18.02 Multivariable Calculus Fall 2007

Size: px

Start display at page:

Download "18.02 Multivariable Calculus Fall 2007"

Karen Ford
5 years ago
Views:

1 MIT OpenCourseWre Multivrible Clculus Fll 2007 For informtion bout citing these mterils or our Terms of Use, visit:

2 6. Vector Integrl Clculus in Spce 6A. Vector Fields in Spce 6A-1 ) the vectors re ll unit vectors, pointing rdilly outwrd. b) the vector t P hs its hed on the y-xis, nd is perpendiculr to it 6A-4 A vector field F = M i + N j + Pk is prllel to the plne 32-4y + z = 2 if it is perpendiculr to the norml vector to the plne, 3 i - 4j + k : the condition on M, N, P therefore is 3M - 4N + P = 0, or P = 4N - 3M. The most generl such field is therefore F = M i +N j + (4N - 3M) k, where M nd N re functions of x, y, z. 6B. Surfce Integrls nd Flux xi +yj +zk 6B-1 We hve n = ; therefore F.n =. Flux through S = I P P F.n ds = (re of S) = 47r 3 Is 6B-2 Since k is prllel to the surfce, the field is everywhere tngent to the cylinder, hence the flux is 0. i+j+k 1 6B-3 is norml vector to the plne, so F.n = -. l.6 l.6 re of region 4 (bse)(height) (4)( 4 ) Therefore, flux -1 = B-4 flux = n = /k xi +yj +zk l.6 l.6 l.6 2' ; F.n = -. y2 Clculting in sphericl coordintes, sin34 sin2 0 dmde. l" l" 2 3ln 6" ds = 4 sin3 4 sin2 0 dm do = L Inner integrl: sin2 O(- cos 4 + $ cos3 4) 1: 1 " Outer integrl: %3($8- $ sin 28) J, = = 2i7r3. $ sin2 8; ll,

3 2 E EXERCISES F.n- z 6B-5 n= i+j+k. '. z dxdy 1 dx dy 1-Y 1 Innerintegrl: =x-~x 2 -xy ]I-. =$(1-~)~. 0 Outerintegrl: = B-6 z = f (x, y) = x2 + y2 ( prboloid). By (13) in Notes V9, ds = (-2xi - 2yj + k ) dxdy. (This points generlly "up", since the k component is positive:) Since F = x i + yj + z k, where R is the interior of the unit circle in the xy-plne, i.e., the projection of S onto the xy-plne). Since z = x2 + y2, the bove integrl The nswer is negtive since the positive direction for flux is tht of n, which here points into the inside of the prboloidl cup, wheres the flow xi + yj + z k is generlly from the inside towrd the outside of the cup, i.e., in the opposite direction. 6B-8 On the cylindricl surfce, n xi+yj F.n=-. y2 = In cylindricl coordintes, since y = sin 8, this gives us F. ds = F. n ds = 2sin28 dz db. lri2 Jdk Flux = 2sin28 dz db = 2h sin28d8=2h( - --~i;28)~/~= -2h. s -r/2 -r/2 2 6B-12 Since the distnce from point (x, y, 0) up to the hemisphericl surfce is z, JJs ds verge distnce = - JJs ds ' In sphericl coordintes, /LzdS = Jd2rJdri2cos).2sin)d)d8. r/2 s~f] :/2. : Inner: = 3Jd sin)cos)d)=3(- -- Outer: = d8 = s3. s3 Finlly, ds = re of hemisphere = 2s2, so verge distnce = s2 2' Jd2r

4 6. VECTOR INTEGRAL CALCULUS IN SPACE 6C. Divergence Theorem 6C-2 Using the product nd chin rules for the first, symmetry for the others, x2 + y2 +z2 dding these three, we get div F = npn-l + 3pn = pn(n+3). P Therefore, div F = 0 # n = -3. 6C-3 Evluting the /k triple integrl first, we hve div F = 3, therefore 2 div F dv = 3(vol.of D) = 3-7r3 = 27r3. 3 To evlute the double integrl over the closed surfce S = S1+S2,the respective norml vectors re: xi +yj +zk nl = (hemisphere S1), n2 = - k (disc 5'2); using these, the surfce integrl for the flux through S is since x2 +y2 +z2 = P2 = 2 on S1, nd z = 0 on S2. SO the vlue of the surfce integrl is which grees with the triple integrl bove. (re of S1)= (27r2) = 27r3, 6C-5 The divergence theorem sys /k~.ds=mdiv~dv. Here div F = 1, so tht the right-hnd integrl is just the volume of the tetrhedron, which is i(bse) (height) = i (i)(1) = i. 6C-6 The divergence theorem sys / k ~. d ~ = / k d i v ~ d ~. P- Here div F = 1, so the right-hnd integrl is the volume of the solid cone, which hs height 1nd bse rdius 1; its volume is i(bse) (height)= 7r/3. 6C-7 Evluting the triple integrl first, over the /k cylindricl solid D, we hve div F = 2x + x = 3s; 3xdv = 0, since the solid is symmetric with respect to the yz-plne. (Physiclly, ssuming the density is 1, the integrl hs the vlue 3(mss of D), where 3 is the x-coordinte of the center of mss; this must be in the yz plne since the solid is symmetric with respect to this plne.) To evlute the double integrl, note tht F hs no k-component, so there is no flux cross the two disc-like ends of the solid. To find the flux cross the cylindricl side,

5 4 E EXERCISES since the cylinder hs rdius 1nd eqution x2 + y2 = 1. Thus 6C-8 ) Reorient the lower hemisphere S2 by reversing its norml vector; cll the reoriented surfce Sb. Then S = S1+ Sb is closed surfce, with the norml vector pointing outwrd everywhere, so by the divergence theorem, since by hypothesis div F = 0. The bove shows /J,;.ds=-/~,:~.ds=/Js,~.ds, since reversing the orienttion of surfce chnges the sign of the flux through it. b) The sme sttement holds if S1 nd S2 re two oriented surfces hving the sme boundry curve, but not intersecting nywhere else, nd oriented so tht Sl nd Sb (i.e., S 2 with its orienttion reversed) together mke up closed surfce S with outwrd-pointing norml. 6C-10 If div F = 0, then for ny closed surfce S, we hve by the divergence theorem b----4 Conversely: /lf. ds = 0 for every closed surfce S 3 div F = 0. For suppose there were point Po t which (div F)O # 0 - sy (div F)o > 0. Then by continuity, div F > 0 in very smll sphericl bll D surrounding Po, so tht by the divergence theorem (S is the surfce of the bll D), /lf.d~=//ldivfdv > 0. But this contrdicts our hypothesis tht F.dS = 0 for every closed surfce S. 6C-11 flux of F = /lf.dn = //Ldiv FdV = //L3dV = 3(vol. of D). 6D. Line Integrls in Spce 6D-1 ) C : x = t, dx = dt; y = t2, dy = 2tdt; z = t3, dz = 3t2 dt; y dx + z dy - x dz = h' (t2)dt+ t3(2t dt) - t(3t2 dt) 1 = 11(t2+2t4-3t3)dt =

6 6. VECTOR INTEGRAL CALCULUS IN SPACE 5 d) C:x=cost, y =sint, z=t; zxdx+zydy+xdz Jc t cos t(- sin t dt) + t sin t(cos t dt) + cos t dt = costdt = 0. 6D-2 The field F is lwys pointed rdilly outwrd; if C lies on sphere centered t the origin, its unit tngent t is lwys tngent to the sphere, therefore perpendiculr to the rdius; this mens F - t = 0 t every point of C. Thus Sc F.dr = Jc F t ds = 0. Sc b) (i) Directly, letting C be the helix: x = cost, y = sint, z = t, from t = 0 to t = 2n, M ~ X+ ~d~ + ~ d z = 1'"" 2 cos t(- sin t)dt + 2 sin t(cos t)dt + 2t dt = 2t dt = (2n)'. b) (ii) Choose the verticl pth x = 1, y = 0, z = t; then J,M ~ X+ ~d~ + pdz = 2t dt = (2n) 2. 1'"" b) (iii) By the First Fundmentl Theorem for line integrls, 6D-5 By the First Fundmentl Theorem for line integrls, where C is ny pth joining P to Q. The mximum vlue of this difference is 1- (-1) = 2, since sin(xyz) rnges between -1 nd 1. For exmple, ny pth C connecting P : (1,1, -/2) to Q : (1,1,/2) will give this mximum vlue of 2 for Jc F dr. 6E. Grdient Fields in Spce 6E-1 ) Since M = x2, N = y2, P = z2 re continuously differentible, the differentil is exct becuse N, = Py= 0, M, = P, = 0, My = N, = 0. b) Exct: M, N, P re continuously differentible for ll x, y, z, nd N, = Py= 2xy, M, = P, = y2, My = N, = 2yz. c) Exct: M, N, P re continuously differentible for ll x, y, z, nd 6E-2 curl F = d, dy d, = (xz2- y) i -yz2j - 22 k. x2y yz xyz2

7 6 E EXERCISES 6E-3 ) It is esily checked tht curl F = 0. b) (i) using method I: = lxl x dx + iyl ydy + 1'' zdz = -xl YI z X Therefore f (x, y, z) = $(x2 + y2 + z2) + C. c2 (ii) Using method 11: We seek f (x, y, z) such tht f, = 2xy + z, fy = x2, f, = x. Therefore f (x, y, z) = x2y + xz + c. (iii) If f, = yz, fy = xz, f, = xy, then by inspection, f (x, y, z) = xyz + c. 6E-4 Let F = f - g. Since V is liner opertor, VF = Vf - Vg = 0 We now show: VF = 0 + F = c. Fix point Po : (xo, yo, 20). Then by the Fundmentl Theorem for line integrls, Therefore F(P) = F(Po) for ll P, i.e., F(x, y, z) = F(xo,YO, zo)= C. 6E-5 F is grdient field only if these equtions re stisfied: N,=Py: 2xz+y=bxz+2y M,=P,: 2yz=byz My=Nx: z2=z2. Thus the conditions re: = 2, b = 2. Using these vlues of nd b we employ Method 2 to find the potentil function f: therefore, f (x, y, z) = xyz2 +y2z + c. 6E-6 ) Mdx +Ndy +Pdz is n exct differentil if there exists some function f (x, y, z) for which df = Mdx + Ndy + Pdz; tht, is, for which f, = M, fy = N, f, = P. b) The given differentil is exct if the following equtions re stisfied: N, = Py: (/2)x2 + 6xy2z+ 3byz2= 3x2 + 3cxy2z + 12yz2; M, = P, : xy + 2y3z = 6xy + cy3z My = N, : xz + 3y2z2= xz + 3y2z2. Solving these, we find tht the differentil is exct if = 6, b = 4, c = 2.

8 c) We find f (x, y, z) using method 2: 6. VECTOR INTEGRAL CALCULUS IN SPACE 7 f, = 6xyz + y3z2 + f = 3x2yz+ xy3z2+ g(y,z); f, = 3 ~ + 23xy2z2+gy ~ = 3 ~ +3xy2z2+dyz3 2 ~ + gy = 4yz3 = 2y2z3+ h (~) f,= 3xZy+ 2xy3z + 6y2z2+ hl(z) = 3xZy+ 2xy3z + 6y2z2 + hl(z) = 0 + h = C. Therefore, f (x, y,z) = 3xZyz+ xy3z2+ 2y2z3+ C. 6F. Stokes' Theorem 46, F. dr = xdx + ydy + zdz = 0, A 6F-1 ) For the line integrl, 46, C since the differentil is exct. For the surfce integrl, V x F = x 8, 8, = 0, nd therefore /J,vxF.~s = 0. X Y Z b) Line integrl: ydx + zdy + xdz = ydx, since z = 0 nd dz = 0 on C. 46, Using x =cost, y = sint, I""- sin2 t dt = dt = -n-. -12":ps2t - Surfce integrl: curl F = 8, 8, 8, = - i - j - k ; Y Z X /k~xf).nd~=- n = x i + yj + z k To evlute, we use x = r cos 0, y = r sin 6, z = p cos q5. r = p sin q5, ds = p2 sin q5 d4d0; note tht p = 1on S. The integrl then becomes - lo lo [sin q5(cos 0 + sin 0) + cos $1 sin q5 dq5 db Inner: - [ (cos 0 + sin 6)(+ - $ cos 2q5) + $ sin2q5]:i2= -[(cos~+sin~i+i I ; Outer: J (-A- 2 cos 0 - sin 0) db = -IT. 6F-2 The surfce S is: z = -x - y, so tht f (x, y) = -x - y. (Note the signs: n points upwrds, nd therefore should hve positive k-component.) curl F = x dy 8, = -i - j - k Y Z X Therefore /L curl F. nds = - 3dA = -3, where S' is the projection of S, i.e., the interior of the unit circle in the xy-plne. As for the line integrl, we hve C : x = cost, y = sin t z = - cost - sin t, so tht

9 8 E EXERCISES hydxfzdy +xdz = - sin2 t - (cos2 t + sin t cos t) + cos t(sin t - cos t)] dt =~2"(-sin2t-cos2t-cos2t)dt= 2 6F-3 Line integrl: yz dx + xz dy + xy dz over the pth C = Cl C4:!c = 0, since z = dz = 0 on Cl; Jc?=lll.ldz=l, sincex=l, y=l,dx=o, dy=oonc2; ydx+xdy= I0 xdx+xdx=-1, sincey=x,z=1,dz=oonc3; k = 0, since x = 0, y = 0 on C4. Adding up, we get F.dr = Ll+ + + = 0. For the surfce integrl, curl F = d, dy d, = i (x - x) - j (y - y) + k (z - z) = 0; thus yz xz xy 6F-5 Let S1 be the top of the cylinder (oriented so n = k ), nd S2the side. /L ) We hve curl F = d, dy dz = -2x j + 2 k. -y X x2 h /11 For the top: /k, curl F.n ds = 2 ds = 2(re of Sl) = 2s2. C xi +yj For the side: we hve n =, nd ds = dz. db, so tht 1'" /11 Adding, /lcurl F.dS = +/12= 2s2. 1:" curl F.n d = ~ 12"lh+dzdB= -2h(cosB)(sinB)dB=-h2sin2e =0. b) Let C be the circulr boundry of S, prmeterized by x = cos 8, y = sine, z = 0. Then /l using Stokes' theorem, (2 sin2 B + 2 cos2 0) db = 2s2. curl F.dS = -y dx + x dy +x2 dz = 12" 6G. Topologicl Questions 6G-1 ) yes b) no c) yes d) no; yes; no; yes; no 6G-2 Recll tht p, = xlp, etc. Then, using the chin rule, curl F = (npn-'z Y --npn-l y --) z i + (npn-'z x z x Y -- npn-lx -) j + (npn-ly -- npn-'x -) k. P P P P P

10 6. VECTOR INTEGRAL CALCULUS IN SPACE 9 Therefore curl F = 0. To find the potentil function, we let Po be ny convenient strting point, nd integrte long some pth to Pl : (xl, yl, zl). Then, if n # -2, we hve dr = Lo ]Po Pl pn(x dx + y dy + z dz) = = pn+ldp= - + c, since Po is fixed. n+2 n+2 n+2 pn+2 Therefore, we get F = V- n + 2' if n # -2. If n = -2, the line integrl becomes = lnpl + c, so tht F = V(1np). 6H. Applictions nd Further Exercises 6H-1 Let F = Mi + N j + P k. By the definition of curl F, we hve If ll the mixed prtils exist nd re continuous, then Pxy= Pyx, etc. nd the right-hnd side of the bove eqution is zero: div (curl F) = 0. 6H-2 ) Using the Jlcurl F. ds = //L div curl F dv = JLO ~ V divergence theorem, nd the previous problem, (D is the interior of S), b) Drw closed curve C on S tht divides it into two pieces S1 nd S2 both hving C s boundry. Orient C comptibly with S1, then the curve C' obtined by reversing the orienttion of C will be oriented comptibly with S2. Using Stokes' theorem, JJ,curl F. = JJ,, = 0. curl ~.ds+/js?~~~l F.~S= hf'.dr +h,f.dr = 0, since the integrl on C' is the negtive of the integrl on C. Or more simply, consider the limiting cse where C hs been shrunk to point; even s point, it cn still be considered to be the boundry of S. Since it hs zero length, the line integrl round it is zero, nd therefore Stokes' theorem gives JJ, curl F. ds i = F. dr = 0. 6H-10 Let C be n oriented closed curve, nd S comptibly-oriented surfce hving C s its boundry. Using Stokes' theorem nd the Mxwell eqution, we get respectively 1E JJ,vx~.ds= B.dr nd J J, v X B. ~ S = J J, - -. ~ S t = ~ ~ J J, E c. ~ dt S. 1 d Since the two left sides re the sme, we get h B dr = ;iljl E. ds. In words: for the mgnetic field B produced by moving electric field E(t), the mgnetomotive force round closed loop C is, up to constnt fctor depending on the units, the time-rte t which the electric flux through C is chnging.

l.6 l.6 l.6 l.6 l.6 2' L ds = a4 sin3 4 sin2 0 dm do = sin34 sin2 0 dmde. $ sin2 8; 6. Vector Integral Calculus in Space

l.6 l.6 l.6 l.6 l.6 2' L ds = a4 sin3 4 sin2 0 dm do = sin34 sin2 0 dmde. $ sin2 8; 6. Vector Integral Calculus in Space 6. Vector Integrl Clculus in Spce 6A. Vector Fields in Spce 6A-1 ) the vectors re ll unit vectors, pointing rdilly outwrd. b) the vector t P hs its hed on the y-xis, nd is perpendiculr to it 6A-4 A vector