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Unformatted text preview: 14.1 Vector pitterentiation and Differential Operators 913 Mﬁtﬁshm HISTORICAL NOTE I The Jacobian matrix is named :n honor of the German mathematiCian Car'
GuaavJJacobMiBOtnBSU.whonmdeanumberohmponwﬂconnmuhons
to nineteenth—century mathematics. He was one of the first to give a systematic
theory of determinants and dealt with detJ,. which is called the Jacob/an
determinant, rather than With the matrix J, per se. He also worked with
mﬂerennalequanOns. f
Pierre~Simon Laplace did extenswe work in physrcs, astronomy, and
chemistry. as well as mathematics. He used the equation Vli = 0 in his work
. on the gravitational attraction between planets in the solar system. This
work helped him explain the regularity and stability of planetary motion. Exercises 14.1 In Exercises 1—4. ﬁnd limxﬂ F(.t) for the given function. Is In Exercises 13—16. compute the curl of the given ﬁeld F at
the function continuous at the given point x = (1? Why or why x and at a.
not?
l3. Fix): 3x3i + xyj + :k121 = (—3, 4. 2)
x4 — i“ , . .1 7 . 3 .
I. ﬁx. )1.) = , 7 ‘ ‘ i + (x + .VJlJ. + Q‘si“; a = (0‘0, 14. The I‘ and a of EXLI’CISC 9
X‘ + _l" W . . ,. . . .
15. Fix): 9‘ stnycoszi + e costi'coszj — e* smysm : k;
. x3+l ._,1 7';
2. Fix. y) : tun nxyi <— e“'3"j + .2 7] k; a = i2. 3) J ‘ ~11..th 6’ ‘ 7 _
3’ + 10. Fix) = iyej"z + Znyw‘”, xe‘“, Eye”? 3 = (I. 2. 3)
x3 — \‘l .
3. 17x, v = ~ 'A i + it:e""' + ixl * i'zz‘ + r“: k: . , .
( ' ) x i y ’ ‘I ‘ ( ) lii Lxermses 17—24, Show that the given functions are harmonic.
a = i0, 0, 2) l7. fix, y) = In \;3+ 3'2: ix, y) ¢ ii), 0) 4. Fix, y, :) of Exercise 3; n = i1. — l. 0) 7 j
18. fix. y) z e“ cosy + 3x — 3)" ln Exercises 5—8. ﬁnd JF(a) for the given f and :i. '9. fix. )‘I = sin x sinh y 20 fix. H = COSh 2X COS 3y
5. Fix. y. :) = (x2 + 6‘" ‘. zili + isinxy i )1erle 2" fix. ‘y‘ :l = x1 “L y: — 2::
u = (I, n. — l) 22. fix. ,i‘. :) : 3x1) 2): — 5:3
6. Pix. y, 2 = ln(.\"’ + y: + ::“")i + Semj; a = (l. 0. l) 23. fix, y. z) = ”Viv: gt y: + :7; (x. y. :l s5 (0. 0.0)
x cos 3.): — y: sin 7.x 24. fit, y. :) = sin .\' sinhy + cos xcosh :
7 Fl~Y,,1'.L')= Vii: + 4)‘2 + (‘2 ; :i = (l, 1,0) 25. Laplaciiin in Polar Coordinates. Suppose thzit : = fix. y). . TC. . _ 2‘ .2 _ ,
tan ’0 ' 4“ and we introduce polar coordinates x = roost) and A") — : _i' = rsin 0. Then : = firms 0. rsin 0) = gir. 0). Show that 8. Fi.\‘.y. :) = x2 — :3 :11 = i1. — l. li‘Z) V’ [39 1 (lg 15g ‘
‘ ‘*‘*' ’5=:» 773*“:
tomxi‘ (r' T r’ (0' r tr _ _ ' (Hint: See Example 4, Section 12.6.)
In Exermses 9—12. compute the divergence of the given vector _ . ‘
ﬁeld F at x and the given a. 26. Laplucmn in Cylindrical Coordinates. Suppose that
W = fix, y. :J. iiiid we introduce cylindrtcal coordinates
9. Fix) = xi + (x: + 3)j + (y: + x)k: n = (2. —2. 3) v = roost), y : I‘Sil’l 0. and : = :. Then w = fir cos 0. rsin 0. r) = gir. 0. :l. Show that , 10. Fix) = ix,2 + x1" + x31 : ,\‘f.x,.\'3.\'3.\',., XIX: — xij,
x,.\'4 — xle); a = (l. — l, 2. 3) ‘ If?! 520 153,0 5.4 7 7 V3ftX.y.:)=r r'+'1+' .eﬁrq,»
ll. l“(x)=xyl:2i+:‘sinyj+x‘e’k;a=(l.x. —'_’) rr'r “' r‘ (’0' ‘3'
, x + : r _ x _ _ r (Hint: Sec Example 4, Section 12.6.)
12F‘(Xi=3—7‘§i+j" 1 a.i+—u' a} 1k; . . v. . . ,
x + y' :‘ X + i" + :“ ." + y' + :‘ 27. Lupluciun in Spherical Coordinates. Suppose that a : (1,0. — 2) W = fix. ‘i'. :l. and we introduce spherical coordinates 914 Chapter 14 Vector Calculus 28. x = psingbcos 9. y = psin (psinB. and : = pcostp. Thcn
w = f(p sin (1: cos 0, psin (,bsin t), p cos 4)) = g(p. (p, 0).
Show that N l
(b) Use (a) to ﬁnd curl (—— x) without further calculation. IXI In Exercises 29—33. prove the indicated result. 29.
31.
33. 34. Theorem l.3(b)
Theorem .5la)
Theorem l.9(b) 30. Theorem 1.3tc)
32. Theorem 1.9la) Prove that if all second partial derivatives of the coor
dinate functions fhfz, and f3 of F are continuous, then curllcurl F) = thiv F) — Vzli‘.
where VZF = (szlﬁ + (Vlfzh' + (Vlfl) k. . Prove Theorem 1.9lc). Use Exercise 34 to compute curl(cur I") if
le,y, 2) = (x: + ylli + (xy + :3)j + lxz + y:)lt. . Verify the formula for divlF x G) in Exercise 35. if F(x, y, z) = xii + ylj + :3k and
Glx,» 2) = yzi + :“‘j + xlk. 38. 39. 40. The trace of a 3by3 matrix it” “t: “is
A: an an (123) , 53g 2 5g 1 (=19 c054) 5g 1 Pg “3% a3: "33.
VL‘v:—‘ﬁi'l'——V +i'l 7 + 7 ~7—TVVV+_T" 177 “71" ' ‘ ‘ '
8p" p 5p p“ (tidy p“ sm 4) (345 p'sin' ¢ (34,» IS defined to be all + an T1133, the sum of the mam
l diagonal entries. If F: R3 —’ R3, then what matrix discussed
(a) Show that V(]x) = — x if x = xi + yj + z k ;é 0. in this section has trace div F? Refer to Exercise 38. Iff(x. y) has continuous second par
tial derivatives, then express V'Vf as the trace of a matrix
we have met. A ﬂuid with velocity ﬁeld F is called incompressible if
div F(.\) = O. A ﬂuid flow is called solenoidal if l"(x) = curl Gtxl for some G. What relationship can you
conclude exists between incompressible and solenoidal
llows? (Assume appropriate differentiability hypotheses.) ln Exercises 41—44, sketch the given vector ﬁeld as in Figure 1.1.
41. F(x,y) = —é_\'i + lxj
l
42 F., ‘ = ":7 .
(r J) (YAFINLNTU)
—l
43. FLY, ). I) = 7 14.; (xi + yj + :1“
\,/.—\'Z + y' :
44. l<‘(x. y. :l = iyi + .rj + k 14.2 Line Integrals The double and triple integrals of scalar functions f: R" —> R, where n = 2 or
3. were developed in the last chapter as analogues of the ordinary deﬁnite intc‘
gral [2 f (x) dx of a function f over an interval [0, b]. They are deﬁned over
regions D having the same dimension—n—as the domain of f: R" —» R. This
corresponds to the fact that a closed interval [(1, b] on the real line has the
same dimension—l— as the domain of the function f in [g ftx)dx. In this section we deﬁne a different kind of generalization of I: f(x)dx, this
time for vector functions F: R" —> R” over a one—dimensional subset ofR". namely,
a parametrized curve C in R”, for n = 2 or 3. Such a curve is a generalization
of a onedimensional closed interval [(4. b] in R‘. We deﬁne the new integral
in such a way that when n = l and C = [a. b]. we get the ordinary deﬁnite
integral [2 f(x)dx. Before constructing this new integral, we brieﬂy review the main facts about
parametrized curves that are needed for its construction. (You may also wish
to review Sections 8.5 and 11.1.) A parametrized curve C in R2 or R3 is a
continuous vector function g: R a R” that is deﬁned on some closed interval
[(1, b] of the real line. We restrict attention to curves that are piecewise smooth.
This means that [a, b] can be partitioned into a ﬁnite number of subintervals
[ab b] on each of which g is smooth. that is, g’ exists and is continuous on Exercises 14.2
In Exercises 1—14. evaluate the given line integral. 1. ic F ‘ dx, where F(x. y) = xzyi + (x3 — y)j. and C is the
curve .\'(I) = (t, l — t). 0 s t S l. 2. ic F ~ Tds. where F(x, y) = (x + 2yli + (x: — y1)j, and C
is the triangular path from (O. 0) to (1,0) to (1. It to (0, OJ. 3. 1" F ' dx, where F(x, y) : vxz + y: i + l — xzj, and C is the upper half of the unit circle .\'3 + y2 = l. traversed
from (1,0) to (— LO). 4. ic F  dx, where F(x, y) = J1 — yli :~ Vii — xzj. and C is
the curve of Exercise 3. 3. (C xydx + (y: + lldi'. where C is the curve y: = x from
(0, 0)IO(1.1). 6. L(xl — y‘l)dx + 2nd}; where C is the curve y = x2 front
(I, —l) to (2, 4).
 v x . .
7. g: ' . dx + , , 11y. where C is the Circle
‘ .\" + y' X‘ + y'
x3 + y: = 9 traversed counterclockwise from (3, 0). (This
exercise is referred to in Section 4.) . + = x — ~ . .
8. f j ), dx — 7, 'i, dv. where C is the path in
‘ .\" + y" x‘ +1" ' Exercise 7. 9. (C F ~ dx, where l<'(x. y. z) ; yi + :j + x k, and C is the
line segment joining (O. 0, 0) to (2. 4, — 1). 10. ic F ' Tds, where F(x. y. :) = xzi + ylj + :I k. and C is
the curve x(t) = (t. {1. t3). 0 s I s 1. 11. If x dx — ydy + :d; where C is the helix
x(t) = (cos i, sin i, Uri), 0 s t s 27:. 12. if F  tlx where Fixia x2, x3, 35.x) = (x1 _ x;. x1 _ x3, 324 " x], xi‘zxsH), and C is the curve xltl = (I, :1. r", :4), 0 s z s 1. l3. [C(x: + y3)dx + (y: — ledy + (.\'y  :3)t1:. where C is
the line segment from (1, 2. 3) to (4, 3. 5) followed by the
segment from (1. 3. 5) to (l, l, I). 14. L» ydx + : dy + x d:. where C is the curve of intersection
'of the plane x + y = 2 and the sphere
(x — l)2 + (y — 1)] + :3 = 2, traversed in the
counterclockwise sense viewed from the point (2. 2, 0). In Exercises 15—20. verify Theorem 2.4 by calculating
if F  dx for the vector function F and curve C of the indicated
exercise. 15. Exercise 1 16. Exercise 3 17. Exercise 5 18. Exercise 7 19. Exercise 9 20. Exercise 11 21. 24. 26. 27. 28. 30. ”:2 Eve 'ntsére‘s 923 Find the work done by a force 1"(x, y) = (2 — y)i + xj
in moving a particle along one arch of the cycloid given
by .\‘(r) = (r —sint, 1 — costl.0 31$ 2:. . Find the work done by a force 1‘(x. y) = (l — y)i + xj in moving a particle along one arch of the cycloid
x(t) = 2(t # sin I, I — cost). 0 s t s 2n. . The repelling force between a charged particle P at the origin and an oppositely charged particle Q at (x, y) is I F. =.——w.— .'+ ‘.
(w) (Xz+yz),,2[“ .vi] Find the work done by F as it moves Q along the line
segment from (1,0) to (—1, 2). Refer to Exercise 23. Suppose that F moves Q along the
polygonal path from (1.0) to (l. l) to (— l. 21. What is
the work done? The line integral ofa scalar function f: R" —> R over
a curve C is deﬁned by (Clixids = ﬂ J‘ixiniix'tnl dr. (Recall that ds = x'(z)dt.) lff(x(t)) gives the density of a
wire in the shape of C at the point x(t). then the mass of
the wire is deﬁned to be (cﬂxids. Formulate a definition
of the center of mass of the wire. Formulate a deﬁnition of the moment of inertia of a wire
in the shape of x = x0) relative to an axis [at distance
dlx. y. :1 from (x, y, :) on C. (Assume that the wire has
density fix, 3'. 3).) Use Exercise 25 to compute the mass ofa helical wire
x(1) = (cos I, sini, t), 0 st 5 27:, if its density is given by 1 (5(_\‘, y, z) = X2 + y" + :'.
Find the center of mass of the wire in Exercise 27. Find the moment of inertia of the wire in Exercise 27
about the :axis. A force 1“ acts on a particle to move it from x(tol to still
along the path x = .\'(1). Show that the work done is the
change in kinetic energy. i.e.. w’ : leth .)2 — imivirolil' (This is one form of the law of conservation of energy.)
[Hintx Compute (d/dt)(v(t)3) and use F = mx“(t).
Newton's second law] . If C is a smooth. but not necessarily simple. closed curve that does not pass through the origin, then the winding
number of C is deﬁned as (l,/21r)j'(~ F  dx. where
i r x Fix. .v) = , 7,7» i + . .
x + i" x‘ + _r' j. It is shown in advanced calculus that the winding
number is always an integer. which gives the number of
times C winds around the origin. If it is positive. then 924 Chapter 14 Vector Calculus C encircles the origin in a counterclockwise sense. It. it is 32. If F(.\'. y) = —y' mi + x’ “lj then compute [CF  dx
negative, then C encircles the origin in a clockwise sense. over the hypocycloid of four cusps x3” + y" 3 z 1
Compute the winding numbers of parametrized in the counterclockwise sense (see
Cii XI!) = (3 cos I. 23m“ 0 :1 t S 67:. Exercrse 41. Section 8.5).
and
('3: x0) = (cost, —sint), 0 s I 5 2r
14.3 Green's Theorem
This section takes its name from a result that extends the fundamental theorem
of calculus to twodimensional vector tields. To describe the extension precisely,
it is helpful to state Theorem 4.2 of Chapter 4 as follows:
lff is a smooth function (that is. f’ is continuous) on [(2, b], then
b , .
(ll Lftxth=ftb)J(a).
Qualitatively. this says that the deﬁnite integral of the derivative of f over FIGURE 3.1
." 3_.1
DEFINITION [11, b] is completely determined by the mines of f on the boundary of [0, b], that
is, on the twopoint set {(1, b}. The various forms of Fubini’s theorem in the
last chapter (starting with Theorem 1.9) constitute an extension of this result
to scalar functions of two or three variables. In particular, partially antidiffer
entiated functions, such as jf(x. y)d_v, are evaluated just on the boundary of
the region D ofintegration. such as the curves y = gl(x) and y = gz(x) in Figure
3.1. Green’s theorem is the first of several analogues of(l) to be presented in this
Chapter. It says roughly that the double integral of a certain kind of derivative
of F over a standard type of region D is given by the line integral of F over
the boundary of D. This lets us evaluate some complicated double integrals as
line integrals. and sometimes makes it possible to compute a complicated line
integral by working out a lessinvolved double integral. To state the theorem
precisely, we must specify the “standard type” of region and the “certain kind"
of derivative mentioned above. We begin with the type of region. A plane region I) is simple if it is the part of R3 enclosed by a piecewise smooth
simple closed curve that intersects any vertical or horizontal line either in at
most two points .4 and B or in one line segment AB. As the name suggests, simple regions are geometrically uncomplicated. Fig
ure 3.2 shows three examples of simple regions. The region in Figure 3.2(a) is
bounded by a smooth curve, the circle x3 + yl = 1. The regions in Figure 3.2(b)
and (c) are bounded by piecewise smooth curves C. Points where x’ fails to be
continuous are labeled. In all three cases, vertical and horizontal lines intersect
C in 0, l, or 2 points. Figure 3.3 shows another type of simple region, in this
case a square. Notice that some lines parallel to a coordinate axis (such as
x = a and y = c) intersect C in at most two points, while others (including
x = — I and y = 1) meet C in entire line segments. Figure 3.4 shows a nonsimple region. Although some lines (like x = a and
y = c) intersect C in at most two points. there are others (like y = d) that
intersect C in four points. We try to picture a general simple region in Figure 3.5. The regions D and
R in Figure 3.6 fail to be simple. because the lines shown in Figure 3.7 intersect 14.3 Green's Theorem 933
EF by
.\'(I) = (x, y) = (I), t). a S t S [3. Then dx/dt = 0, so 11x 2 0. We therefore get I” Ptx, _v)d.\' = 0. Similarly, I,” Ptxt y)dx = 0. Thus from Deﬁnition 2.3, ('l 1) JC P(x, _y)dx = In: P(.\‘, y)tlx + I” Ptx, y) dx. Since a vertical line crosses each of AE and BF exactly once. these arcs arc
the graphs of functions y = gl(x) and y = gltx) for x e [(1, b]. So they can be
parametrized as curves C. for i = 1, 2, by x = I, y = gitt). It is natural then to
use x as a parameter instead of I. Since C2 is traced out as x goes from a to b,
C; joins B to F. Since C traverses the boundary of D in the counterclockwise sense, C proceeds from F to 8. Thus we have from (1 1) and Theorem 2.4 J; Ptx. y) dx = Thus (9) is proved. f6] P(x. y) dx + fc; Ptx, y)d.\'
Ll P(x, y) dx — .i‘cz P(x, y) dx
L" Plx. gitxndx ~ f ax, 91mm — I 1" [Pm yam — Ptx. 91mm»: b y=g2(.x) . _ b 92“} (3})(1‘. Y)
_ f; P(x.y)]y=gl(xld.\ _ . [W] a), .1) dx
5P
_ff 4:7 d4. by Fubini's theorem
1) (y The approach to (10) is quite similar and is left as Exer~ cise 37. Addition of (9) and (IO) gives (2). [Qtan Exercises 14.3 In Exercises l—li. use Green’s theorem to evaluate the given
integrals. 1. *5c 21de + 3.\‘ (1y, where C is the polygonal path from
(—1, 2) to (3, 2) to (3.5) to (—l. 5) to t—l,2) h)  {tr (312 — ISMV + (xi + yl)dy. where C is the unit circle
X: + y2 = l traversed counterclockwise (a) . {3( F . (Ix, where I“ is as in Exercise 2, but C is the circle
x1 + y2 = 9 traversed clockwise 4. {T —ye‘dx + xe" dy, where C is the polygonal path from
(I, 1) to tl. 1) tot—l. —l) to (t, —1) toll. 1) U! . {if yzeﬁlx + 2yt"dy. where C is the path of Exercise 1 6. If (8" + y — 2.\‘l)d.\' + (7x — sin y) dy. where C is the
triangular path from (O, 0) to (0, l) to (l. 0) to (0.0) 7. [C(xz + y)d.\’ + (2x — y)r1y, where C = Cl u C3. Here Cl
is the polygonal path of Exercise 1, and C2 is the unit
circle parametrized in the clockwise sense. 8. j} (x3 + sin x + y)dx + (23‘ — sinycosyhly, where C is as
in Exercise 7 9. §(sin y — x3y)d.\' + (x cos y + xy): rly. where C is the curve
of Exercise 2 10. {6 tr — xy)dx + (y3 + l)tl_v, where C is the polygonal
path from (I, 0) to (2.0) to (2. l) to (l, l) to (1.0) —v x . .
11. ¢ , ‘ , dx + —, 1(1y, where C 15 the ellipse
f .1" + y' x‘ + y‘ 4x: + y2 = 16, traversed counterclockwise . _ V
11g). , ' ,dxl— , ,
L x‘ + y' x‘ + y' 5x2 + y: = 25, traversed clockwise (1y. where C is the ellipse x t‘ ..
11f , ,dx+ ,' —;cly,where( =C1uC:.Here
C x’ + y' .\" + y’ Cl is the parabolic path from (— 1,0) to (,2. 3) along
y = x3 — I, and Cl is the line segment from (2. 3) to
(— 1,0). 934 Chapter 14 Vector Calculus I4. x r
I , , dx 4 . , ' , dy. where C = Cl u C1. Here
C X‘ + y' X" + y‘ ‘ Cl is the parabolic path from (—2, 0) to (3, 5) along
j.‘ = x1 — 4, and ('2 is the line segment [3, 5) to ( — 2, 0. In Exercises 15—20. ﬁnd the area of the given region. IS. l6. 17.
18.
I9.
20. 2]. l0
DJ 24. 26. 27. The region D enclosed by the ellipse 4x3 + ",2 = lb The region D enclosed by the hypocycloid x13 + y“ = I.
(See Exercise 32. Section 2.) The region D enclosed by the hypocycloid x1 3 + y2 3 = 4
The triangle with vertices (l, 2). (3. 2). and (3. 5) The annular region between x: + y:
(Hint: Think before resorting to (8)!) 2 l6 and x3 + y: = 9. The annular region between 3:3 + y: = 4 and x3 + y: = 16 Show that if th. y) = y‘lexi + 2ye*j is the velocity field
ofa ﬂuid ﬂow. then there is no tendency for a vortex
around the origin. . Repeat Exercise 2! for F(x. y) = (x3 + y3 —1 y sin xli + (3xyI — cosxlj. [f D is a region satisfying the hypotheses of Theorem 3.3.
then ﬁnd a formula for f and y". the coordinates of the
centroid of D. as a line integral. Generalize Example 5. namely. let (‘P LQ
F(x) = Ptxti + thlj. where e; — — for x 1 0.
t‘)‘ t X
Suppose that Cl and C2 are smooth simple closed curves
with the same orientation, which enclose regions D; such that 0 6 D' 2 D2. Then show that (tic! F ‘ dx =g5t“ I" ~ tfx. . Generalize Exercise 11. namely. prove that if y . N ____.V._ I .3. Fix}: . ,
x’+y‘ «.l, x2 + y"
then for any smooth simple closed curve C ifO is enclosed by C
ifﬂ is not enclosed by (I ‘3 g) F ¢1\—_{" lfC is asmooth curve parametrized by x = xi!) = mt). y(r))
for l in [(1. b] then show that 1 di dx
N‘ = ., e e
l!) Mill (d1 d1 1) is a unit normal vector to C at every point on the curve Use Exercise 26 to show that iii ’ de + de =(ﬁ [7  Nds. If F: P(x y)i + Q(x y)j is the velocity ﬁeld ofa ﬂuid
ﬂow in R' this integralis called the ﬂux across C
Explain how it measures the amount of ﬂuid flowing
across C. (This is referred to in Sections 6 and 8.) 28. 29. Use Exercise 27 and Theorem 3.3 to show that, under
suitable hypotheses. if div 1? (1.4 = {D (In Section 8 we will see how Green‘s theorem in this
form can be extended to threedimensional vector ﬁelds
F: R3 —v R3.) 1“  N (Is. Prove Green' s fit st identity an analogue of integration by parts If f: R“ —> R is diﬂerentiable on a simple region
Dand all second partial derivatives of g: R’ —» R are continuous on D, then use Exercise 27 to show ffrixiv’igod4=f jmivg Ny—ds £1fo ngA =j:.,,f(xiDth)dx—Hwygda,
D where D\(g) is the directional derivative of g in the
direction of N . ln Exercise 29 assume that all second partial derivatives of f and g are continuous on D. Then prove Green '5
second identity. fﬂfhlengl—gixwzflxl] dA=fw f(\')1Vg ' Nids
D — f,” yth . Nids
{LDUDtxlyl—gDstfllds. What can you conclude if f and g are harmonic
(Deﬁnition 1.7)? . Suppose that a ﬂowing ﬂuid in the plane has density 5(.\', y, r) at the point (x. y) at time I. Show that ‘ic 6h: y. Ill‘ ~ Nds is the rate of loss of mass per unit
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