'"70, -'

A = tiB x cj = illX2- y4) x (x- y-z4)1= !lx(-4)(-4.\.Lv(_(?\(_L1\\.LtT(,)(_1\_[ ..1\1\1

•.. ' '. /, -' ..•.. \, \.- •.•, "I' -,-\ -/ \. ').1.)1

CHAPTER.

= ~lx16+ y8 +221 = ty'162 + 82 + 22 = ~v'324= 9,

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

A . C = (X2 - y3 +z) . (x4 + y2 - z2) = R- fi -? = ()B·C= (X2-Y+Z3)·(x4+Y2-z2)= 8-2-6 =0.

A = x(:l.- 1)+ y\ -1 - ( -1» + z(0 - (- 2)) = X+ z2,IAI = .../1 ..j.. 4 = 7 74 .•• . " - I

~ A x+z2 ~045 ~089a=-,.,=--:-=-:-=X. +Z ..IAI .L.L.'+

Problem'.! Vector A ~t::lrt<::::It nn1"t (1 _ 1 _ ')\ ",..,A o~A~ M ~~:_. r,., _, r'\ ",. ,_ -J.:.----.\-' -i _/ .•..•·.•.•.o,)~\,..t'vJ...,I,..i. •• \-, .J.,Vj_J..LU,U

a unit vector in the direction of A.

Solution:

rrOO1em -,.,j In CartesIan coordmates, 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.

Solution: Let B = p--;A = x2 - y4 and C = PI A = x - y - z4 represent two sides ofLilt: mangle. Since the magmtude ot 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 thetnangle:

"o!Uuon:

........:-..--

Givl"n A =y?_v~.L71 ,,,,ri-n -~Q I;'''',;;V.--- •• - . -- -.- -- - • •.........x I.J'" L.;k.J':,

T".oI1'>r;r.. •..• h ....fou •....•......••..•D ...._...J T> :.r .•. =_ . __ t· 1 •.••.•..

••.....••..•.•..•••...u \..I ••••." •••••.•.. ""'JJ JJx U,.uu..LJ;: 11 .•.""A 1':>pCl.1JCUUl~Ula.r lU D.

CHAPTER 3

"ru -4-.c.VJU __Bx = 3V14 - 3

Given vectors A = ~-L y2 - Z3, B = x3 - 5'4. and C = 5'3 - z4, find

or

From the x-component,

There are an innnilt:: number of vectors whkh could b~ B <mdbe perpendicular to A,

but their x- and z-components must satisfy this relation.This result could have <Usabeen obtained by assuming OAti = 90° <l..'1d calculating

IAIIBI = IA x BI·

v14 - V4+B; +B;which can only be solved for the minus sign (which means that A and B must pointl'n opposit •..ui···· .. - L" __ ~ .••• _- ~~ t..~ ~~_"non <.'nl"T-nO'f...,r r:<2-L r:<2"" ret;uUH:.") IVI Ulc;.Ul LV u\.t pUL'-LLl.v.1.}.• v_ .••. .o.ub --XI - Z'

-3From the y-component,

~_1 __ ~ .•..•._.L:)VJ.U.LIV~.I.

and, from the z-component,

A/IAI = ±B/IBI,

(a) If A is parallel to B, then their directions are equal or opposite: aA = ±aB, or

-2B,=-.- 3

This is consistent with our result for B; + B;.1'hese resuits couid also hav~ b~til obtained by asSUIT'.i~g 8AE was 00 or 1800 a..T1d

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),

80

81

• ~.rn ..,r<\_ A. ••.,,{':;'1~ I .:';1') I ~O'\-v~L1_~,7_7?()•.'""1.A \.JJ A '-'} - .•.""'1 A-\.')..LV T J.L- J L;/; - ..-•••....•••• :""" "

Eq. (3.30) could also have been '..1sed in the ~o!1Jtion_ Also, Eg. (3.29) could be usedin conjunction with the result of part (d).

(1) By repeale.d application of Eq. (3.27),

Bcos6BC= B·C _ -12r - --'-' 5

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

and, from Bq. (3.5),

(c) From Eq. (3.21),

". __ ",,<"-1 A· C _ ,.,..",-1 6 + 12 = ~()S-1 18 = 15.80.VAl.. - ~w AC - --- "\;'14"\~ - .- 5V14

(d) From Eq. (3.27),

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

(a) A <:I.LHl a,(b) the component of B along C,(c) 8Ac,(d) A x C,(e) A ·(B x C),(0 A x (B x C),(g) Xx B, and\11) (AxY)·z.

Solution:

(a) From Eq. (3.4),

CHAPTER 3

. "Eg. (3.33) could also have been used.

CHAPTER 3

~ ~A ~ ';/f

- ...• - J,,::::,,'-£<-'- ~ -xo.17 -YO.70+z0.70.33

Problem 3.8 By expansion in Cartesian coordinates, prove:(<» thp rp 1"tirm f"r thp c;r::t1:lr trin1f>nror!nrt p"iven bv n 291 and'-.I --~----_·---------~--------r--r--·---·c- - --., ,,'.- -"'~

(b) the relation for the vector triple product given by (3.33).

Solution:

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

A ..•,,"1) f~"".:'. I ~"2\ '-/ f~'J ~..."~ ,C'1A ~ I \ ~ - J T LJ) A \./'J-J - ~.t-)

C = :t:61A x BI = :t:61(X2 - Y + i3) x (x3 - i2)1I x2+y13+z3 I ~ ~ ~= :t:6--;:::===:::::=:::: :t:(xO.89+v5.78+zL33).

V22 + 132+ 32 -

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

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

pt:rpt:Iluicuhu to both of the origiual v~t0rs. TwO v~t015 cxist 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.

(A Y V1. 7 = (y~ -1--71. 7. = 1\--".I} -. \P_' -) - -.

Problem 3.6 Given vectors A = x2 - y+z3 and B = x3 - 22, find a vector C whose

magnitude is 6 and whose direction is perpendicular to both A and B.

Solution: The cross product of two vectors produces a new vector which is

Fn n ?91 :mn Fn n ?'i) ~(jHlrl~1~(jhaVf~been used in the solution.--: •.: \.-~-_./ ---- --....t.- ,----- •• - -. -.- --

82

1 .•.1 1':_ ~_tw~ = .....,..: _ •......•. :1..•....•..:_ ...t.:r-o+ ...•.•..•".,..o •....•,,,,..,.,,, .f;..l""\~ i-'ho r\~t"T~1""\

lC;UbUl~ lUt....H.::;~v .1...LU~a.llJ ill ¥.1.VPVJ...UVU LV u..L\"'-.U.. "-U •.:H.~..U.""'''''''~H t;..r..j ".VUJ. .•.u_ •••.• '-' .•....I.b..l.u~

3.17 When sketching or demonstrating the spatial variation of a vectorWI": oftl":n n;:f'; arrows, as in Fig. 3-25 (P3.17). wherein the length of the arrow

made to be proportional to the strength of the field and the direction of the arrowthe S3..'11eas that of t.~e field's. The sket{:h shown in Fig. P3.17, whil'h r~pre;:p.nts

vector field E = fr, consists of arrows pointing radially away from the origin and

89

~_r.."'I"""f, ~_A C£. ~t'\"'A= AU.oJ / T J V •.JV T £V./'"T.

B = x( -2) +y(2+ 3) - z(l) = -x2 + y5 - z,A B -x2+yS -z -x2+y5 - zb - - - ---;;:=== - ----

- 'H! - ../4+25+ 1 - "/"30

. _ (~64\. EM ~, • (128 / .,\ 512 512 r\

A' C = \1)' 3") T I..:)L. • v J T \.3' \.-'+J ) = 3 - 3 = v/ ,(11\ / 1,(\ /1')(1 \ ':;1'1 '<:;1')

B· C = lo, v3-.)+ l32. ~3V)+ ~~;v . (-4)) = ~;- - ~;- = 0

arrow representation, sketch each of the following vector fields:

Vprifv th:<t r ;;:orthoonn::!l to A and R----.I ------ - -- - _n c..,)

4. C = X ~4 +Y 32 +Z 1~8

C A 64 , A 32 ..L. A 128A X'fTY ,ZTC--- .- ICI - v'(~)2 +322 + e~8) 2

c points away from the origin as desired.

Solution: At P( 1,0,-1),

CHAPTER 3

CHAPTER 3

/'"~L)

~

Oroblem 3.18)use arrows to sketch each of the following vector fields:(a) EI = Xx- yY,

(b) E2 = -<b.

(e) E3 = y~,(d) E4 = rcos<j).

Solution:

93

94

(a)

(b)

n..-, 10",", • 1? _ ~,~ ~ '0rJ. OQ. • .1.:11- .•••..,A - J J

y

CHAPTER 3

CHAPTER 3

(c)

(d)

y

I 11111 IIi if it t.

;dd!HUHillJ 1111 t tH.,~x'l1TT1l1Tmlt t t t t t t t t t • •.

, t + 11111 It II1I1I ! I '

,1~lilt. 1111'

, t tr I x indicates lEI is infInite~

1\P3.18c: E3= y(l/x)

1\ .P3.18d: E4= rcosej)

95

CHAPTER 3

_. ':. .. ;:, -;:. -1,1,-" ,-,,,,,\ +_~-1(1/1\\P3 - lV 1 -;-1 T L- , L<UI \ V' T ~ I'" j, c~. \' I -)J

= (,a,0.616 rad.n/4 rad) ~ (2.45,35.3°,45.0°).

(" " __ A r\ -_A\ _ (":I, 00 ()O,-\..J,V~Q.U.~VJ.~ __j-\-', .....'i- J-

CG~V(;:tt.~~C00!di~",tf'<: of the following points from Cartesian tospherical coordinates:

(a) h(i,2,O),(b) P2(0, 0,3),(c) P3(1,1,2),

(d) P4( -3,3, -3).

ot:;./v

Note that in both the cylindrical and spherical coordinates. <i> is in Quadrant!.(b) In the cylindrical coordinate system.

Sections 3-2 and 3-3: Cooniinat.t: Sy~tciJ.15

Nme mal 111uuuI lh6 Cy'1iuGri;:;;i ;l."';.d sphe!"iC'?l ('onrrlinHtes. <i> is arbitrary and maytake any value.

(c) 1n the cylindrical L:UUlulua.tc. 5)"5-::": ••••

In the spherical coordinate system,

In the sphencal coordinau:: ::.y::.ic;w.

Solution: Use the "coordinate variables" column in Table 3-2.

!':o!e !!:O'.t jn noth the cylindrical and spherical coordinates. q>is in Quadrant 1.

(a)

97

. / .• ') _~ 1 ,_ ,

f'4=\V\-jf-r-j-,ran '~:>/-:»,-:»= (')"....h '),,'11:/4 rnti -'1,) ~ (4.24.135.0°. -3).

\. - , -, - -, I J "' ' . ",

1"'" T." ,- _ -1-= __ 1 .J;_ .•.~_ .......••.._...._\UJ 111LUC (.;)iHH ..U.h...u.J. ""VUj,vi.d.a..L\... 0) "H.VH.i.,

CHAPTER 3

P((r,<1>,z) = PI (Rsine,<jJ,Rcose) = PI(5sinO,0,5cosO)1"> 11\ n ~\

-, l\V,V,_'}.

(b) P2(X,y,Z) = P2(3cosO,3sinO,0) = P2(3,0,0).

(c) P3(X,y,Z) = P3(4cosn,4sinn,2) = P3( -4,0,2).

Note that in both the cylindrical and spherical coordinates, <1> is in Quadrant II.

Tn the spherical coordinate system.

1) _ (. If ~\2.L ~2..L (_ ":n2 t""..,-I (. I( _'),,)2 -1-?,,2/ _?,) t::.n-I (?,/ - :rnJ.. 4 - \ V\..-":J I •..•••• \ -' J :t .•. _- 'V \ - I • -, - ••I - - " '.' "

= (3.J3, 2.187 rad,3nj4 rad) ~ (5.20,125.3°,135.0°).

Problem 3~O Cony~rt t..'1ecocrdi~2!e£ of the foHc\\ving p011')t~from cylindrical toCartesian coordinates:

(a) Pl(2,n/4,-3),(b) P2(3.0.0).

(c) P3(4,1t,2).

Solution:(a)

P1 (x,y.z) = PI (rcos<jJ, rsin<jJ,z) = PI (2cos ~ ,2sin ~,-3)= PI (1.41,1.41, -3).\. ""'t' -r "

Problem 3.21 Convert the coordinates of the following points from spherical tocylindrical coordinates:

(a) Pl(5,0,0).(b) P2( 5,0, n),

(c) P3(3,nil, n).

CHAPTER 3

(c)

}_2

/

(e)

/

x

y

+

(d)

/Figure P3.22: Surfaces described by Problem 3.22.

(a) (b)

• Nt) = n/3

(~) TTsing Eq. (3.43a).

Problem 3.22 Use the appropriate expression for the differential surface area ds todeterrmne ~.e are~ of each of the following surfaces:

(a) r = 3; 0 ~ <P ~ 11:/3; -2 ~ z ~ 2.(0) 2:5,.:s 5; rr,/2 $. tv ~~; 2: = 0,(c) 2 ~ r ~ 5; <p = 11:/4; -2 ~ z ~ 2,(d) R = 2; 0 ~ e s nj3; 0::; ~ :::;n,(e) 0 $ R $ 5; e = 1[/3; 0 $ <p $ 211:.

Also sketch the outlines of each of the surfaces.

98

Solution:

99

y

(b)

z

t

x

y

(a)

Figure P3.23: Volumes oescribed by Problem 3.23

x

r5 (Tt ... I ., " ,.~ \ pt 2111:

A = J 1. lr) 1:::=0 d<j>dr = I \ ~ r""$ ) j ~=2)j = 4 .r=2 Q=Tt/2 \ Q=TC/2

(c) Using Eq. (3.43b),

(a) From Eq. (3.44),

~? h~ :

A = j - j - (1)1$=Tt/4drdz = ((rz)!;=_z) r- = 12.:::=-2 r=Z r-Z

(d) Using Eq. (3.50b),

leittJ lTC ( /3 ) ITCA = . (R2 sinO) IR=Zd(j>dO = (-4<j>cosO)!~=o . = 211:.s=o 0=0 (1)=0

(0) UbingEq. (3.431:),

Also sketch the outE:nc of each volume.

CHAPTER 3

Solution:

(e) Using Eq. (3.50c),

(5 tTC ( ( 11:\12Tt \ 15 25011:A = (RsinO)IA="."d<j>dR = ~R2(j>sin:::) ! = ~JR=OJ$=O . ," \\- .)JI¢=O/IR=O L

~f1b~ Find the volumes described by(a) 2 ~ r ~ 5; n/2 ~ q, ~ n; 0 ~ z ~ 2,(b) 0 i: R i: 5; 0 i: 0 i: 11:/3; 0 i: <P i: 2n.

~ i=(rO.487+cPO.228+z0.843).

+ A x B = +r(-4(1 + ~yj))(~) -cPG)( i) - z(4(i + 1v3))d.J3)-!A x BI - . /(")(1 ..L 1.h\,,2 ...!... rl.'·/·...!... (':I,...!... I, n)2

V \""'\ ~ , 2 v -' II I ~4 ) I ~~ , - v ~ }

CHAPTER 3

A ~f ..t. I "",_, A {"") ... , If .•..:_A\ • ;;.(_ ,..,_'\ft = 1\\,..VI>'!' T J4} -'I'\.f.ol T ""''''111'1') T LV - L..<'J>

r< T> <- T> A f~"", 2", (R4+Eh-cP)L.=Doa=Do- = \-~-r-o/.:J}' ---

IAI y'16+4+ 1-R-3 11

= - V2f = -2.4.

. B = -rsin<!>+ zcos<!>,

D = B - C = (-R2+4l3) + (R2.09+91.05 -$0.52)

= Ro.09+e 1.05 +$2.48.

(b) Vector component of B in direction of A:

C ~ - - (-2.tJ)C = aC = A- = (R4+ e2 - cp) -­

IA!= -(R2.09+9 1.05-~0.52).

(3) Scalar component of B in direction of A:

(c) Vector component or B perpem.li(,;ularLuA:

Problem 3.27 Given vectors

_1(AoB\ _1(0\eAB = cos - \ AB ) = (,;OS AB) = 900•

(b) At (2, rtj3, 1), A = r~- $4(1 + !v'3) and B = -r~v3+ z~. Since A x H isperpendicular to both A and B, a unit vector perpendicular to both A and B is givenby

find

(a) 8AB at (2,rt/2,O),(b) a unit vector perpendicular to both A and Bat (2, rt/3, 1).

Solution: It doesn't matter whether the vectors are evaluated before vector productsare calculated, or if the vector products are directly calculated and the general resultsare evaluated at the specific point in question.

'.-' A. .••• ,...., _ ,.., n\ A ~ ..io I ~i') r.•.•A n _ ~ "'C.,..."..,..,...., 1:",., ("'J..-""tit) J-\.l 1..."',II../ ~,V},.t'1. - -'t'U Th~ a.uu ~ - - .•.• L .l..VU1..LJ"1" \-", •••••..1./,

102

105

P4 = (2sin(n/2),1t/4,2-:os(n/2)) = (2,45°,0),~ ~ j

D(P4) = r+$".(e)

Pj = ( V\2+(-1)2+22,tan-i (V12+(-li/2) :tan-ie-l/l»)= (.J~~~~o. _4'::;°).,. -;----; - /-

A(PJ) ~ R2.856-82.888 +~2.123.

(b)

A = (Rsinecos$ +ecosecos$ -~sin$)(Rsinesin$)2• /..;::. " 1"\." J. .;:. . .r\ _O • .J. ! i: . __. ..L\/n_-=_f\ ..L\fn f)\

;- \..KSllltlSHlqJ T o cu:;O:;lUIjI TIjIl:U:;\jJ )\.1\ 1>1UO~U:>\jIJ\.l\ \"'u:>uJ

+ (R cos e - e sin e)4

= R(R2sin2esin<j>cos$(sin8sin$+cose) + 4cos e)

+ 8(R:!.sin8cosesin<j>cos<j>(sinesinq, + cos 8) -4sine)

E = (rsin8 +zcos8)cos<l>+ (rcos8 - Zsine) sin$+$sin:.<e,_ f_ n \1-'5 = \5, "2,n) ,

~/~, f~. 1t, ". __1t\ q __ , (~ n <>_:_1t\_: __ • 2_~_21t ~,.!::~\.r5) = \r:;w"2 T;£I,.;U:;"2) I,.;U:;Ji.-r-v· ~u1>2 -L.l>lU2) :>HlILT't':>lU "2 = -. T't'°

(a)

CHAPTER 3

Uroblem 3.3]) Transfonn the following vectors into spherical coordinates and thenevaluate them at the indicated points:

(a) A = xl +yxz+Z4 atPl(1,-1,2),(b) B = y(xl+y2+zZ)-z(X"+y2) atP2(-1,0,2),(c) C = rcos<j>-~sin<j>+zcos$sin$ atP3(2,n/4,2),and

~d) D = xl/(.:2 +1) - y2/(.:2+1) +Z4 at P4(1,-1,2).

Sections 3-4 to 3-i; Gradieut., Divergence, and Curl Operators

Cd)

Prnhl"m ~~" f:;';T1rlthp <yr::~rl;pntnf thF'. fnl1nw1n<Y <:~::!J:;lrfnnr.tion<:'_ •......- •.....•....- -.-- - .--- ---- 0--------- -- ---- ----- ---0 - .. ----- -- --- " -

()T-2/('?,_2)a -! r-r..: ,(b) V = .\)2 z3,

CHAPTER 3

c = (Rsin8+ecas8)cas$-~sin$+ (llcas8 - esin8)cosq,sinq,

= Rcos<j>(sin8+ cos8sin<j>) +ecos$( cos8 - sin8sin<j» -$sin<j>,

P3 = (V22 + 22,tan-1 (2/2), Tt/4) = (2V2,45°,45°),

C(P3) ~ Ro.854+eo.I46-~O.707.

B(P2) ~ -RO.896 +30.449 -$5.

P4( 1, -1,2) = P4 [VI + 1+4, tan-Ie V1+1/2),tan-1( -1/1)]

= P4(../6, 35.26°, -45°),

A A A R2 sin2 8 sin2 <j>D = (Rsin8cos<j>+6cos8cos$-cpsin$) 2 . 2 . 2 . 2R sm 8sm $ + R2 sm 8eosLtj}

A A . A R2 sin2 8eos2 $- (Rsin8sin$+6cos8sin$+4>cos¢)~.., , ,?" ' '? . ~.., • .,,, ..,

t(~ sm- t1sm- <p + t(~ sm- t1cos~ <p

+ (Reos8 - esin8)4

= R(sin8cos¢sin2 $ - sin 8 sin¢cos2 $ + 4cose)

+ a(cos8cos<j>siIf <j>- cose sin<j>cosL<j>- 4 sin 8)

- $(cos3 ¢ + sin3 ¢),

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

+e(cos35.26°cos45°sin245° -cos35.26°sin( -45°)cos245° -4sin35.26°)

_ (j)(cos3 45° + sin3 45°)_ D ~ ",'7 A 1 '7~ _,I,() '7()",!-'&''--'_>./1 -_~.t..." """"''''''',.

106