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Unformatted text preview: 552 VECTOR CALCULUS _'_________,__—_.—.._———————~——— ' 5: . 732 1;. I Find div :2 where
(a) v(r) = 3x2yi + zj + xlk
(b) v{r) = (3x +y)i + (22 + 10} + (z — 2y)k  32} HF = (nyz + 251' + (3x222 —y223)j + (y:3 # x23)k,
calculate div f at the point (—1, 2, 3). everywhere zero, the ﬂow entering any element of the space is exactly balanced by the
outﬂow. This implies that the lines ofﬂow of the ﬁeld F(r) Where divF : 0 must either form closed curves or ﬁnish at boundaries or extend to inﬁnity. Vectors satisfying this
condition are sometimes termed solenoidal. F = (2x2);2 + 22H + (Bury3 — xzz)j + (3.30222 + xy)k
is soienoidal.
:36 ; (Spherical polar coordinates) Using the notation I introduced in Exercise 30, show, working from ﬁrst
principles, that  Vudivv=ii(rzv)+ 1 3(t) sine)
33 Find V(a  r), (a ' V)r and a(V  r), where a is a _ r2 6,, ’ r Sin 9 39 3
constant vector and, as usual, r is the position vector 1 a
r = (16,}: Z} _
+ r sin 68¢(v‘9) The vector v is deﬁned by v : rr‘l, where where v _ y u + v u + U u
r=(x,y,z)andr=r.Showthat _ ’ ’ 9 9 '9 °'
WV _ v) E grad div” 3 _23 r V force ﬁeld F, deﬁned by the inverse square law,
r is given by . F = rer
35 Find the value of the constant 1 such that the vector
ﬁeld deﬁned by Show that V F = 0.
7.3.3 Curl of a vector field
It is clear from observations (for example, by watching the movements of marked corks
_v on water) that many ﬂuid ﬂows involve rotational motion of the ﬂuid particles. Com
D 1 C 7, plete determination of this motion requires knowledge of the axis of rotation, the rate
03 v3 Z'AZ of rotation and its sense (clockwise or anticlockwise). The measure of rotation is thus a
(x’y’ 2) it ' vector quantity, which we shall ﬁnd by calculating its 2:, y and 2 components separately.
A,“ _ _2 gay, _ _,,B Consider the vector ﬁeld v(r). To ﬁnd the ﬂow around an axis in the x direction at the Figure 7.10 Flow
around a rectangle. point r, we take an elementary rectangle surrounding r perpendicular to the x direction, as shown in Figure 7.10.
To measure the circulation around the point r about an axis parallel to the x direction, we
calculate the ﬂow around the elementary rectangle ABCD and divide by its area, giving [222(x, y*, z — Az)(2Ay) + 123(x, y + Ay, 2*)(2Az)
— v2(x, )7, z + Az)(2Ay) — v3(x, y — Ay, §)(2Az)1/(4AyAz) where y*, )7 E (y — Ay, y + Ay), 2*, E E(z  AZ, 2 + A2) and v I vli + vgj + t/Jk.
Rearranging, we obtain —[t}2(x, 3?, z + A2) — v2(x, 32*, z — Az)]/(2Az) + [va(x, y + Ay, 2*)  123(x, y  Ay, 5)l/(2Ay) 13 DERIVATIUES OF A 1rtECTIJR PUINT FUNCTION 555 Henee when any rigid body is in motion, the our! of its linear velocity at any point is
twice its angular velocity in magnitude and has the same direction. Applying this result to the motion of a ﬂuid, we can see by regarding particles of the
ﬂuid as miniature bodies that when the curl of the velocity is zero there is no rotation
of the particle, and the motion is said to be curlfree or irrotational. When the eurl is
nonzero, the motion is rotational. .. .. .. .. .. __ l   ‘ ‘ “39 .361. Isa} Find u = eurl r; when U = (3x33, —ya, .1: + 23). 1ft: 2—yi +xj+xysk is the velocity veetor eta ﬂuid! ﬁnd the local value of the angular velocity at the A vector ﬁeld is deﬁned by r: 2 (ye, .rsg, xy). Show 9'3“th (1* 3r 2) that our] v : 0. __ __ _ If the veloeity of a ﬂuid at the point (it, y? a) is given by Show that ifu : (2x +ya,~ 2y + at, 22 + xy) then H = (in? + byﬁ + (Cl + dylf l : ' f] h th = . ' ' I
C”? i" 0: and ndfm SUE at U grﬂdf ﬁnd the conditions on the eonstants o, b, r: and d in order that
By evaluating eaeh term separately, verify the identity div L! = U, curl I; = 0 V K (jg) Eﬂ‘pr x U) + (19f) K v Verify that in this ease r: = % grit14:10:11:2 + 2hr}: m nyi)
forﬂr) = .r‘l — y and 120*) z (s, 0, —x). H (Spherieal polar eoordinates) Using the notation Find constants o, t) and e sueh that the veetor ﬁeld introdueed in Exereise 30, show that deﬁned b
y V X U = eurl s! F = (4,132 + 5133):? + (tori + Bali + (6x32 + ey)k is irrotational. With these values of (I? h and e, = 1 i i ‘3
determine a scalar ﬁmetion $(I, y, a) such that F=va 7.3.5 Further properties of the vector operator V ' So. farwe have used ways": I a. ﬂ") “cases I . F(_r) a vector ﬁeld I ; (gi— gliil (%_%lj (Elijah, .  For) a veetor ﬁeld ._._l . —.. (d) aurl E H —
_—— Also: gala
3r " C832 '1 .93 : ""s(gf3d f?” k), a 32‘”
1 33 a (ERIE 132 a 3:2 I
"—2 C i j
E E.
C 3.3:: 3y 0 0 1 33¢
I 2
(3)23: 2
[All grad d) >< k 2
(3:2 mi ?.3 DERIVATIVES OF A HECTOR POINT FUNCTION 559 —12C111‘1(k§ £9) + 3» aurl grad d: :92: k
d . .
3; , smaa Cur} grad (9 = 0 by (7.21) it c?
a? 3% j
3x352 sirtaa k is a aanstarit vaatar [(aaaaaaxi] 3x By r— —J—~—~— ('33 '32? sa that wa hava “13:1
VXE— a 3: Shaw that ifg is a ﬂirtatiaa af r : (a, y, 3) than Ida]r
gradg= Fd—ir Daduaa that if a is a vaatar ﬁald than div [(u :s: ﬁg] = (r auri u)g Far Ma, y, a) = xlyzai' and Hart y, a) : rzyr' + xyzaj — yzlk datarrnina (a) 371$ (13) grad divF (a) aurl aurl F Shaw that if a is a aarrstant vaatar and r is tha pasitian vaatar r = (x, y, 3) than div {grad [(r  r)(r  a)]} = 100'  a) 3: )_I[.33¢5 .aip]
_ _ I a "J a
a dyé‘t 5x3: Varify the identity
V21: = grad div v — aurlaurla1 far tha vaatar ﬁald v : argyﬂri + yj + 3k). Verify, by aaiaulatirrg aaah tarrrr saparataly,
tha idarrtitias div(aaa):v'auria—a*aurla aurl(u K v) = adith —.vdiva + (H‘V)H
* (_u*?)v whart at = xyj + Lift? and r: = xyr‘ + yak. If r is tha usual pasitiarr vaatar r = (x, y, z),
shaw that 560 VECTOR CALCULUS ________‘__—______.—.—.———w——— By evaluating V  (Vf), show that the 'Laplacian
in spherical polar coordinates (see Exercise 30) is
given by 13 gar l a . 8f
v2 =——[ —]+ — s 9—]
f r28!" Far rzsin989(m as
1 2
+ a .2 ﬂ:
r‘sm 98¢ (a) div grad[%) = 0 (b) curl [k x gradEJl/ + grad [k  gradv; = 0 If A is a constant vector and r is the position vector
r = (x, y, 2), show that A . r A (A . 1,.) Show that Maxwetl’s equations in free space, namely
(a) grad( .) )=—3—3 5 r
r‘ r r divH 2 0, divE = 0
AXr _2A 3 _lr9_E f_13ii
(b)curl( r3 ]—?+F(AXr)Xr VxH—C t, V><E_ Cat
are satisﬁed by
If r is the position vector r = (x, y, z), and a and b
are constant vectors, show that H = 1 our] Egg
c r
3 er=0
( ) E 2 curl curlZ
(b) (a i V». = a where the Hertzian vector Z satisﬁes
(c) V'X[(ar)b—(br)a1=2(a><b) 7 1322
V'Z = — —
(d)V[(ar)b—(br)a]=0 c :2 __________________,____.._.___—_—————— Figure 7.13 Deﬁnite y integral as an area. Topics in integration In the previous sections we saw how the idea of the differentiation of a function of a
single variable is generalized to include scalar and vector point functions. We now turn
to the inverse process of integration. The fundamental idea of an integral is that of summing all the constituent parts that make a whole. More formally, we deﬁne the
integral of a function f(x) by a all Age—>0 i=1 whereoizx0 < x1< x2 < < xH < xn=b,Ax[=xi—x,_. andxH —‘’= 2?, S x,.
Geometrically, we can interpret this integral as the area between the graph y = f (x), the
x axis and the lines 3: = a and x = b, as illustrated in Figure 7.13. 10 X1 X2 :0 *. =b
xi 564 VECTOR CALCULUS In general, W depends on the path chosen. If, however, Hr) is such that F (r)  dr is an
exact differential, say *dU, then W: fC — dU : UA — UB, which depends only on A and
B and is the same for all paths C joining A and B. Such a force is a conservative force,
and U(r) is its potential energy, with F (r) =  grad U. Forces that do not have this prop
erty are said to be dissipative or non—conservative. Similarly, if v(r) represents the velocity ﬁeld of a ﬂuid then 95C 1) 'dr is the ﬂow
around the closed curve C in unit time. This is sometimes termed the net circulation
integral of v. If sﬁcu dr 2 0 then the ﬂuid is curlfree or irrotational, and in this case v
has a potential function Mr) such that v = — grad gt). Evaluate fy d5 along the parabola y = 2pc from $61 i Find the work done in moving a particle in the force
. 2 2 7 2 7 _  z . n
A(3, 23) to 8(24, 4‘96). [Recth (37:) = 1+ [3—9 ,1 ﬁeld F  3x I + (2x. )0} + k alone (a) the curve deﬁned by )c2 : 4y, 33c3 = 82 from
x : 0 to x = 2; E l t BZxd + t 2d 1 th vauaehl y x (x y) Haong em (b) thestraightlinefrom(0,0,0)to(2,1,3). of the circle x2 + y2 : l in the ﬁrst quadrant from A“; 0) to B“), 1) Does this mean that F is a conservative force?I Give .; _ reasons for Our answer.
_; Evaluate the integral f5 Vdr, where y V: (2yz + 3x2, )22 + 4x2, 222 + 6xy), and C is the .62.
curve with parametric equations x : I], y = t3, 2 = r ' "
joining the points (0, 0, 0) and (1 , l, 1). Prove that the vector ﬁeld F = (3x2 7 y, 2}:2 i x,
2y22) is conservative, but not solenoidal. Hence
evaluate the scalar line integral fp F ~ dr along
any curve Cjoining the point (0, 0, 0) to the lfA = (2y + 3): +x21 + (yzex)k, evaluate fﬁA dr point (13 2, 3) along the following paths C: lfF =xyi —zj+x2k and C is the curve 1: = {2,}; = 2!.
2' :13 from t 2 0 to t: 1, evaluate the vector line (a) x=2t2,y=t,z:t3fromt=0tot=l; (b) the straight lines from (0, 0, 0) to (0, 0, 1), then to (0, l. 1) and then to (2, l, l};
(c) the straight line joining (0, 0, 0) to (2, 1. l). 64_ Prove that F 2 (y2 (:08): + 23)i' + (2}: sinx — 4)} + (3ch2 + z)k is a conservative force ﬁeld. Hence
ﬁnd the work done in moving an object in this ﬁeld
from (0, 1, —1) to (TE/2. fl. 2). 7.4.3 Double integrals integral chx dr. lfA = (3x+y, way—z} and B: (2, —3, 1)
evaluate the line integral 95C (A x B) x dr around
the circle in the (x, y) plane having centre at the
origin and radius 2. traversed in the positive
direction. In the introduction to Section 7.4 we deﬁned the deﬁnite integral of a function ﬁx) of one variable by the limit l f(x)dx= lim EraMa n—>><= ‘
an Ase—>0 ’"1 where a = x0 < x1 < x2 < . . . < x” = b, Ax). = x1. —xH and xH g 3?, g x, This integral
is represented by the area between the curve y =f(x) and the x axis and between x 2 a and x = b, as shown in Figure 7.13. ...
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This note was uploaded on 12/13/2010 for the course ELECTRICAL EECS 145A taught by Professor Chinc.lee during the Fall '10 term at UC Irvine.
 Fall '10
 ChinC.Lee

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