Solution:

where the cross product is evaluated with Eq. (3.27).

93CHAPTER 3

Chapter 3

A = ilB x C/ = t/(i2-y4) x (x-y-Z4)1= !lx(-4)(-4) +y(-(2)(-4» +i(2(-I} - (-4)1)1

= HiI6+y8+Z21 == ~JI62+82+22 = !v324 = 9,

Problem 3.3 In Cartesian coordinates, the three comers of a triangle are Pl (O,2,2),P2(2,-2,2), and P3(1, 1,-2). Find the area of the triangle.

"A·C = (:i2-Y3+z)· (i4+y2-Z2) = 8-6-2= 0,

B·C= (i2-Y+Z3)·(i4+Y2-Z2) =8-2-6=0.

Section 3-1: Vector Algebra

Problem 3.1 Vector A starts at point (1, - 1,-2) and ends at point (2,-1,0). Finda unit vector in the direction of A.

--+ ---+

Solution: Let B =P1P2= i2 - y4 and C = PI P3 = i: - Y - z4 represent two sides ofthe triangle. Since the magnitude of the cross product is the area of the parallelogram(see the definition of cross product in Section 3-1.4). half of this is the area of thetriangle:

A = i(2-1)+y(-1- (-l»+z(O- (-2» =i+Z2,

IAI = vI +4 = 2.24,

A A i+Z2 ~O45 A 9a = IAI = 2.24 = A. +zO.8 -

~blem ~) Given vectors A = :i2- y3+ Z, B = i:2- y+Z3, and C = i4+y2 - n,show that C is perpendicular to both A and B.

Solution:

Problem 3.4 Given A = X2- y3 + zl and B = XBx + y2+ZBz:

(a) find Bx and Bz if A is parallel to B;

(b) find a relation between Bx and Bz if A is perpendicular to B.

or

Bx = -2..;56 _ -43V14 -"3

From the x-component,

2 -B__ - __ . _x

04 - "/56/9'

and, from the z-component,

-2Bz = --3

This is consistent with our result for B; +B;.These results could also have been obtained by assuming €lAB was 00 or 1800 and

solving IAIIBI = ±A· B, or by solving A x B = O.(b) If A is perpendicular to B, then their dot product is zero (see Section 3-1.4).

Using Eq. (3.17),

-3 ±204 - -/4+B;-t--jj;

which can only be solved for the minus sign (which means that A and B must pointin opposite directions for them to be paraHel). Solving for B; +B;,

94

From the y-component,

Solution:

(aj If A is parallet to D, then their directions are equal or opposite: 8A = ±8B, or

B;+B;= (=~v'i4)2-4= ~o.

O=A·B= 2Bx-6+Bz,

There are an infinite number of vectors which could be B and be perpendicular to A,but their x- and z-components must satisfy this relation.

This result could have also been obtained by assuming €lAB = 90° and calculatingIAIIBI = IA x BI·

~blem ~iven vectors A = x +y2 - 23, B = x3 - Y4. and C = y3 - Z4, find~

(d) From Eq. (3.27),

A x C = x(2(-4} - (-3}3) +Y«-3)0- 1(-4}) +z{1(3) -O( -3}} = x+Y4+i3.

95

(c) From Eq. (3.21),

-I A·C _ -I 6+ 12 =cos-I ~ = 15.80.BAc = COS AC - cos .•114h5 Sa

A· (B x C) =A· (x16+ y12+Z9) = 1(16) +2(12) + {-3)9 = 13.

i+y2-Z38A = VI4

(b) The component of B along C (see Section 3-1.4) is given by

B·C -12Bcos6BC = -- = --.C 5

(a) A and a,(b) the component ofB along C,(c) BAC,

(d) AxC,(e) A·(BxC),(f) Ax (Bx C),(g) xxB,and

. (h) (Axy)'z,

CHAPTER 3

(e) From Eq. (3.27) and Eq. (3.17),

Solution:(a) From Eq. (3.4),

and, from Eq. (3.5),

Eq. (3.30) could also have been used in the solution. Also, Eq. (3.29) could be usedin conjunction with the result of part (d).

(f) By repeated application ofEq. (3.27),

A x (B x C) =A x (i16+y12+z9) = iS4-y57 -Z20.

Eq. (3.33) could also have been used.

AxD (i2-y+i3) x (i3-z2)C=±6jAxB/ =±6/(i2_y+z3) x (x3-Z2)/

x2+y13+z3 (A A 78 A 33)= ±6-";=22=+=13=2=+=3=2 ~ ± xO.89 +y5. +zl. .

Problem 3.7 Given A = x(2x+3y) -y(2y+3z) +z(3x-y), detenninea unit vectorparallel to A at point pel, -1,2).

Solution: The unit vector parallel to A = i(2x+3y) -y(2y+3z) +z(3x-y) at thepointP(1,-1,2) is

Problem 3.6 Given vectors A = x2 - y + z3 and B = x3 - z2, find a vector C whosemagnitude is 6 and whose direction is perpendicular to both A and D.

Solution: The cross product of two vectors produces a new vector which isperpendicular to both of the original vectors. Two vectors exist which have amagnitude of 6 and are orthogonal to both A and B; one which is 6 units long inthe direction of the unit vector parallel to A x B, and one in the opposite direction.

CHAPTER 3

-x- "4 "Y +z4 _ -i-y4+z4

V(-1)2+(-4f+42 - v33 ::::::-iO.17-YO.70+z0.70.

(A x y) ·z= (x3+i) ·z= 1.

(g) From Eq. (3.27),

xxB= -.24.

(h) From Eg. (3.27) and Eq. (3.17),

96

Eq. (3.29) and Eq. (3.25) could also have been used in the solution.

A(1,-1,2)IA(I,-1,2)f

Problem 3.8 By expansion in Cartesian coordinates, prove:(a) the relation for the scalar triple product given by (3.29), and(b) the relation for the vector triple product given by (3.33).

Solution:

(a) Proof of the scalar triple product given by Eq. (3.29): From Eq. (3.27),

A x B = x(AyBz - AzBy) + y(AzBx - AxBz) +z(AxBy - AyBx) ,

(d)

B(P2) ~ -RO.896+60.449-~5.

(c)

P4(I, -1,2) =P4 [VI + 1 +4, tan-l (v'f+T/2),tan-l( -Ill)]= P4(V6,35.26°, -45°),

A R2sin2esin241

D = (R sin Beos «I>+ 8 cos Bcos $- • sin CP)R2 sin2 e sin2 4>+ R2 sin2 Beos2 $

A R2sin2ecos2cp- (Rsin esin «I>+9cos9sincp+ 'cos</» R2 sin2 6sin2 4>+ R2 sin2 Bcos2$

+ (Rcose-8sin9)4

=R(sinecoscpsin2<j1 - sin8sincpcos2 cp +4cos B)

+a(cosecos «I>sin2$- cosesincpcos2 </>- 4sin 8)

- ~(cos31j1+ sin3 CP),

C = (Rsine +6cose)cos<jl-$sin<jl + (Reose -asine) coscpsin41

=Rcos$(sinB+cos8sincp) +6coscp(eosB - sin8sin CP) -$sinljl,

P3 = (v'22 + 22, tan-I (2/2),X/4) = (2.J2,45°,45°),

C(P3) :::::RO.854+80.146-~O.707.

D(P4) =R(sin35.26° cos45° sin245° - sin 35.26° sine -45°) cos245° +4cos35.26°)

+6(cos35.26° cos45° sin245° -cos35.26° sine -45°) cos245° -4sin35.26°)

-.$( cos3 45° + sin3 45°)

= R3.67 -91.73 -~O.707.

Sections 3-4 to 3-7: Gradient, Divergence, and Curl Operators

~oblem 332JFind the gradient of the foHowing scalar functions:(a) T = 2/(;? +r).(b) V =xy2z3,

121

(d) From Eg. (3.83),

(c) From Eg. (3.82),

'flU = -r 2rzcostji _+ zsintji zcoslj>(1 + ,.2)2 r(1 +,.2) + 1+,.2 .

(b) From Eg. (3.72),

(e) From Eg. (3.72),

M = Rcos8sinip,

~ aM A I dM A 1 aM A A A COS ipVM = 1(. aR +8 R de +,Rsin8 dip = RcosSsinip-SsinSsinlj>+, tanS'

CHAPTER 3

s=2e-z+r,Vs ~as A as A as M_ -- A2 ,,2-z

=x dX +y dy +z dz =~e ""+y y-z;re .

(0 From Eg. (3.82),

N = ,.2coSIj>,aN A I aN aN A

VN = f3""" +4>- ~ +z~ = iZrcosiP-4>rsinlj>.ur r ucj> uz

(g) From Eg. (3.83),

(c) U = zcos<p/(1 +?),(d) W = e-R sin e.(e) S=x2e-z+r, ,(f) N = ,.2 cos <P.

(g) M =Rcosesinip.

Solution:

(a) From Eg. (3.72),

..:.:::'.: .-----------------------------iZ52ProbJem 3.3SJFor the scalar function V = xy - z'l. detennine its directional

derivative along the direction of vector A = (i - yz) and then evaluate it atP(I, -1,2).Solution: The directional derivative is given by Eq. (3.75) as dV/dl =VV .a/. wherethe unit vector in the direction of A is given by Eq. (3.2):

CHAPTER 3

_i(_ar x +_aT__ Z X_+-iiT--=L)- aR --";-;::x=z=+=y2=+=z=zae xZ +y2 +Z2 v.~xz-+-y2- a<jJx2 +y2

. CT. y aT z y aT x )+ Y -aR -v.-;::x=2=+=y2=+=Z=2+ -ae x-;2:-+-y2~+-Z-;;:2·-v.--;x=2=+=y2=+ -aq,x-2-+-y2-

. CT z aT -1 iiT )+ z aR vi xl + y2 + Z2+ ae x2 +y2 +Z2v'Xl +y2+ aq,0

A (aT Rsinecos$ aT Rcose Rsineeos<jJ aT -RSineSin<jJ)=x -.----+-.------+-----

aR R ae R2 Rsine a<jJ RZsinZe

A (aT Rsinesinq, aT Rease Rsin 8 sin q, aT RSinecosq,)+y -aR--R-- + -de -R-2 ---R-s-in-e- + -aq, -R-z-s-in-2-e-

A (aT Reose aT -RSine)+z aR R + ae R2

_ A (aT . a '" aT cosacosej> aT -sinej»- x aR sm eos'j'+ aa R + a$ Rsine

A (aT . e . '" aT cosesinq, aT cos<jJ)+y aR SID SID'j'+ ae R + afj) Rsine

A (aT aT -sine)+z aRcose+ ae -R-(A. e '" A. e . '" A e) aT= xsm cOS",+YSID SID",+ZCOS aR

f' e '" A e' '" A· e) 1 aT+ XCOS COS'j'+YCOS SID'j'-ZSID -::\R va

(A. '" A ) I aT+ -XSID'j' + YCOsq. Rsine aep

aT A 1 aT A 1 aT

=l'iaR+8liae +·Rsineacp'

•..•.which is Eq. (3.83) .

x-yz81 = vI +Z2'

123

Problem 3.36 For the scalar function T = e-r/5 cos $, detennine its directionalderivative along the radial direction r and then evaluate it at P(2, n/4,3).

Solution:

Problem 3.37 For the scalar function U = k sin2 e, detennine its directionalderivative along the range direction R and then evaluate it at P(4, nj4, n(2).

Solution:

CHAPTER 3

~e-r/5 sin</>r '

,,_ Y7J ~"ill= v'1+z2·dV

(dV)/ -3dl (I,-1,2) = ..j5.

T =e-r/5 cos $,

VT =r aT +~.!.aT +zdT = -r e-r/5cos$dr r d4J dZ 5

dT A e-r/5 cos$-=VT·r=-----dl 5 '

dT I e-2/5 cos 1!_ =_ 4 = -9.48 X 10-2•dt (2,11:/4,3) -

U = kSin29,

V _ A au 6.!.au -t._1_ dU __ A sin29 -6 2sin9cos9U - R aR + R de +'I' R sine a$ - R R2 R'

dU = VU. R =_ sin2 edl R2 '

dU I = - sin2{1t/4) = -3.125 x 10-2.dl (4,1[/4,1[/2) 16

124

and the gradient of V in Cartesian coordinates is given by Eg. (3.72):

VV =xy+Yx-z2z.

Therefore, by Eq. (3.75),

AtP(I,-1,2),

and

125CHAPTER 3

where ER IS a function of R only.

E = fUR.

. Problem 3.38 Vector field E is characterized by the following properties: (a) Epoints along Ii, (b) the magnitude of E is a function of only the distance from theorigin, (c) E vanishes at the origin. and (d) V . E =6, everywhere. Find an expressionfor E that satisfies these properties.

Solution: According to properties (a) and (b), E must have the form

E= HER

Hence,

~blem 3.3VF<>,r the vector field E = ixz - yyz:. - ixy, verify the divergence. eorem by computing:(a) the total outward flux flowing through the surface of a cube centered at the

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:

(a) For a cube, the closed suriace integral has 6 sides:

f E . ds = /<top +Fbottom +Fright +Fieft +FfroDt +Fback,

126 CHAPTER 3

Problem 3.40 For the vector field E =rIOe-r - Z3z, verify the divergence theorem

for the cylindrical region enclosed by r =2. z =0, and z = 4_

Solution:

ZiProblem 3.4J..,) A vector field D = i,3 exists in the region between two concentric

g(cylindricaI surfaces defined by r = I and r'-:' 2, with both cylinders extending.,>between z =0 and z =5. Verify the divergence theorem by evaluating:

--i(a) "D.ds,....... JS

127

fE.dS = [2 [~~(iIOe-r -Z3z). (-zrdrdlfl))/z=oJr=o1f=O

+ f21t r (ilOe-r-Z3z)-{rrdlj>dz»1r=21,=01=0

12 L2~+ (ilOe-r -Z3z)·{zrdrdlfl»1 =4r=O ¢=O =121C14 12127t= 0+ . lOe-22dlj>dz+ - 12rdrdcp

¢=o =0 r=0 4'=0

= 1601te-2 -481t:::=-82.77,

rfr V.Edv = ~ r2 r21t (lOe-r(1- r) 3) rdlj>drdziii 1z=oir=oilr=o r

=81t f2 (lOe-r(I-r)-3r)drir=o.,

=81t( -lOe-r + lOe-r(l+r)- 3;)I~o= 1601te-2-481t:::=-82.77.

(b)

CHAPTER 3

CHAPTER 31,,)QkO

,. ,.27t r1t r2 / t ::I \

j V·Dd'V'= I I I l--1-Ju (R2(3R2») R2sinedRded«jJ'll J.p=oJe=o1R=l R QR

= 21t( -cos8)1~=o (3/t) 1~=1= 18011:.

(b) 10/ V·Dd'll.Solution:

(a)

Therefore, fJD' ds = 15011:.

(b) From the back cover. V·D = (lJr)(dJdr)(rr3) = 4,.2. Therefore.

Problem 3.42 For the vector field D = R3R2, evaluate both sides of the divergencetheorem for the region enclosed between the spherical shells defined by R = 1 andR=2.

Solution: The divergence theorem is given by Eq. (3.98). Evaluating the left handside:

CProblem 3.~ For the vector field E =Xxy - y(r + 2r). calculate

(a) i E e dI around the triangular contour shown in Fig. P3.43(a), and

(b) 1(V x E) .ds over the area of the triangle.

Solution: In addition to the independent condition that z = 0, the three lines of thetriangle are represented by the equations y = O.x = I, and y = x, respectively.

129

x2I

(b)

y

The right hand side evaluates to

CHAPTER 3

FigureP3.43: Contours for (a) Problem 3.43 and (b) Problem 3.44.

(a)

fEedl =LI +L2 +L3,

LI = ({Xxy-y(x'l+2j2».(idx+Ydy+zdZ)J

=Lo (xy)ly=o,z=odX-!o (.?+2~n:=ody+ l~o(O)j>=odz=O,

Therefore,

1E.dI=O-~+~=-I.'j 3 3(b)FromEq. (3.105), VxE=-z3x, so that

f[ VxE.ds= [1 r «-z3x). (zdydx»lz=oi 1 lx=:oiy=o

11ix 11 J=- 3xdydx= - 3x(x-0)dx= - (~)Io =-1.

x=0 y=O x=:0

CHAPTER 3

£2 =I (ixy-j(2- +21-))· (idx+ydy+zdz)

=11(xy}lz=odx- [1 (2-+2l)/.r==I,z=Ody+ [0 (O)lx=:IdZx=: 1 ly=o Jz=O

=0- (y+ 2r)r +0= -5,3 y=o 3

L3 =/ (ixy-y(.x2 +21-»· (idx+ydy+zdz)

= fO (xy)l}~ z=Odx- fO (2-+ 2l) ! =ody+ fO (O)ly=xdZix=1 ' iy=1 =y, z- lz=o

= (~)J:I- (1)/:1 +O=~.

130

Problem 3.44 Repeat Problem 3.43 for the contour shown in Fig. P3.43(b).

Solution: In addition to the independent condition that z = 0, the three lines of thetriangle are represented by the equations y =0, y =2 - x, and y = x, respectively.

(a)

fE ·dl =£1 +L2 +L3,r

LI = J (ixy-Y(~+2l»·(idx+Ydy+zdz)

= [2 (XY)/y=o,z=odx- fO (2-+21-)lz=Ody+ [0 (0) 1)=0dz = 0,lx=:o ly=o lz=o

£z =J (ixy- Y(2- + 21-)}. (idx+ydy+zdz)

= [I (XY)/z=O,Y=2-xdx- t (.x2+2l)lx=z_"z=ody+ 10(O)/y=z_xdz1x=2 ~=o J' =0

= ( .,:-_ x3)/l _ (<111- 1,,2 4-113) II -1-0 = -11\" 3 ) IX=2 \ 'J -J • J ly:i:O' ~ 3'

CProblem 3.40Verify Stokes's theorem for the vector field B = (rrcosej> ++sinej»by evaluating:

(a) i B· dI over the semicircular contour shown in Fig. P3.46(a), and

(b) Is (V x B) .ds over the surface of the semicircle ..

Solution:(a)

fB.dl= { B·dl+ r B·dJ+ 1B·dl,iLl 1L2 L3

B ·dl = (rrcosej>++sinej». (rdr+if.rdej>+ zdz) = rcosej>dr+ rsin<j>dcj>,

1B·dl= (12rcos<Pdr)/ + (fO rSin<Pd$)IL) r=0 ~=o, z=O J~=o z=o')

= (t?)I~o+O=2,

131

f 11 2E·dI=O- 3+3 = -3.

(b) From Eq. (3.105), V xE = -Z3x, so that

{{ VxE.ds= {I r «-i3x). (zdydx»lz=o11 1:=0/;=0

{2 r2-x+ lX=11y=0 ((~Z3x).(zdYdx»lz=O

11leX 12l2-x=- 3xdydx- 3xdydxx=O y=O .;=1 y=0

=_rl~~_~dx_f.~~_~_~~1.;=0 1:=1

=- (X3) I~- (3r -~) I~l= -3.

CHAPTER 3

Therefore,

Problem 3.46 Repeat Problem 3.45 for the contour shown in Fig. P3.46(b).

Solution:

CHAPTER 3

Figure P3.46: Contour paths for (a) Problem 3.45 and (b) Problem 3.46.

r B.dl= (12 rCOS$dr)1 + (L: rSin$d$)/JL2 Jr=2 z=o $-0 r=2, z=o

=0+ (-2cos<j»/;=o=4,

r B·dJ = ( rO rcos<j>dr) I + ( r: rSin$dtp) IJL3 Jr=2 $=rc,z=O J~_1C ;:;::0

= (-~,-2)1~2+0= 2,

!B.dJ=2+4+2=8.(b)

132

VxB = Vx{rrcos$+$sin$)

=r(.!.~o-~(Sin<j») +~(~(rcos<j» - ~o)r a$ dZ dZ ar

+z~ (;.cr{Sin<j») - ad$(rCOS$»)

= i'O+~O+z~(sin<l>+ (rsin<l>)) = zsin<l>(I+ ~) ,

IIVxB ·ds = l;oLo (zsin<l>(1 +~)) .(zrdrd<p)

= i:o/~osin <j>(r+l}drd<j>= ((-cos<j>(!,.z+r))I~=o) 1:=0 = 8.

For the hemispherical surface, ds = ftR2 sin e de d<j>.

CHAPTER 3134

A =R cos e +,sin e = RAR +9Ae +~A~.

Hence, AR = cose. As = 0, A~ = sine.

~ 1 (a. ) A 1 a ~1 aARVxA=R-- -(A~sme) -6--(RA$)-cp--RsinS ae R aR R ae

A I a . 2 ~l a . Al a=R---(sm e)-e--(Rsme)-q.--(cos9)Rsine ae R aR R as

A 2cose e sin e .i. sine=R-R-- T+"R'

Problem 3.47 Verify Stokes's Theorem for the vector field A = RcosS+.sin8 byevaluating it on the hemisphere of unit radius.

Solution:

The contour C is the circle in the x-y plane bounding the hemispherical surface.

1.A .dl = {2ft (Rcose+4tsine) '~Rd<j>le=7t/2 = Rsine (2ft d<j>/e=7t/2 = 21t.JC 1~=o R=I 10 R=l

c::¥roblem 3.45:> Determine if each of the following vector fields is solenoidal,conservative, or both:

(a) A =x2.ty- Yr,(b) B = x.xz - " +z2z.(c) C = f(sin<j»/? +~(cos<j»/?,(d) D=R/R,(e) E=r(3-1~J+Zz,(()F= (iy-yx)/(x2+r),(g) G =x(x2 +z2)+y(r+x2)+zC1+z2).(h) H = R(Re-R).

The field C is neither solenoidal nor conservative.

135CHAPTER 3

Solution:

(a)

V·A = v·(lliy-Yl) = ~2xy- ~1=2y-2y = 0,ax dy

VxA = Vx(i2xy-yy2)

=i (~O-~(-?»)+Y( ~(2xy)- ~O)+z( ~(-r)-~(2xy»). Oy dZ az ax ax dy

= iO+YO-z(2x).

The field A is solenoidal but not conservative.

(b)

V·B= V.(~-Yl+z2z) = ~r-~I+~2z=2x-2Y+2,ax dy dz

VxB = Vx (xr-yy2+ Z2z)

=i (~(2z) - ~(-.r»)+Y (i-(.xZ) - ~(2z») +z ( ~(-.r) - ~(xZ»)dy dz az ax ax cry

=xo+yo+zo.

The field B is conservative but not solenoidal.

(c)

Hence, E is conservative, but not solenoidal.(f)

CHAPTER 3

(d)

(' R ') 1 d ( ., (] )) 1 a. 1 d 1\7·D=\7. - = -- R~ - +---(Osme)-+- ---0=-

R R2(jR R Rsineae . Rsinea<jJ R2'

VXD~VX(~)

~ 1 (a a) A1( 1 a (1) a )=R-- -COsine) - -0 +8- -- - - -(R(O))

R sin e ae· a<l> R sin e a<jJ R dR

Al(d d(l)) A A+<I>R aR (R(O)) - ae R = fO+80+$O.

136

The field D is conservative but not solenoidal.

(e)

CHAPTER 3

VXF=:X(O-O)+Y(O-O)+Z[~(~)-~("Y .,)]dx x•.+ y- dy x~+ y~

( " 2" )~ 1 2.\- 1 y-

=z .,- .,---.,+., ')x2 + y (x2 + y2)~ xl + y- (x~+ y2)~

~ 2CT -~)

= z (2 ")2:f O.X +y-

Hence, F is neither solenoidal nor conservative.

(g)

G = :X(~ +z2) +y(i +~)+z(1+Z2),<1.,., <1.,., d2 2

V· G= ~(x- + Z-) + ~(y +x-) + ~(y + Z )oX oy. oZ

= 2x+2y+2z:f 0,

~(a ., ., <1.2 .,) ~ (d " ., d "2))V XG = x oy (y~ + z-) - dZ (y +x-) + y dZ (x- + r) - ax (y + r .

+z (;x (1+~) - ;y (~+C))= x2y+y2z+z2x# O.

Hence, G is neither solenoidal nor conservative.(h)

H=R(Re-R),

V· H = ~2 a~(R3e-R) = ~2 (3R2e-R - R3e-R) = e-R(3 - R) # 0,

V x H = O.

Hence, H is conservative, but not solenoidal.

Problem 3.49 Find'he Laplacian of the folJowing scalar functions:(a) \l = xTz3,(b) V =.xy+yz+zx,(c) V = 11(~ +1),(d) V=5e-rcos<!>,

(e) V = lOe-R sin e.

(a) From Eq. (3.110), V2(xlz3) = 2xz3 +6.xrz.

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