ) ( s, t ) t = 1 ove the plane egon D. Thus, f d f s t s s t ( ) = ( (, )) (, ) ( s, t) da. t D Example Let's use ou new-found knowledge to fnd the aea of a sphee of adus a. Obseve that the aea of a suface s smply the ntegal d. In the pevous secton, we found a vecto descpton of the sphee: 15.8

9 ( s, t) = acoss snt + a sns snt j + a cost k, t π and s π. Compute the patal devatves: s t = a sn s snt + acos s sn t j, and = a coss cost + a sn s cost j a snt k Then s j k = a sn s sn t coss snt t coss cost sn s cost -snt = a [ coss sn t sn s sn t j snt cos t k] Next we need to fnd the length of ths vecto: = a [cos s sn t + sn s sn t + sn t cos t] s t = a [sn t + sn t cos t] = a [sn t (sn t + cos t)] = a / 4 1 / 1 / sn t ence, Aea = d = da = a t da s t sn D D 15.9

10 = a π π sn t dsdt π = π a sntdt = 4π a Anothe Example Let's fnd the centod of a hemsphecal shell of adus a. Choose ou coodnate system so that the shell s the suface x + y + z = a, z. The centod ( x, y, z) s gven by x = xd d y = yd d and z = zds d. Fst, note fom the symmety of the shell that x fom the pecous example that d = π a evaluate: = y =. econd, t should be clea. Ths leaves us wth just ntegal to zd. Most of the wok was done n the example befoe ths one. Ths hemsphee has the same vecto descpton as the sphee, except fo the fact that the doman of s the π ectangle s π, t. Thus 15.1

11 π / π zd = a a cost s t dsdt π / π π / 3 3 = a cost sn t dsdt = π a cost snt dt π / 3 = π a sn t = π a 3 And so we have z 3 π a a = =. Is ths the esult you expected? π a Yet One Moe Example Ou new defnton of a suface ntegal cetanly ncludes the old one fo plane sufaces. Look at the "suface" descbed by the vecto functon ( θ, ) = cosθ + snθ j, wth defned on some subset D of the θ plane. Fo what we hope wll be obvous easons, we ae usng the lettes θ and nstead of s and t. Now consde an ntegal f ( x, y) d ove the suface descbed by. We know ths ntegal to be gven by f ( x, y) d = f ( cos θ, sn θ ) θ D da. Let's fnd the patal devatves: 15.11

12 θ = snθ + cos θ j, and θ θ j = cos + sn. Thus, θ j k = snθ cosθ = k, cosθ snθ and we have θ =. ence, f ( x, y) d = f ( cos θ, sn θ ) da = f ( cos, sn ) da θ θ θ D D. Ths should look famla! Execses 11. Fnd the aea of that pat of the suface z = x + y that les between the planes z = 1 and z =. 1. Fnd the centod of the suface gven n Poblem Fnd the aea of that pat of the Eath that les Noth of lattude 45. (Assume the suface of the Eath s a sphee.) 14. A sphecal shell of adus a s centeed at the ogn. Fnd the centod of that pat of t whch s n the fst octant. 15.1

13 15. a)fnd the centod of the sold ght ccula cone havng base adus a and alttude h. b)fnd the centod of the lateal suface of the cone n pat a). 16. Fnd the aea of the ellpse cut fom the plane z = x by the cylnde x + y = Evaluate ( x + y + z) d, whee s the suface of the cube cut fom the fst octant by the planes x = a, y = a, and z = a. 18. Evaluate x y +1 d, whee s the suface cut fom the paabolod y + 4z = 16 by the planes x =, x = 1, and z =