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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 £>(.rR3sinedad¢)=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 ZE—f zds‘
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 Aa=ﬂw
1
jAdl=joﬁdr=% Ergmﬂm‘3ﬁ)m(1.mé,m’2
r=l, 9=1n‘3;dr=0,d9=0  .1: . {3' . Ad‘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 IAdl=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 jAdl=j(:ﬂdr=—% l ..¢CAdl= + 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 Gigsin 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 _ xcomponent ofVXA=F(4x+yy+2z)—5;(ﬁx3yZ)=)’+1
a a ycomponent of‘V XA=a—(x+2y'+az)—a;(4x+yy+22)=a—4
z . I a 6 zcomponent 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 ‘Fis conservative but nonsolenoidal (717 #0), then
the scalar potential 95 would not satisfy V245 = 0. To be precise, if
VF = 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 twodimensi 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
(VF E 0), then the associated potential «15 is such that VF=VVr15=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—yyzzzxyﬂz 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—yyzzzxy)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= (ixzyyzz—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 VEdv: / V(ixz—iyzzWWZCIJ’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) fVVde. 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, ffDds = 15017..
(1)) From the back cover, VD = (1/r)(3/3r)(r13) = 412. Therefore, — 5 27‘: 2 7 I i  4 2 21:
ff VDda/_/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'
[WVDaw— ¢=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 xKd—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 yyl ' ' '
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)dxdzf
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)2a2] (yl+b) TlleiQZ—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
.  51 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) ' jéEdl=L1+L2+L3, L1=[($01—i(x2+2y2))(idx+§dy+2dz) a, ’3  0 0
= [i=0 (w)ly=o,z=o dx — [F0 (x2 + 2%) 11:0 dy+ [2:0 (0)13,==o dz = 0, La: few—i<x2+2y2>)(2dx+§czy+2dz)
' 1 1 0
= ./,.—=2V(xy)‘z=dJ=2—xdx_ F0 + zyz) L:2—y,z=0 dy+ 1:0 (cubedx dz _.~._..._....... .~ L3=fCiVy?(xz+2y2))'(idxif’dyiﬁdz) O D ' 0 = 51 (xy)ly=x,z=oir =1 (x2+2y2)1,=,,z=ody+ Focomﬂdz
x3 0 o 2 = (E) r1“ We”? Therefore,
' 11 2
7€E.41_o—?+§_—3.
(b) r . VxE=+23x,sothat ff VxEds = y; ((—23x) . (idydx?)z=o
+ /; fy::((—23x)'(idydx))lz=o 1 x 2 2—: .
= —/ / chiydxef f 3xdydx
x=0 y=0 z=l y=0 , 2 =/103x(x—0)dx/ 3x(<2—x)—0)dx x: =1 = —(x3)Ié—.(329—9)E=1=—3 ...A
N Q9 ® (frogsd) +<§Sin¢) (a) 7:»; B  d! over the semicircular contour shown >  I
‘ 3’
(13) AC? x B) . :15 over the surface of the semicircle. ? ‘ .h_.... @ Verify Stokes’s theorem for the vector ﬁeld B =
by evaluating: = rcos¢dr + rsin¢d¢, o 
F0, z=o+ (Azorsmqadqgl z=0
z 3
= 2 Tc/
B d1: rcos¢dr)L + rsm¢d¢>
L9 1:2 2:0 ¢=0 1:2 2:0 2.0 + (fa/2rsin¢d¢> =0+(—cos¢)l§=n/2=11
jéBa'l=g+2+0~1=§. ' 2
(b) 1:1, 2:0 ‘— VXB = Vx(frcos¢+$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
VxARm(%(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
ﬁAa’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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 Summer '08
 Ferguson
 Electromagnet, Surface, Emoticon, closcd surface

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