Ece3443f05hw4.pdf

93

CHAPTER 3

Chapter 3 Section 3-1: Vector Algebra Problem 3.1 Vector A starts at point a unit vector in the direction of A.

(1, - 1,-2) and ends at point (2,-1,0). Find

Solution:

A = i(2-1)+y(-1IAI = vI +4 A

a

A

= IAI =

(-l»+z(O-

(-2» =i+Z2,

= 2.24, i+Z2 2.24

~O45

= A.

A

+ zO.8

~blem ~) Given vectors A = :i2- y3+ Z, B show that C is perpendicular to both A and B.

9 -

= i:2- y+Z3, and C = i4+y2 -

n,

Solution:

"A·C = (:i2-Y3+z)· (i4+y2-Z2)

= 8-6-2=

0,

B·C= (i2-Y+Z3)·(i4+Y2-Z2) =8-2-6=0.

Problem 3.3 P2(2,-2,2),

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

---+

Solution: Let B = P1P2 = i2 - y4 and C = PI P3 = i: - Y - z4 represent two sides of the 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 the triangle: 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, where the cross product is evaluated with Eq. (3.27). 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.

94

Solution:

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

From the y-component,

-3

±2

04 - -/4+B;-t--jj; which can only be solved for the minus sign (which means that A and B must point in opposite directions for them to be paraHel). Solving for B; + B;,

B;+B;=

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

From the x-component, __

2

-

__

-B . _x

Bx

04 - "/56/9'

= -2..;56 _ -4 3V14 -"3

and, from the z-component,

-2

Bz

= --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),

O=A·B= 2Bx-6+Bz, or

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 calculating IA x BI· IAIIBI

=

~blem

~~iven

vectors A

= x + y2 -

23, B

= x3 -

Y4. and C

= y3 -

Z4, find

95

CHAPTER 3 (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, Solution: (a) From Eq. (3.4),

and, from Eq. (3.5),

i+y2-Z3

=

8A

VI4

(b) The component of B along C (see Section 3-1.4) is given by Bcos6BC

B·C

-12

= --C = --.5

(c) From Eq. (3.21),

BAc

= COS

-I

A·C _ AC -

-I

6+ 12

cos .•114h5

=cos-I ~

Sa =

15.80.

(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.

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

A· (B x C)

= A· (x16+

y12+Z9)

= 1(16) +2(12)

+ {-3)9

= 13.

Eq. (3.30) could also have been used in the solution. Also, Eq. (3.29) could be used in 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.

CHAPTER 3

96 (g) From Eq. (3.27),

xxB= -.24. (h) From Eg. (3.27) and Eq. (3.17),

(A x y) ·z= (x3+i) ·z= 1. Eq. (3.29) and Eq. (3.25) could also have been used in the solution. Problem 3.6 Given vectors A = x2 - y + z3 and B = x3 - z2, find a vector C whose magnitude is 6 and whose direction is perpendicular to both A and D. Solution: The cross product of two vectors produces a new vector which is perpendicular to both of the original vectors. Two vectors exist which have a magnitude of 6 and are orthogonal to both A and B; one which is 6 units long in the direction of the unit vector parallel to A x B, and one in the opposite direction.

AxD

(i2-y+i3) =±6/(i2_y+z3)

C=±6jAxB/

=

x2+y13+z3

x (x3-Z2)/ (A

±6-";=22=+=13=2=+=3=2 ~

Problem 3.7 Given A = x(2x+3y) parallel to A at point pel, -1,2). Solution:

x (i3-z2) A

± xO.89 + y5.

78

+ zl. 33). A

-y(2y+3z) +z(3x-y), detenninea unit vector

The unit vector parallel to A is

= i(2x+3y) -y(2y+3z) +z(3x-y)

at the

pointP(1,-1,2)

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

-x- Y"4 +z4" V(-1)2+(-4f+42

_ -i-y4+z4 -

v33

::::::-iO.17-YO.70+z0.70.

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), AxB

= x(AyBz

- AzBy)

+ y(AzBx

- AxBz)

+ z(AxBy

- AyBx) ,

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

(c) C

P3

= (Rsine +6cose)cos<jl-$sin<jl + (Reose -asine) = Rcos$(sinB+cos8sincp) +6coscp(eosB - sin8sin = (v'22 + 22, tan-I

coscpsin41 CP)

-$sinljl,

= (2.J2,45°,45°),

(2/2),X/4)

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

(d) R2sin2esin241

A

D

= (R sin Beos «I>+ 8 cos Bcos $- • sin CP)R2 sin2 e sin2 4>+ R2 sin2 Beos2 $ R2sin2ecos2cp 'cos+ R2 sin2 Bcos2$ A

- (Rsin esin «I> +9cos9sincp+ + (Rcose-8sin9)4

= R(sinecoscpsin2<j1 + a(cosecos

sin8sincpcos2

cp

+4cos B)

«I>sin2 $ - cosesincpcos2 - 4sin 8)

- ~(cos31j1+ sin3 CP),

= P4 [VI +

P4(I, -1,2)

1 +4, tan-l (v'f+T/2),tan-l(

= P4(V6,35.26°,

-45°),

= R(sin35.26°

cos45° sin245° - sin 35.26° sine -45°)

+6(cos35.26°

cos45° sin245° -cos35.26°

D(P4)

-Ill)]

cos245° +4cos35.26°)

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

(a)

(b)

332JFind

the gradient of the foHowing scalar functions:

= 2/(;? +r). V = xy2z3, T

121

CHAPTER 3

= zcos

(c) U

(d) W (e)

= e-R sin e.

,

S=x2e-z+r,

(f) N = ,.2 cos

(g) M


= Rcosesinip.

Solution: (a) From Eg. (3.72),

(b) From Eg. (3.72),

(c) From Eg. (3.82), 'flU

= -r

2rzcostji ,.2)2

(1

+

_+

zsintji r(1 +,.2) +

z coslj>

1 +,.2 .

(d) From Eg. (3.83),

(e) From Eg. (3.72),

s=2e-z+r, Vs

=x

~as dX

A

as

A

as

M_

+y dy +z dz =~e

--

A2 ,,2-z ""+y y-z;re .

(0 From Eg. (3.82), N

= ,.2coSIj>,

I

aN A aN aN VN = f3""" +4>~ +z~ ur r ucj> uz

A

= iZrcosiP-4>rsinlj>.

(g) From Eg. (3.83),

M VM

= Rcos8sinip, ~ aM

= 1(. aR

A I dM +8 R de

+,Rsin81 A

aM dip

A

A

= RcosSsinip-SsinSsinlj>+,

A

COS ip

tanS'

123

CHAPTER 3

aR --";-;::x=z=+=y2=+=z=z ae xZ a<jJx2 +y2 _i(_ar x +_aT __+y2Z + Z2 v.~xz-+-y2X_+-iiT--=L)

+ Y. CT. -aR -v.-;::x=2=+=y2=+=Z=2 + -ae + -aq, y aT x-;2:-+-y2~+-Z-;;:2·-v.--;x=2=+=y2= z y aT x-2-+-y2x ) + z. CT aR =x A

vi xl

+zy2 + Z2 + aT ae x2 +-1 y2 + Z2 v'Xl + y2 + iiT aq,0)

-.----+-.------+-----

(aTaR

+y

Rsinecos$ R

(aT -aR--R-Rsinesinq,

A

aT ae Rcose R2 Rsineeos<jJ Rsine

+ -de aT

aT a<jJ-RSineSin<jJ) RZsinZe

-R-2 Rease---R-s-in-eRsin 8 sin q, + -aq, aT RSinecosq,) -R-z-s-in-2-e-

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

-_ xA (aT aR sm . a eos'j'+ '"

aa aT

R + cosacosej>

+y (aT aR SID . e SID'j'+ . '" aT ae cosesinq, R A

+z A

=

aRcose+ (aT

a$ aT Rsine -sinej»

+ aT afj) Rsine cos<jJ)

-R-

ae -sine) aT

'" A. e SID",+ZCOS . '" A e) aT xsm e cOS",+YSID aR

(A.

f'

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

+

I

'" A ) aT -XSID'j' + YCOsq. Rsine aep aT A 1 aT A 1 aT (A.

=l'iaR+8liae

+·Rsineacp'

•..•. which is Eq. (3.83) . :.:::'.:

..

----------------------------.

iZ52ProbJem 3.3SJFor the scalar function derivative

P(I,

along the direction

V of vector A

-1,2).

= xy - z'l. = (i - yz)

detennine its directional and then evaluate it at

Solution: The directional derivative is given by Eq. (3.75) as dV /dl the unit vector in the direction of A is given by Eq. (3.2):

x-yz

81

= vI +Z2'

= VV . a/. where

CHAPTER 3

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

VV =xy+Yx-z2z. Therefore, by Eq. (3.75), dV

,,_

J

Y7

~"

ill = v'1+z2· AtP(I,-1,2), (dV)/ dl

(I,-1,2)

-3 = ..j5.

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

VT

= e-r/5

cos $,

=r aT +~.!. aT +z dT = -r dr r d4J dZ

e-r/5cos$ 5

~ e-r/5r sin

'

dT e-r/5 cos$ -=VT·r=----dl 5 ' A

_

dT dt

=_

I

(2,11:/4,3)

4

e-2/5 - cos 1!

= -9.48 X 10-2•

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

= kSin29,

V _

A

au

U - R aR

dU dl dU dl

au -t._1_ dU + 6.!. R de + 'I' R sin e a$

= VU. R = _ sin2 e

I(4,1[/4,1[/2) = -

R2

'

sin2{1t/4) = -3.125 x 10-2. 16

__

-

sin29 R R2 A

-6

2sin9cos9 R'

125

CHAPTER 3

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

=

Solution:

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

E = HER where ER

IS

a function of R only.

Hence, and

E = fUR.

=

ixz - yyz:. - ixy, verify the divergence . ~blemeorem by3.3VF<>,r computing:the vector field E (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

= /
126

CHAPTER 3

127

CHAPTER 3

(b)

= rI Oe-r -

Problem 3.40 For the vector field E for the cylindrical region enclosed by r

Z3z,

verify the divergence theorem

= 2. z = 0, and z = 4_

Solution: fE.dS

= Jr=o1f=O [2 [~~(iIOe-r

-Z3z).

(-zrdrdlfl))/z=o

+ 1,=01=0 f21t r (ilOe-r-Z3z)-{rrdlj>dz»1r=2 +12 (ilOe-r -Z3z)·{zrdrdlfl»1 r=O L2~ ¢=O ==4 = 0+ 121C.14 lOe-22dlj>dz+ ¢=o =0

- 12rdrdcp 12 r=0 127t 4'=0

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

V.Edv

= 1z=oir=oilr=o ~ r2 r21t (lOe-r(1-r =81t ir=o f2 (lOe-r(I-r)-3r)dr =81t( -lOe-r

+ lOe-r(l+r)-

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

r)

3) rdlj>drdz .,

3;)I~o

ZiProblem 3.4J..,) A vector field D i,3 exists in the region between two concentric g(cylindricaI surfaces defined by r and r'-:' 2, with both cylinders extending .,>between z 0 and z 5. Verify the divergence theorem by evaluating:

=

--i(a) .......

JS "D.ds,

=

=I

CHAPTER 3

kO

1,,)Q

(b) 10/ V·Dd'll. Solution: (a)

=

15011:. Therefore, fJD' ds (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 divergence theorem for the region enclosed between the spherical shells defined by R = 1 and

R=2.

Solution: side:

The divergence theorem is given by Eq. (3.98). Evaluating the left hand ,.

,.27t

r1t

r2

j 'll V·Dd'V'= J.p=oJe=o1R=l I I I = 21t( -cos8)1~=o

/ t

::I

\

R QR (R2(3R2») R2sinedRded«jJ l--1-Ju

(3/t) 1~=1= 18011:.

129

CHAPTER 3 The right hand side evaluates to

i

CProblem 3.~ (a) (b)

For the vector field E

= Xxy - y(r + 2r). calculate

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

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 the I, and y = x, respectively. triangle are represented by the equations y = O. x

=

y

I

2

x

(b) FigureP3.43:

Contours for (a) Problem 3.43 and (b) Problem 3.44.

(a)

fEedl LI

=LI +L2 +L3,

= ({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,

130

CHAPTER 3

=

I

=

11

£2

(ixy-j(2-

x=: 1

=/

2r)r 3

y=o

(ixy-y(.x2

= ix=1 fO

[1 (2-+2l)/.r==I,z=Ody+ ly=o

(xy}lz=odx-

=0- (y+ L3

+21-))· (idx+ydy+zdz) [0 (O)lx=:IdZ Jz=O

+0= -5, 3

+21-»· (idx+ydy+zdz) (2-+ 2l) !=y, z-=ody+ lz=o fO (O)ly=xdZ

(xy)l}~ ' z=Odx- iy=1 fO

= (~)J:I - (1)/:1

+O=~.

Therefore,

'j1E.dI=O-~+~=-I. 3 3 (b)FromEq.

(3.105), VxE=-z3x,

r

1

[1 «-z3x). lx=:oiy=o

i f[ VxE.ds=

so that (zdydx»lz=o

= - 11 3xdydx= x=0 ix y=O

-

113x(x-0)dx= x=:0

-

(~)IoJ

=-1.

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 the triangle 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

lx=:o

£z

=J

(XY)/y=o,z=odx- ly=o fO (2-+21-)lz=Ody+

(ixy-

Y(2-

+ 21-)}.

[0 (0) 1)=0 dz lz=o

= 0,

(idx+ydy+zdz)

t

= 1x=2 [I (XY)/z=O,Y=2-xdx-

~=o (.x2+2l)lx=z_"z=ody+ J'

= (\".,:_ x3)/l _ 3 ) IX=2

1,,2 4-113) II -1-0 -J • J ly:i:O' ~

(<111\ 'J

= -11 3'

10 =0 (O)/y=z_xdz

131

CHAPTER 3

Therefore,

f E·dI=O-

3+3 11 2 = -3.

= -Z3x, so that

(b) From Eq. (3.105), V xE

{{ VxE.ds= 1:=0/;=0 {I r «-i3x). 11 {2 r2-x + lX=11y=0

(zdydx»lz=o

((~Z3x).(zdYdx»lz=O

= - 11 3xdydxx=OleX y=O

3xdydx 12 .;=1 l2-x y=0

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

=CProblem 3.40Verify

i

(X3)

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

Stokes's theorem for the vector

field B

= (rrcosej> ++sinej»

by evaluating:

(a) (b)

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

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

Solution: (a)

fB.dl= iLl{ B·dl+ 1L2 r B·dJ+ B ·dl

= (rrcosej>++sinej».

1B·dl= L)

')

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

L3

(rdr+if.rdej>+ zdz)

(12rcos
1B·dl, z=O

= rcosej>dr+

+ (fO J~=orSin
rsin<j>dcj>,

CHAPTER 3

132

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

JrL2 B.dl=

(12 J r=2 rCOS$dr)1

=0+ JrL3 B·dJ

z=o

+ (L: $-0 rSin$d$)/

r=2, z=o

(-2cos<j»/;=o=4,

= ( JrO r=2

rcos<j>dr)

I

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

$=rc,z=O

I

+ ( Jr: ~_1C rSin$dtp)

;:;::0

2,

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

= Vx{rrcos$+$sin$)

= r(.!.~or a$ ~(Sin<j») dZ +~ (~(rcos<j» dZ +z~ (;.cr{Sin<j»)

= i'O+~O+z~(sin+

II

VxB ·ds = l;oLo

(zsin(1

- ad$(rCOS$») (rsin))

Solution:

= zsin (I +

~) ,

+ ~) ) . (zrdrd
= i:o/~o sin <j>(r+l}drd<j> = Problem 3.46

~o)

- ar

((-cos<j>(!,.z+r))I~=o)

1:=0

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

= 8.

CHAPTER 3

134

Problem 3.47 Verify Stokes's Theorem for the vector field A evaluating it on the hemisphere of unit radius.

= RcosS+.sin8

by

Solution: A

= cose.

Hence, AR

As

= R cos e +,sin e = RAR + 9Ae + ~A~. = 0,

VxA=R-~ RsinS 1

A~

= sine.

-(A~sme)) -6--(RA$)-cp-ae (a.ae A R 1 aR a ~R1 aAR a . 2 ~l a . Al a

I =R---(sm Rsine ae A

e)-e--(Rsme)-q.--(cos9) R aR R as

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

A

For the hemispherical surface,

ds

= ftR2 sin e de d<j>.

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

1. A .dl JC

= 1~=o {2ft (Rcose+4tsine)

'~Rd<j>le=7t/2 R=I

= Rsine

(2ft 10

d<j>/e=7t/2 R=l

= 21t.

c::¥roblem 3.45:> Determine if each of the following vector fields is solenoidal, conservative, or both: (a) A = x2.ty- Yr, (b) B (c) C

= x.xz - " + z2z. = f(sin<j»/? +~(cos<j»/?,

(d) D=R/R, (e)

E=r(3-1~J+Zz,

(g)

G =x(x2 +z2) +y(r+x2)+zC1+z2).

(()F= (iy-yx)/(x2+r),

(h) H = R(Re-R).

135

CHAPTER 3 Solution: (a) V·A

VxA

= v·(lliy-Yl) = ~2xyax = Vx(i2xy-yy2)

~1 = 2y-2y = 0, dy

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

ax ~(-r)-~(2xy») dy

= iO+YO-z(2x). The field A is solenoidal but not conservative.

(b)

V·B= V.(~-Yl+z2z) VxB

= Vx (xr-yy2+

= ~r-~I+~2z=2x-2Y+2, ax dy dz Z2z)

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

=xo+yo+zo. The field B is conservative but not solenoidal.

(c)

The field C is neither solenoidal nor conservative.

cry ~(xZ»)

CHAPTER

136 (d)

\7·D=\7. -

R (' R ')

= --

R~ -

R2(jR

1 d (

., (] R ))

+---(Osme)-+---0=Rsineae . Rsinea<jJ a.

1

1

R2'

d

1

VXD~VX(~)

= R-~ R sin 1 e +R Al(d

-0

-COsine) a (aae· a)

aR (R(O))

-

ae d(l))R

+8-- a (1)-R - -(R(O)) R sin1e a<jJ dR A1( a = fO+80+$O.

The field D is conservative but not solenoidal. (e)

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

A

A

)

3

CHAPTER

137

3

dx x•.+ yVXF=:X(O-O)+Y(O-O)+Z[~(~)-~("Y

~

=z

1

.,-

( x2 + y

2.\-

x~+ y~.,)]

1 y.,---.,+., ')) xl + (x~2" +

+ "y2)~

(x2

dy

y-

y2)~

~ 2CT -~)

=z

(2 X +y")2:f

O.

Hence, F is neither solenoidal nor conservative. (g) G

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

V· G= ~(xoX

<1.,.,

+ Z-)

d2

2

+ ~(y oy. +x-) + ~(y oZ

= 2x+2y+2z:f

)

0,

V X G = x~(aoy (y~.,+ z-) ., - dZ<1.2(y

+z (;x (1+~) = x2y+y2z+z2x#

+Z

+ x-).,)

+ y~ (ddZ (x-" + r)., - ax +r d (y"2))

;y (~+C)) O.

Hence, G is neither solenoidal nor conservative.

(h)

H = R(Re-R), V· H V

x

H

= ~2 a~ (R3e-R) = ~2 (3R2e-R

-

R3e-R)

= e-R(3 -

= 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.

R)

# 0,

.