. . . . . .
.. Chapter 15 Vector Calculus.
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15.1 Vector Fields
15.2 Line Integrals
15.3 Fundamental Theorem and Independence of Path
15.3 Conservative Fields and Potential Functions
.
......
Warning: Students should come to lecture as the contents are not easy tounderstand, and many notations are involved.
. . . . . .
.. Vector Fields.
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Definition. Let D ⊂ R2. A vector field on D is a vector-valued function F thatassigns to each point (x, y) in D a two dimensional vector F(x, y), i.e.F(x, y) = P(x, y)i + Q(x, y)j = ( P(x, y), Q(x, y) ).
.
......
Definition. A vector field F on a region E ⊂ R3, assigns each point (x, y, z) in Eto a three dimensional vector F(x, y, z), i.e.F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k.
. . . . . .
.
......
Definition. If f : Rn → R is a differentiable function, then ∇f is a vector field onRn, and it is called the gradient vector field of f given by∇f (x, y, z) = ( fx(x, y, z), fy(x, y, z), fz(x, y, z) ).
.
......
Example. Prove (i) ∇(f · g) = f∇g + g∇f .(ii) ∇(af + bg) = a∇f + b∇g, where a, b ∈ R.
Proof. (i) ∇(f · g) = ( (f · g)x, (f · g)y, (f · g)z ) =
( f · gx + gfx, f · gx + gfx, f · gx + gfx) = f (gx, gy, gz) + g(fx, fy, fz) = f∇g + g∇f .We leave the proof of (ii) to reader.
. . . . . .
.
......
Definition. Let F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k be a vector field,with differentiable coordinate functions. Then the curl of F is defined by curlF(x, y, z)
= ∇× F =i j k∂
∂x∂∂y
∂∂z
P Q R= (Ry − Qz, Pz − Rx, Qx − Py).
.
......Example. Find the curl of F = (x2 − z)i + xezj + xyk.
Solution. curlF = ∇× F =
i j k∂
∂x∂∂y
∂∂z
x2 − z xez xy=
(∂∂y (xy)− ∂
∂z (xez))
i +(
∂∂x (xy)− ∂
∂z (x2 − z)
)j +
(∂∂x (xez)− ∂
∂y (x2 − z)
)k =
x(1 − ez)i − (y + 1)j + ezk........Example. Let F(x, y, z) = xzi + xyzj − y2kk. Find curl F.
Solution. curlF = ∇× F =
i j k∂∂x
∂∂y
∂∂z
xz xyz ez= −(2y + xy)i + xj + yzk.
. . . . . .
.
......
Theorem. If f (x, y, z) is a scalar function and has continuous 2nd order partialderivatives, then curl (∇f ) = 0.
Proof. curl (∇f ) =i j k∂
∂x∂∂y
∂∂z
fx fy fz
=
( (fz)y − (fy)z)i + ( (fx)z − (fz)x)j + ( (fy)x − (fx)y)k = 0..
......
Let a, b ∈ R, and f be a differentiable function, then(i) ∇× (aF + bG) = a∇× F + b∇× G;(ii) ∇× (f F) = f∇× F + (∇f )× F.
We skip the proof.
. . . . . .
.
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Definition. Let F(x, y, z) = ( P(x, y, z), Q(x, y, z), R(x, y, z) ) be a differentiablevector field, i.e. the functions P, Q and R are differentiable functions on thedomain of F. The divergence of F is defined to bedivF(x, y, z) = ∇ · F(x, y, z) = ∂P
∂x + ∂Q∂y + ∂R
∂z .
.
......
Example. Find the divergence and the curl of the vector fieldF(x, y, z) = xeyi + z sin yj + xy ln zk.
Solution. div F(x, y, z) =∂
∂x(xey) +
∂
∂y(z sin y) +
∂
∂z(xy ln z) = ey + z cos y+
xyz
.
curl F =i j k∂
∂x∂∂y
∂∂z
xey z sin y xy ln z
= ( (xy ln y)y − (z sin y)z, (xey)−(xy ln z)x, (z sin y)x − (xey)y )
= (x ln z − sin y)i − y ln zj − xeyk..
......
Example. Let F, G be a differentiable vector field defined on a domain D, anda, b be some constants. Prove that (i) ∇ · (aF + bG) = a∇ · F + b∇ · G; (ii)∇ · (f G) = f∇ · G +∇f · G.
. . . . . .
.
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Example. Recall div (P, Q, R) = ∇ · (P, Q, R) = Px + Qy + Rz,and curl (P, Q, R) = ∇× (P, Q, R) = (Ry − Qz, Pz − Rx, Qx − Py).Prove the following identities:
...1 ∇ · (f · F) = f∇F +∇f · F;
...2 ∇(fg) =
g∇f − f∇gf 2 ;
...3 ∇ · (F × G) = G · (∇× F)− F · (∇× G).
...4 div(curlF) = 0;
...5 div∇(f · g) = f div∇(g) + gdiv∇(f ) + 2∇f · ∇g;
...6 div(∇f ×∇g) = 0.
...7 ∇ · (F1 × F2) = F2 · (∇ · F1)− F1 · (∇× F2);
...8 ∇× (F1 × F2) =
(F1 · ∇)(F2) + (F2 · ∇)(F1) + F1 × (∇× F2) + F2 × (∇× F1),
in which ((P, Q, R) · ∇)(A, B, C) = P∂
∂x(A, B, C) + Q
∂
∂y(A, B, C) + R
∂
∂z(A, B, C)
= (PAx + PBx + PCx) + (QAy + QBy + QCy) + (RAz + RBz + RCz).
. . . . . .
.
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Definition. A curve C is called a piecewise continuously differentiable, if thereexists a parametrization r : [a, b] → Rn (n = 2, or 3) such that C is the image ofr([a, b]) and the the coordinate functions (x(t), y(t), z(t)) = r(t) arecontinuous, and the first order derivatives are continuous except finitely manypoints in [a, b].
Example. One can consider the the square, in which the curve is continuousbut the tangent vectors do not exist at the four vertices.Remark. One can replace the interval [a, b] by open intervals, or unboundedinterval.
. . . . . .
.
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Definition. Let r(t) (a ≤ t ≤ b) be a parametrization of a curve C in Rn, thecurve is oriented by the order of R, then the curve parameterized by thes(t) = r(b + a − t) where a ≤ t ≤ b is the curve C with reverse direction, anddenoted by −C.
Example. Let r(t) = (cos t, sin t) (0 ≤ t ≤ 2π) be a parametrization of theunit circle of radius 1 in counterclockwise direction.Then thes(t) = ( cos(2π − t), sin(2π − t) ) = ( cos(−t), sin(−t) ) = (cos t,− sin t)where 0 ≤ t ≤ 2π is a parametrization of the unit circle of radius 1 in clockwisedirection.
. . . . . .
.
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Definition. Let C be a piecewise continuously differentiable curve in a domainD, parameterized by r(t) = x(t)i + y(t)j + z(t)k for a ≤ t ≤ b, and T(t) be theunit tangent vector of C at r(t). Then the arc-length element ds isds = ∥r′(t)∥dt =
√x′(t)2 + y′(t)2 + z′(t)2 dt.
.
......
Definition. Let f be a scalar function defined in a domain D containing C, the
line integral of f with respect to arc-length of C is∫
Cf ds =
∫ b
af (r(t)) ∥r′(t)∥ dt.
.
......
Example. Evaluate the line integral∫
Cxy ds, where C is a segment from A(1, 2)
to B(9, 8).
Solution. Parameterize the segment C byr(t) = (x(t), y(t)) = (1 − t)
−→OA + t
−→OB = (1 + 8t, 2 + 6t) where 0 ≤ t ≤ 1. By
definition,∫
Cxy ds =
∫ 1
0x(t)y(t)
√x′(t)2 + y′(t)2 dt
=∫ 1
0(1 + 8t)(2 + 6t)
√82 + 62 dt = 10
∫ 1
0(2 + 22t + 48t2) dt = 290.
. . . . . .
.. Line Integrals of Scalar Functions along Coordinates.
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Definition The line integral of f with respect to the coordinate axes are defined
to be∫
Cf dx =
∫ b
af (r(t)) x′(t) dt,
∫C
f dy =∫ b
af (r(t)) y′(t) dt and∫
Cf dz =
∫ b
af (r(t)) z′(t) dt respectively.
Remarks....1
∫−C
f dx = −∫
Cf dx,
∫−C
f dy = −∫
Cf dy,∫
−Cf dz = −
∫C
f dz, where −C denotes the curve C with reversed
orientation....2
∫−C
f ds =∫
Cf ds.
. . . . . .
.
......
Example. Evaluate the line integral∫
Cydx + xdy, where C is a segment from
B(9, 8) to A(1, 2).
Solution. Parameterize the segment C byr(t) = (x(t), y(t)) = (1 − t)
−→OB + t
−→OA = (9 − 8t, 8 − 6t) where 0 ≤ t ≤ 1. By
definition, we have∫
Cydx + xdy =
∫ 1
0(y(t)x′(t) + x(t)y′(t)) dt =∫ 1
0( (8 − 6t)(−8) + (9 − 8t)(−6) ) dt = 70.
. . . . . .
.. Line Integrals of Vector Fields.
......
Let F = (P, Q, R) be a continuous vector field defined on a region D, and C bea piecewise continuously differentiable curve in a domain D, parameterized by
r(t) = x(t)i + y(t)j + z(t)k for a ≤ t ≤ b, and T(t) =r′(t)
∥r′(t)∥ be the unit
tangent vector of C at r(t).
.
......
Definition. F · T ds = F · dr = (P, Q, R) · r′(t)∥r′(t)∥ ∥r′(t)∥ dt
= ( P(r(t))dxdt
+ Q(r(t))dydt
+ R(r(t))dzdt
) dt= Pdx + Qdy + Rdz
.
......
Definition. The line integral of F along the curve C is defined to be∫C
F · T ds =∫ b
a( P(r(t))x′(t) + Q(r(t))y′(t) + R(r(t))z′(t) ) dt.
Remark. The line integral is sometimes called the work done of F along thepath C.
. . . . . .
.
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Example. Find the work done by the force field F(x, y, z) = yi + zj + xk inmoving a particle from (0, 0, 0) to (1, 1, 1) along the twisted cubicC : r(t) = (t, t2, t3) where 0 ≤ t ≤ 1.
Solution. Note that r(t) = (x(t), y(t), z(t)) = (t, t2, t3) for 0 ≤ t ≤ 1.
So work done W =∫
CF · dr =
∫C
F · T ds =∫
Cydx + zdy + xdz
=∫ 1
0
(t2 d
dt(t) + t3 d
dt(t2) + t
ddt(t3)
)dt =
∫ 1
0(t2 + 2t4 + 3t3) dt =
8960
.
.
......
Example. Find the work done by the force field F(x, y, z) = yi + zj + xk inmoving a particle from A(0, 0, 0) to B(1, 1, 1) along the segment AB.
Solution. Note that r(t) = (x(t), y(t), z(t)) = (t, t, t) for 0 ≤ t ≤ 1.
So work done W =∫
CF · dr =
∫C
F · T ds =∫
Cydx + zdy + xdz
=∫ 1
0
(t
ddt(t) + t
ddt(t) + t
ddt(t)
)dt =
∫ 1
0(t + t + t) dt = 3
∫ 1
0t dt ==
32
.
Remark. The integral∫
C F · T ds depends on the path C, not just the end points.
. . . . . .
.
......
Example. Find the work done by the force field
F(x, y, z) =krr3 =
k(xi + yj + zk)(x2 + y2 + z2)3/2 in moving a particle along the straight line
segment C from (0, 4, 0) to (0, 4, 3).
Solution. Let r(t) = (x(t), y(t), z(t)) = (0, 4, t) where 0 ≤ t ≤ 3, thenr′(t) = (x′(t), y′(t), z′(t)) = (0, 0, 1).It follows that work done is given by the line integral
W =∫
CF · dr =
∫ 3
0
k(x(t)x′(t) + y(t)y′(t) + z(t)z′(t)(x(t)2 + y(t)2 + z(t)2)3/2 dt =
∫ 3
0
kt(42 + t2)3/2 dt
=k2
∫ 3
0
d(42 + t2)
(42 + t2)3/2 =k2
[(42 + t2)−
32+1
− 32 + 1
]3
0
=
[− k√
16 + t2
]3
0=
k20
.
. . . . . .
.
......
Example. The line integral∫
CF · dr will change by a minus sign if we reverse
the direction of the path C.
Proof. Parameterize C from A to B, by r(t) = x(t)i + y(t)j + z(t)k (a ≤ t ≤ b),and let C be the curve from B to A with the direction reversed. ParameterizeC : q(t) = x1(t)i + y1(t)j + z1(t)k =
x(b + a − t)i + y(b + a − t)j + z(b + a − t)k (a ≤ t ≤ b). We are going to prove
that∫
CF · dq = −
∫C
F · dr. In fact,∫C
F · dq =∫ b
t=a( P(q(t)) · x′1(t) + Q(q(t)) · y′1(t) + R(q(t)) · z′1(t) )dt =∫ b
a[ P(r(b + a − t)) · (−x′(b + a − t)) + Q(r(b + a − t)) · y′(b + a − t) + R(r(b +
a − t)) · (−z′(b + a − t)) ]dt(∗)=∫ a
b[ P(r(ω)) · x′(ω) + Q(r(ω)) · y′(ω) + R(r(ω)) · z′(ω) ]dω =
−∫ b
a[ P(r(ω)) · x′(ω) + Q(r(ω)) · y′(ω) + R(r(ω)) · z′(ω) ]dω = −
∫C
F · dr,
where in (*) we replace ω = b + a − t, dω = −dt.The last equality follows from by reversing the upper and lower limits, andreplace ω by t.
. . . . . .
.
......
Example. Evaluate the line integral∫
C(x2 + y2) ds, where C is the straight line
segment from A(0, 0) to B(3, 4).
Solution. Let r(t) = (1 − t)−→OA + t
−→OB = (3t, 4t) for t ∈ [0, 1] be the convex
combination of vectors−→OA and
−→OB.
Then ds =√
x′(t)2 + y′(t)2dt =√
32 + 42dt = 5dt, and hence∫C(x2 + y2) ds =
∫ 1
0( (3t)2 + (4t)2 ) · 5dt = 125
∫ 1
0t2 dt =
1253
.
.
......
Example. Evaluate the line integral∫
Cy2dx + x2dy. where C is the part of the
graph of y = x2 from (−1, 1) to (1, 1).
Solution. Let x = t where t ∈ [−1, 1], then y = x2 = t2, and hence r(t) = (t, t2)
be a parametrization of y = x2. It follows that∫
Cy2dx + x2dy =∫ 1
−1
(y2(t)
dxdt
+ x2(t)dydt
)dt =
∫ 1
−1( t4 · (1) + t2 · 2t)dt =
[t5
5+
2t4
4
]1
−1=
25
.
. . . . . .
.
......
Example. Evaluate the line integral∫
Cxyz ds, where C is the path from
A(0, 0, 0) to D(1, 2, 3) consisting of three line segments joining together, thefirst one is parallel to the x-axis, the second parallel to the y-axis, the thirdparallel to the z-axis.
Solution. It follows from the description of the path that it starts first fromA(0, 0, 0) to B(1, 0, 0), next from B(1, 0, 0) to C(1, 2, 0); and finally from C(1, 2, 0)to D(1, 2, 3). Note if a path is parallel to one of the coordinate axes, so theremaining two coordinates are constant, in particular, their derivatives withrespect to the time-parameter t are zero. With the description, one has∫
Cxyzds =
(∫−→AB
+∫−→BC
+∫−→CD
)xyz ds =∫ 1
0t · 0 · 0dt +
∫ 2
01 · t · 0dt +
∫ 3
01 · 2 · tdt = 2
∫ 3
0tdt =
[t2]3
0= 9.
. . . . . .
.
......
Fundamental Theorem for Line Integrals. Given a conservative field F = ∇f ,then the line integral (or work done of F along C) is∫
CF · Tds =
∫C
F · dr =∫
C∇f · dr = f (r(b))− f (r(a)), where C is parameterized
by r(t) (a ≤ t ≤ b).
Proof. The proof is just a matter of notation and application of fundamentaltheorem of calculus. Recall that∫
C∇f · Tds =
∫ b
a
(∂f∂x
· dxdt
+∂f∂y
· dydt
+∂f∂z
· dzdt
)dt =
∫ b
a
ddt(f (r(t))) dt =
[f (r(t))]ba = f (r(b))− f (r(a)).
Remark. One should remember that∫
C∇f · Tds = f (B)− f (A) where A and B
are the starting point and terminal point of the curve C, as it is independent ofthe parametrization of C.
. . . . . .
.
......
Example. Evaluate the line integral∫
CF · T ds, where F(x, y, z) = xi + yj + zk,
and C is the curve parameterized by r(t) = (x(t), y(t), z(t)) = (e2t, et, e−t) for0 ≤ t ≤ ln 2.
Solution I. First x′(t) = 2e2t, y′(t) = et, z′(t) = −e−t.∫
CF · T ds =∫ ln 2
0
(e2t · 2e2t + et · et + e−t · (−e−t)
)dt =
∫ ln 2
0(2e4t + e2t − e−2t) dt = · · · .
Solution II. Note that F(x, y, z) = xi + yj + zk = ∇(
12(x2 + y2 + z2)
). Hence,
it follows from fundamental theorem of line integral that∫C
F · T ds =∫
C∇
(12(x2 + y2 + z2)
)· T ds =
12(x(ln(2))2 + y(ln(2))2 + z(ln(2))2)− 1
2(x(ln(0))2 + y(ln(0))2 + z(ln(0))2) =
12(e2×2 ln 2 + e2×ln 2 + e2×(− ln 2))− 1
2(e2×2×0 + e2×0 + e2×(−0)) =
12(24 + 22 +
122 − 1 − 1 − 1) =
12× (17 +
14) =
698
.
. . . . . .
.
......
Example. Let F(x, y, z) =k(xi + yj + zk)(x2 + y2 + z2)3/2 be a vector field defined in
R3 \ { (0, 0, 0) }. Determine the work done∫
CF · dr, where C is a straight line
from A(0, 4, 0) to B(0, 4, 3) by means of fundamental theorem of line integral.
Solution. Let f (x, y, z) = −kr= − k√
x2 + y2 + z2, then its gradient vector field
∇f (x, y, z) =k(xi + yj + zk)(x2 + y2 + z2)3/2 = F(x, y, z), so it follows from the fundamental
theorem of line integral that
W =∫
CF · dr =
∫C∇f · dr = f (0, 4, 3)− f (0, 4, 0) =
−k5
− −k4
=k
20.
Remark. In general, given a vector field F = (P, Q, R) defined on a domain D,how can one determine if F(x, y, z) = ∇f (x, y, z) for some scalar function fdefined on the domain D of the vector field F? That is the same to find thesolution f of the equation (P, Q, R) = (fx, fy, fz).However, we have a necessary condition: 0 = ∇×∇f = ∇× F. In terms ofcoordinate functions, we have (Ry − Qz, Pz − Rx, Qx − Py) = (0, 0, 0) i.e.(Ry, Pz, Qx) = (Qz, Rx, Py).
. . . . . .
.. Line Integral Independent of Path.
......
Definition. Let F be a vector field defined in a region D. The line integral
I =∫
CF · T ds is said to be independent of path in region D, if the following
condition holds:For any two points A and B in D, the line integral I has the same value alongevery piecewise smooth curve or path in D from A to B.
In this case, we may write∫
CF · T ds =
∫ B
AF · T ds, because the value of the
integral depends only on the location of end points A and B of C, not on theparticular choice of the path joining them.Remark. In fact, the concept of conservative vector field F depend on F andthe domain D as well.
. . . . . .
.
......
Example. Suppose that F(x, y, ) = (−2y, 2x) = −2yi + 2xj. Let C1 be the curveof the upper semi-circle from A(10, 0) to B(−10, 0) with center at (0, 0); and C2be the straight line from A(10, 0) to B(−10, 0) Determine the work done of Falong the path Ci (i = 1, 2).
Solution. Let C1 : r(t) = (10 cos t, 10 sin t) for 0 ≤ t ≤ π. Hence the work done∫C1
F · Tds =∫
C1
−y dx + x dy =∫ π
0(−y(t)x′(t) + x(t)y′(t)) dt =∫ π
0100(sin2 t + cos2 t) dt = 100π.
Let C2 : r(t) = (10 − t, 0) for 0 ≤ t ≤ 20. Hence the work done∫C1
−y dx + x dy =∫ 20
0(−y(t)x′(t) + x(t)y′(t)) dt =
∫ 20
00 dt = 0. Remarks. (i)
In fact this exercise is to show that the vector field F is not conservative.(ii) Though the curves had the same starting and terminal points, but in
general the work done∫
CF · T ds also depends on the curve. Moreover, this
method can be used to disprove the vector field F is conservative.
. . . . . .
.
......
Proposition. Let F be a continuous vector field defined on a region D, provethat the line integral of F is independent of path if and only if
∫C F · T ds = 0 for
any piecewise smooth closed curve C in D.
Proof. Suppose that line integral of F is independent of path, then let C be anyclosed curve with the same starting and terminal point A, then the constantpath C′ with A for all t is also a curve with the same starting and terminal pointA. It follows from the path independence of the line integral of F that∫
C F · T ds =∫
C′ F · T ds =∫
C′ F · 0 ds = 0.Conversely, suppose C1 and C2 are two paths, both of them starts from thesame point A, and terminates at point B. Let C = C1 ∪ (−C2) be a closed pathfrom A to B via C1, and back from B to A via −C2 (in reverse direction of C2).Then C is a piecewise smooth closed curve in D, hence one has
0 =∫
CF · T ds =
∫C1
F · T ds +∫−C2
F · T ds =∫
C1
F · T ds −∫
C2
F · T ds. Hence∫C1
F · T ds =∫
C2
F · T ds. So the line integral of F is independence of path.
. . . . . .
.
......
Theorem. The line integral∫
CF · T ds of the continuous vector field F is
independent of path in the plane or space region D ⇔ F = ∇f for somefunction f defined on D.
Proof. Suppose that F = ∇f for some continuously differentiable functiondefined on D. It follows from the fundamental theorem of line integral that∫
CF · T ds = f (B)− f (A) where C is a path from point A to point B.
For the converse, we assume that the line integral of F is path-independent,then one can define a potential function f (x, y, z) =
∫ (x,y,z)(x0,y0,z0)
F · T ds. It remainsto show that ∇f = F on D. Here we just verify that fx(x, y, z) = P(x, y, z), andthe other two can be done similarly. As D is an open, so one may choose aball centered at B of positive radius ρ such that ball lies inside D. In particular,for any h ∈ (0, ρ), we have E′(x + h, y, z) ∈ D. Then joining from B to E with asegment
−→BE parameterized by r(t) = (x + t, y, z) where 0 ≤ t ≤ h. Note that
r′(t) = (1, 0, 0), and−→BE lies in the ball above.
. . . . . .
.
......
Theorem. The line integral∫
CF · T ds of the continuous vector field F is
independent of path in the plane or space region D ⇔ F = ∇f for somefunction f defined on D.
Proof. By the path-independence of line integral,
f (x + h, y, z)− f (x, y, z) =∫ B(x,y,z)
A(0,0,0)F · T ds −
∫ E(x+h,y,z)
A(0,0,0)F · T ds =
∫−→BE
F · T ds =∫ h
0(P, Q, R) · (1, 0, 0) dt =
∫ h
0P(x + t, y, z) dt. It remains to show that
limh→0
f (x + h, y, z)− f (x, y, z)h
= P(x, y, z). For this, recall that P is continuous on
the domain D, so for any ε > 0, there exists 0 < δ < ρ, such that|P(u, v, w)− P(x, y, z)| < ε for all (u, v, w) in the ball centered at B(x, y, z) of
radius less than δ. It follows that∣∣∣∣ f (x + h, y, z)− f (x, y, z)
h− P(x, y, z)
∣∣∣∣=
1h
∣∣∣∣∫ h
0( P(x + t, y, z)− P(x, y, z) ) dt
∣∣∣∣≤ 1
h
∫ h
0| P(x + t, y, z)− P(x, y, z) | dt <
1h
∫ h
0ε dt = ε. Hence fx = P.
. . . . . .
.
......
Remark. Suppose that we know that the line integral of the vector field F isindependent of path, it is still hard to find the scalar function f such thatF = ∇f .
.
......
Example. Find a function f (x, y) such that∇f (x, y) = F(x, y) = (6xy − y3)i + (4y + 3x2 − 3xy3)j.
Solution. F is defined on D = R2, so the line segment C =−→AB from A(0, 0) to
B(x1, y1) lies in D, which is parameterized by r(t) = (tx1, ty1) for 0 ≤ t ≤ 1.Hence one can apply the fundamental theorem of line integral to evaluatefunction f (x, y) i.e.
f (x1, y1) =∫ B(x,y)
A(0,0)F · T ds =
∫−→AB
(6xy − y3) dx + (4y + 3x2 − 3xy2) dy =∫ 1
0(6x1y1t2 − y3
1t3)x1 dt + (4y1t + 3x21t2 − 3x1y2
1t3) y1dt =∫ 1
0(4y2
1t + 9x21y1t2 −
4x1y31t3) dt =
[2y2
1t2 + 3x21y1t3 − x1y3
1t4]1
0= 2y2
1 + 3x21y1 − x1y3
1. Let
f (x, y) = 2y2 + 3x2y − xy3, and one can easily check that fx(x, y) = 6xy − y3,and fy(x, y) = 4y + 3x2 − 3xy2.
. . . . . .
.
......
Definition. Vector field F defined on a region D is conservative, if there exists ascalar function f defined on D such that F = ∇f at each point of D. In thiscase, f is called a potential function of F.
Let F = (P, Q, R) = ∇f for some differentiable function f on R3, where P, Q andR are functions defined on R3. Assume that O(0, 0, 0) ∈ D. For any pointE(a, b, c) in D, one can travel from O(0, 0, 0) to A(x, 0, 0) along the x-axis, thenfrom A(a, 0, 0) to B(a, b, 0), and finally from B(a, b, 0) to E(a, b, c). In this way, wecall the path C. It follows from the fundamental theorem of line integral and
F = ∇f that∫
CF · T ds =
∫C∇ · T ds = f (D)− f (O) = f (a, b, c)− f (0, 0, 0).
On the other hand, as the line integral of F is independent of path, it followsfrom the definition of line integral that f (a, b, c)− f (0, 0, 0)
=∫
CF · T ds =
∫−→OA
F · T ds +∫−→AB
F · T ds +∫−→BD
F · T ds
=∫ a
0P(x, 0, 0)dx︸ ︷︷ ︸
depends on aonly
+∫ b
0Q(a, y, 0) dy︸ ︷︷ ︸
depends on b and aonly
+∫ c
0R(a, b, z) dz︸ ︷︷ ︸
depends on a,b,c
. (♡).
. . . . . .
.
......
Theorem. Let F(x, y) = P(x, y)i + Q(x, y)j be a continuously differentiablevector field in an open rectangle R in the xy-plane. Then F is conservative in R,
if and only if at each point of R,∂P∂y
=∂Q∂x
. In this case, the vector field F has a
potential function f (x, y) defined on R
Proof. If F is conservative, then F = ∇f for some scalar function f , and then(0, 0, Qx − Py) = curl(P, Q, 0) = curlF = curl(∇f ) = (0, 0, 0), it follows thatPy = Qx on R.For the converse, it follows from Green’s theorem which is discussed in thenext section.
. . . . . .
.
......
Example. Let P(x, y) = 6xy − y3, and Q(x, y) = 4y + 3x2 − 3xy2.(a). Prove that Py = Qx on R2,(b) Find a potential function f (x, y) by means of fx(x, y) = P(x, y) andfy(x, y) = Q(x, y).
Solution. (a) Py =∂
∂y(6xy − y3) = 6x − 3y2 =
∂
∂x(4y + 3x2 − 3xy2) = Qx, on
any rectangle inside R2, so F = (P, Q) is conservative on any rectangle, inparticular, the line integral of F is path independent on any rectangle, andhence on R2.(b) We first integrate fy(x, y) = Q(x, y) with respect to y, while treating x asconstant (comparing to ♡) so that
f (x, y) =∫
fy(x, y) dy =∫(4y + 3x2 − 3xy2) dy = 2y2 + 3x2y − xy3 + h(x), ♣
where h(x) is a function to be determined later. Next we differentiate ♣ withrespect to x, and obtain 6xy − y3 = P(x, y) = fx(x, y) = 6xy − y3 + h′(x), i.e.h′(x) = 0, and it follows from the fundamental theorem of calculus of onevariable that h(x) is a constant. In particular, the potential functionf (x, y) = 2y2 + 3x2y − xy3 + C for some constant C.
. . . . . .
.
......
Definition. A subset D in R2 (or R3) is said to beopen if for any point P in D, there is a disk (ball)with center at P that lies entirely in D, i.e. D doesnot contain any boundary points.
.
......
Definition. A subset D in R2 (or R3) is said to beconnected if any two points in D can be joinedby a path that lies in D.
. . . . . .
.
......
Definition. A subset D in R2 (or R3) is said to be connected if any two points inD can be joined by a path that lies in D.
.
......Definition. A simple curve is a curve which does not intersect itself.
.
......
Definition. A simply connected region in theplane is a connected region such that everysimple closed curve in D encloses onlypoints that are in D.
Remark. The actual definition of simply connectedness is more complicated ifthe underlying space is not a plane. However, the definition above is moreintuitive.
. . . . . .
.
......
Definition. A simply-connected region in theplane is a connected region such that everysimple closed curve in D encloses onlypoints that are in D.
.
......
Theorem. Let F(x, y) = P(x, y)i + Q(x, y)j be a vector field on an opensimply-connected region D ⊂ R2, where P and Q have continuous partial
derivatives in D. If∂P∂y
=∂Q∂x
, then F is conservative on D, i.e. F = ∇f for some
function f defined on D.
Remark. This is a consequence of Green’s Theorem.
. . . . . .
.
......Example. Determine F(x, y, z) = y2i + (2xy + e3z)j + 3ye3zk is conservative onR3.
Answer. F is conservative on R3, as ∇(xy2 + ye3z) = F(x, y, z) on R3.Remark. Do you know how to find the potential function xy2 + ye3z for thevector field F(x, y, z)?
. . . . . .
.
......
Example. Suppose that F(x, y, z) = y2i + (2xy + e3z)j + 3ye3zk. Determine F isconservative on R3. If so, find a function f such that ∇f = F on R3.
Solution. (a) One can first check curl F(x, y, z) = (0, 0, 0) on R3, and it followsfrom the fact that R3 is simply connected that F is conservative on R3. Thedesired functions f (x, y, z) = xy2 + ye3z + c, where c is arbitrarily constant.(b) Let F = (P, Q, R) = ∇f = (fx, fy, fz). Integrate Q(x, y, z) with respect to y, sof (x, y, z) =
∫(2xy + e3z) dy = xy2 + ye3z + g(x, z) ♡ for some function g(x, z).
Then differentiate ♡ with respect to z, so3ye3z = R(x, y, z) = fz(x, y, z) = 3ye3z + gz(x, z), and hence gz(x, z) = 0 ♣.Integrate ♣ with respect to z, and g(x, z) = 0 + h(x) for some function h(x), i.e.f (x, y, z) = xy2 + ye3z + h(x).Differentiate with respect to x, so y2 = P(x, y, z) = fx(x, y, z) = y2 + h′(x), i.e.h(x) = c for some constant c. It follows that f (x, y, z) = xy2 + ye3z + c.
. . . . . .
.. Summary of Line Integral of Vector Field
Let F be a vector field defined on a region D. Then we can summary ourimportant result as follows:.
......
F is conservative on D ⇐⇒ F = ∇f for some function f⇕ ⇓(−)xy = (−)yx∮
CF · T ds = 0
for any closed path in D
Simply connected⇐=Stokes Thm
∇× F = 0 on D
Remark. We have proved all the blue arrows, except the red arrow. One caneasily checked that F(x, y) = −yi+xj
x2+y2 on D = R2 \ {(0, 0)} satisfies curl F = 0
on D, but there is no function f (x, y) defined on D such that F = ∇f on D.
