Hmwk 3 Solutions

Hmwk 3 Solutions - CD The solution will proceed...

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Unformatted text preview: CD The solution will proceed in‘spherical coordinates because both the surface and the field are described that way. Ou'the surface at radius R0 the field is E C/Ré and the element of surface is VRCZ, sinB d6 drb. Note that the specific radius of the sphere R}, has been substituted for the general radius R because that describes the particular surface over which we are integrating. Tne solution is then seizes S ‘21: 9a A A ' =J (r £>-(.rR3sin-edad¢)=2nC(—cos6)l%° ¢=0 6=O = 27rC(l — cos 60). @ We now have a mixed coordinate system problem. The field is in spherical coordinates and the disk is described most easily in cylindrical coordinates; It is easiest to convert the field to cylindrical coordinates and then proceed From the geometry of the figure, the disk is at height 20 =.R0 cos 90' above the xy plane. The spherical radius r is giVen in terms of the cylindrical radius}? as Y = 423 +597! ' —- C A=Tfrfi =5(“.sin6+’§cos9) C“ , r- Milka/12, sine=_=_.&__; $ cosg=l= fa . r za+f2 Au element of area on the disk is d? = $9 dd) (1}) The disk has an outer radius f0 = R0 gas 60. The flux through the surface is 5M ' _. _. . a 2 W=JA1£S ' a 3% S , = 27rC(1 — cos 60). The result 'is the same as ‘ . _ for the previous example. Later on we will learn how to test whether the result depends on the particular surface, or just on the outer boundary of the surface. Note that the integral of the field E over the closcd surface consisting of the polar cap and the circular disk is- 39 real Z-E—f z-ds‘ 5’ cap disk =0. The second integral is subtracted because for the disk we took the normal to the area to be directed upward 3,, = E, For the closed surface the outward normal over the disk is downward, 5,, = - 6,, and the direction of positive flow reverses. O A’=r2i,.—rws Gig+rsin95¢ From (0, 0, to 7V3, 0), ‘ ....-___._.V 6=1d3,¢=0;d6=0,d¢=0 a=wg A-a=flw 1 jA-dl=jofidr=% Ergmflm‘3fi)m(1.mé,m’2 r=l, 9=1n‘3;dr=0,d9=0 - .1: . {3' . A-d‘1=(1sin§)%d¢=%d¢ [A'dl_=Ifz%d¢=% Exam (1, 1:73, 2:12) ud(1,0,1rf2), r= l,¢=m‘2;dr=0,d¢=0 dl = 1 (1939: deig 'i' Aodl=—1 cos 6d9=~cos Gde IA-dl=J:B (—cos ®d9=-i2§- From (1, 0, :02) to (o, o, 0), e=o,¢=<r,de'=o,d¢=o m=wg A°dl=r2dr jA-dl=j(:fldr=—% l .-.¢CA-dl= + U) (a) V ° (yzzix + 221dy + xzyiz) = 0 V 1‘ 071i: + 221d, + xzyiz) ix i, i, = g g % yz, z2Jr xzy (b) V'(Sin¢i,+cos¢i¢) ll Q) Q) Q) 1 .—Vx[:(cos Gig-sin 9m] iE i2 :2 fixing .rsine r = a a _ 6) fi ‘ We require curl A to be identically zero, which implies that each component of it must be zero. 6 6 _ x-component ofVXA=F(4x+yy+2z)—5;(fix-3y-Z)=)’+1 a a y-component of‘V XA=-a—(x+2y'+az)—-a;(4x+yy+22)=a—4 z . I a 6 z-component ofVXA=5—(flx—3y—z)—5;(x+2y+azz)=;3—2 xi . 1 Hence a=4,B=2 and y: —l, and we obtain A:(x+2y+4z)i+(2x—3y—z)i+(4x—y+2z)fi The corresponding scalar potential (P is such that ad’ _ —a—=x+2y+4z or <P=1x2+2xy'+4x2+f(y,z) x , a—=2x—3y—z or -d>=2xy—%y2—yz+g(x,z). y l 045 a_=4x—y+ 22 or <P=4xz—yz+zz+h(x,y) 2 Hence <15: 12~(x2 — 3y2 + 222) + 2xy —yz + 4x2 + a constant - ' The required line integral is given by @(1, 2, l) — <D(l, 0, 0) = 5‘— § = 1. Evaluating the line integral along the specific path x = 1, z = %y gives « J{(2—3y—z)dy+(4—Y+2Z)dz} C 2 =[ {(2—3y—iy)dy+i(4—y+y)dy} 0 2 . I =J (4—iy)dy=1 0 (9 a) In Cartesian coordinates, xi+yi xi+y§~ =———— so F: ‘3 x/(xz+y2) x2+y2 i i E a a 6 Thus VXF= 6x By 62 x 2 2 7y 2 0 x+y x-+y - 6x x2+y2 6y x2+y2 6x 262 + y2 59 p2 6x p3 p p“ Similarly, noting the symmetry, 6( x >__ ny 6y xl+y2 p4 -___.. a_._._..._.____—._ .. _,. I - f. VXFEO that is, F is indeed conservative. Putting F = Vii), therefore, and equating components, we obtain 645— x‘ and 5—9: 'y , . (q) V5;_x2+y2 6y x2+y2 Integrating the first of these equations with respect to x, xdx I 2xdx ‘ 45= =- =llnx2+ 2 + Ixz yz zsz yz 2 ( y) Difl'erentiating with respect to y, 6q5__ y 6y x2+y2 + g’(y) so, by comparisOn with the second of equations (4 ), g’(y) = 0. Therefore, g(y) is a constant c, and the required potential is <P=1ln(x2+y2)+c=llnp2+c=lnp+c ‘0 Clearly, if ‘F-is conservative but non-solenoidal (7-17 #0), then the scalar potential 95 would not satisfy V245 = 0. To be precise, if V-F = f(x, y, z) in general, then V245 =f(x, y, z) This partial differential equation is known as Poisson’s equation. Now consider Vzti) In two dimensions, 52115 (3245 V2 =__—+.._ 6x2 (3y2 2 Here, ., x ), using the first of equations (4) x2 ax x‘+y ' _x a < 1 >+ 1 0x x2+y2 x2+y2 —2x2 1 - + Similarly, again noting the symmetry, _ 624) _ x2 _ yl I 6y2 (y2 + x2)2 . - , Adding equations ’) and ( 6 ‘) tells us that 7245 = 0 here, so that thfggtential <15 is a solution of the two-dimensi a1 La lace e nation. \_ / This is just'o‘nFEiEample of a general result of vector field theory ' is ‘ namely that if F is a conservative vector field (V K F E U) and, in addition, is a solenoidal vector field (V-F E 0), then the associated potential «15 is such that V-F=V-Vr15=0, thatis V2¢=0 ‘73 For the vector field E = ixz— $3122 — ixy, verify the divergence theorem by computing: 1 (a) the total outward flux‘ flowing through the surface of a cube centered at the g7 origin and with sides equal to 2 units each and parallel to the Cartesian axes, and (b) the integral of V - E over the cube’s volume. Solution: (3) For a cube, the closed surface integral has 6 sides: 5 7§E - ds = Fmp +Fbom + Flight + Fm + Fm. + Fback: z:=— =-l 1 1 1 x2 1 =-/ 1i>1 x=—1 y:— 4 __1 —_1 1 1 A * A Fbottom=f (xxzr—yyzz-zxyflz 1 (-zdydx) x=—1 y=—l 1 1 '1 1 l l = / xydydx = . = 07 x=—l y=—l 4 ___1 x=_1 1 1 A" A anht=/ / (Hz—WZZ—WMFI (dedx) x=—l z=—l - ‘ - 1 1 1 1 -4 =— szde=— =T1 x=—1 z=—l A 3 z=_1 x=_1 1 1 A A erfi=/ / (m—yyzz-zxy)ly=_1-(-ydzdx) x=—1- z=—1 l 1 1 1 ._4 = — zzdzdx= -— ‘—' '—7 1 x=—l z=-l 3 z_—_._1 x=_1 1 1 A A Ffi'ont: (11062—59122—zxy)|x=1 (xdzdy) Y=-1 z=-l - ,7 I 1 1 1 l n: . zdzdy: ‘ > :0, y=-l z='-l 2 z=_1 __1 y— 1 1 1 Fback= (ixz-yyzz—ixyflx 1(—xdzdy) y=—1 z=-1 . 1 1 1 1 , = / zdzdy: =0, y=—l z=—l 2 z=_1 __1 —4 —4 —8 = _ _ o = _, ‘ jén ds o+o+ 3 + 3l+o+ 3 1 1 1 1 A ff V-Edv: / V-(ixz—iyzz-WWZCIJ’dX I x=-1 )1 —l z=-1 /:_1(z—zz)dzdydx ll 1h ‘fi ~11 II A /"\ A Q /‘\ NW | ml’i. V v ill H 1/ C?) 57 AVector field B =_ fr3 exists in the region between two concentric cylindrical surfaces defined by r ; 1 and r = 2, with both cylinders emnding between z = 0 and z = 5. Verify the divergence theorem by evaluating: (a) jéo - d5, ' 5 VW (b) fVV-de. solution: (a) f/D'ds = Finner+Fouter+Fhotmm+Ftopa 2:: 5 A A A Fm: [M «new—redone] 2:: 5 4 ‘ . - = the/=0 (—‘r dzdmlml =40?“ 2:1: 5 A A Four, = M fmo (crotrrdzdonez 27: 5 4 _ Mfg) (r dzd¢)[,=2= 1602:, 2 2:: _ ' Em =/,_1/ (<fe>-<—2rd¢dr»lz=o=0, =0 Fmp = f; ((fr3)-(ird¢dr))l.z=5 = 0. Therefore, ffD-ds = 15017.. (1)) From the back cover, V-D = (1/r)(3/3r)(r13) = 41-2. Therefore, — 5 27‘: 2 7 I i - 4 2 21: ff V-Dda/_/z=o/¢=o =14r’rdrdtpd..— (((r )lml) qE0) M @. For the vector field D = R3122, evaluate both sides of the divergenee theorem for the region enclosed between the spherical shells defined by R = 1' and R = 2. 5 =1507c. 2:0 Solution: The divergence theorem is given ~ Evaluating the left hand side: _ 21!: 1: 2 2' [WV-Daw— ¢=ofe=of=1<RzaR(R(3R2)))R 51116dede = 2n:(—cos9)_[§=o (3R4)[:=1 = 1301:. ~ The right hand side evaluates to finds: (£O(R3R2)-(—RR2sin9ded¢)> - + (A: e:00‘23122)-(1‘mzsineded¢)) H 11: I 1: = —21: zsmede+21cfe O48sined9 = 1801:. 9:0 > = R=1 ' GE) = (x + a)(‘f + b)“ ‘1' “Ex v. X WI L‘"‘8~. _ 3A"__ 3A_ A=-.“i +nx1 oy z a: y = (x + a){-(z + c)E; + (y + b)§§] _ x1‘ 0 2 = f Axax + f Axcx x=0‘ x=xl y=71 Y=y2 z: z: 2 x1 2 o = (x + a) (y + b)c + LE—i—él— (y + b)c 2 l 2 2 x=0 x=x l guyl .+ b)[(xl + a)2 - a2] — (yz + b>t<xl + a)2 - azh %[(x1 + a)2 - a12][3'l - 3'2] fv xK-d—s= If (v xX)zdxdy- flat surface at z=0 foX- rectangular cylinder 5 = x1 YZ - c 2 2 f f —c(x + a)dxdy = - E (yz - Y )[(x + a) ‘ a J _ _ l l x—0 y-yl ' ' ' 0 y=72 surface (v x X) dxdz — y y=Y1 surface (V x ./ . (v x A)ydxdy + f A)xdydz r x'xl surface i§0 I W; X) xdydz + x=0 surface 21"1 2 I f(x+a)(y2+b)dxdz-f 0 x=0 o I Z=Zl surface _ (22x'X)zdxdy x1 f'(x-+ 1 a)(y1 + b)dxdz l x y H \N l . I (21 + C)(x + a)dxdy 0 yyl Z 2 (3'2 + b) '2i[(xl+a)2-a2]- (yl+b) Tllei-QZ—az]: - (z + CH? - Y ) . L .7; 1 2 1 “xi + a)2 _ a2J ; %[5l—VJ{(K,H\) ~ «.1 _————“ 2 GP _ For the vector field E = ixy — fir"- + Zyz), calculate (a) E mil around the triangular contour shown . - 5-1 y (b) [5(7 x E) - ds over the area of the triangle. \ ._—.—_.\ .__..H_ _ 1 Solution: In addition to the independent condiu'on that z = 0, the three lines of the triangle are represented by the equations y = O, y = 2 — x, and y = x, respectively. (a) ' jéE-dl=L1+L2+L3, L1=[($01—i(x2+2y2))-(idx+§dy+2dz) a, ’3 - 0 0 = [i=0 (w)ly=o,z=o dx — [F0 (x2 + 2%) 11:0 dy+ [2:0 (0)13,==o dz = 0, La: few—i<x2+2y2>)-(2dx+§czy+2dz) ' 1 1 0 = ./,.-—=2V(xy)‘z=dJ=2—xdx_ F0 + zyz) L:2—y,z=0 dy+ 1:0 (cubed-x dz _.~._..-.-_....... .~ L3=fCiVy-?(xz+2y2))'(idx-i-f’dy-i-fidz) O D ' 0 = 51 (xy)ly=x,z=oir- =1 (x2+2y2)1,=,,z=ody+ Focomfldz x3 0 o 2 = (E) r1“ We”? Therefore, ' 11 2 7€E.41_o—?+§_—3. (b) r . VxE=+23x,sothat ff VxE-ds = y; ((—23x) . (idydx?)|z=o + /; fy::((—23x)'(idydx))lz=o 1 x 2 2—: . = —/ / chiydxef f 3xdydx x=0 y=0 z=l y=0 , 2 =-/103x(x—0)dx-/ 3x(<2—x)—0)dx x: =1 = —(x3)Ié—.(329—9)E=1=—3- ...A N Q9 ® (frogsd) +<§Sin¢) (a) 7:»; B - d! over the semicircular contour shown > - I ‘ 3’ (13) AC? x B) . :15 over the surface of the semicircle. ? ‘ .h_..-..-- @ Verify Stokes’s theorem for the vector field B = by evaluating: = rcos¢dr + rsin¢d¢, o - F0, z=o+ (Azorsmqadqgl z=0 z 3 = 2 Tc/ B d1: rcos¢dr)L + rsm¢d¢> L9 1:2 2:0 ¢=0 1:2 2:0 2.0 + (fa/2rsin¢d¢> =0+(—cos¢)l§=n/2=-11 jéB-a'l=g+2+0~1=§. ' 2 (b) 1:1, 2:0 ‘— VXB = Vx(frcos¢+$sin¢) ' ‘ 1 a 3 . a a a = 1- (:EO— +11) (Eflcow) — 50) +2; <§(r(sin¢)) — %(rcos¢)) = $0 +é0+2%(sin¢+ (rsin¢)) = isimp (1 + , [/me = "/2/2 (251mb (1+ -(2rdrd¢) 43:0 r=l 7' r = fr:lsin¢(r+1)'§rd¢ ' = (('_cos¢(%rz+r))l::.l) 5 2 . ® Verify Stokes’s Theorem evaluating it on the hemisphere of imit radius. Solution: - A = Rcose+¢sine = MR+éAe+$A¢. Hence, AR = cos 6, A9 = 0, A4, = sine. _.1 a . .13 .IaAR VxA-Rm(%(A¢sm6)) —6§fi(RA¢)— .13”2 .13 ', -13 —RRSine gall: _fi2cose_é§i_l;9_ Asini ‘ R R R‘ For the hemispherical surface, ds = RRZ sin9d6d¢. , 21: 1t/2 / f (V x A) ~ds ¢=O 9=0 2:: 1c/2 9 A - k - A =f (Leose_esm6 +¢sm9).Rstineded¢ ¢=o e=o R R R R=1 . 2 3/2 = 41:12 51“ e = 21:. ° R=1 The contour C is—the circle in the x—y plane bounding the hemispherical surface. . 27: A A A . ‘ 21: 2 fiA-a’l: A=O(Rcose+¢sm9)-¢Rd¢ 9:342 =Rsm6/o dq) amt/2 _ 7:. 12:1 for the vector field A = Recs 6 +$sin6 by ...
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Hmwk 3 Solutions - CD The solution will proceed...

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