. Chapter 15 Vector Calculus . 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 to understand, and many notations are involved. .

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. Vector Fields . Definition. Let D ⊂ R2 . A vector field on D is a vector-valued function F that assigns 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 E to 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.

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. Definition. If f : Rn → R is a differentiable function, then ∇f is a vector field on Rn , 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.

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. 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 curl F(x, y, z) i

j

k

= ∇ × F = ∂x∂ ∂y∂ ∂z∂ = (Ry − Qz , Pz − Rx , Qx − Py ). P Q R . . 2 z .Example. Find the curl of F = (x − z)i + xe j + xyk. i

Solution. curlF = ∇ × F = (

)

(

x2

j

∂ ∂x

−z

∂ ∂y xez

∂ ∂ ∂ ∂ z 2 ∂y (xy) − ∂z (xe ) i + ∂x (xy) − ∂z (x x(1 − ez )i − (y + 1)j + ez k.

k

∂ ∂z

=

xy

) ( ∂ − z) j + ∂x (xez ) −

) − z) k =

∂ 2 ∂y (x

. Example. Let F(x, y, z) = xzi + xyzj − y2 kk. Find curl F. . i

Solution. curlF = ∇ × F =

j

∂ ∂x

∂ ∂y

xz

xyz

k

∂ ∂z ez

= −(2y + xy)i + xj + yzk. .

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. Theorem. If f (x, y, z) is a scalar function and has continuous 2nd order partial derivatives, then curl (∇f ) = 0. . i

Proof. curl (∇f ) =

∂ ∂x

j

∂ ∂y

k

∂ ∂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 differentiable vector field, i.e. the functions P, Q and R are differentiable functions on the domain of F. The divergence of F is defined to be ∂Q ∂R divF(x, y, z) = ∇ · F(x, y, z) = ∂P ∂x + ∂y + ∂z . . . Example. Find the divergence and the curl of the vector field y .F(x, y, z) = xe i + z sin yj + xy ln zk. Solution. div F(x, y, z) = curl F =

i ∂ ∂x xey

j ∂ ∂y z sin y

∂ ∂ ∂ xy (xey ) + (z sin y) + (xy ln z) = ey + z cos y + . ∂x ∂y ∂z z

k ∂ ∂z 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 − xey k. . Example. Let F, G be a differentiable vector field defined on a domain D, and a, 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; . ∇( f ) = g∇f − f ∇g ; g f2 .3 ∇ · (F × G) = G · (∇ × F) − F · (∇ × G). 4. div(curlF) = 0; 2

. div∇(f · g) = f div∇(g) + gdiv∇(f ) + 2∇f · ∇g; 6. div(∇f × ∇g) = 0. 7. ∇ · (F × F ) = F · (∇ · F ) − F · (∇ × F ); 5

1

2

2

1

1

2

. ∇ × (F × F ) = 1 2 (F1 · ∇)(F2 ) + (F2 · ∇)(F1 ) + F1 × (∇ × F2 ) + F2 × (∇ × F1 ), ∂ ∂ ∂ in which ((P, Q, R) · ∇)(A, B, C) = P (A, B, C) + Q (A, B, C) + R (A, B, C) ∂x ∂y ∂z .= (PAx + PBx + PCx ) + (QAy + QBy + QCy ) + (RAz + RBz + RCz ). 8

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. Definition. A curve C is called a piecewise continuously differentiable, if there exists a parametrization r : [a, b] → Rn (n = 2, or 3) such that C is the image of r([a, b]) and the the coordinate functions (x(t), y(t), z(t)) = r(t) are continuous, and the first order derivatives are continuous except finitely many .points in [a, b]. Example. One can consider the the square, in which the curve is continuous but the tangent vectors do not exist at the four vertices. Remark. One can replace the interval [a, b] by open intervals, or unbounded interval.

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. Definition. Let r(t) (a ≤ t ≤ b) be a parametrization of a curve C in Rn , the curve is oriented by the order of R, then the curve parameterized by the s(t) = r(b + a − t) where a ≤ t ≤ b is the curve C with reverse direction, and denoted by −C. . Example. Let r(t) = (cos t, sin t) (0 ≤ t ≤ 2π ) be a parametrization of the unit circle of radius 1 in counterclockwise direction. Then the s(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 clockwise direction.

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. Definition. Let C be a 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) be the unit tangent vector √of C at r(t). Then the arc-length element ds is ′ ′ 2 ′ 2 ′ 2 .ds = ∥r (t)∥dt = x (t) + y (t) + z (t) 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 .

∫ b

f ds =

C

a

f (r(t)) ∥r′ (t)∥ dt.

. ∫ Example. Evaluate the line integral xy ds, where C is a segment from A(1, 2) to . B(9, 8).

C

Solution. Parameterize the segment C by −→ −→ r(t) = (x(t), y(t)) = (1 − t)OA + tOB = (1 + 8t, 2 + 6t) where 0 ≤ t ≤ 1. By ∫ ∫ 1 √ definition, xy ds = x(t)y(t) x′ (t)2 + y′ (t)2 dt C 0 ∫ 1 ∫ 1 √ = (1 + 8t)(2 + 6t) 82 + 62 dt = 10 (2 + 22t + 48t2 ) dt = 290. 0

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. Line Integrals of Scalar Functions along Coordinates . Definition The line integral of f with respect to the coordinate axes are defined ∫

∫ b

f dx =

to be

C ∫ b

∫

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C

f dz =

a

a

f (r(t)) x′ (t) dt,

−C

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2

∫ b

C

f dy =

a

f (r(t)) y′ (t) dt and

f (r(t)) z′ (t) dt respectively.

Remarks. ∫ ∫ 1. f dx = − f dx, ∫ −C

∫

f dz = −

∫C C

∫ −C

f dy = −

∫ C

f dy,

f dz, where −C denotes the curve C with reversed

orientation. ∫ ∫ f ds = f ds. −C

C

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∫

Example. Evaluate the line integral C

B . (9, 8) to A(1, 2).

ydx + xdy, where C is a segment from

Solution. Parameterize the segment C by −→ −→ r(t) = (x(t), y(t)) = (1 − t)OB + tOA = (9 − 8t, 8 − 6t) where 0 ≤ t ≤ 1. By ∫

∫ 1

definition, we have ∫ 1 0

C

ydx + xdy =

0

(y(t)x′ (t) + x(t)y′ (t)) dt =

( (8 − 6t)(−8) + (9 − 8t)(−6) ) dt = 70.

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. Line Integrals of Vector Fields . Let F = (P, Q, R) be a continuous vector field defined on a region D, and C be a piecewise continuously differentiable curve in a domain D, parameterized by r′ (t) be the unit r(t) = x(t)i + y(t)j + z(t)k for a ≤ t ≤ b, and T(t) = ′ ∥r (t)∥ tangent vector of C at r(t). . .

r′ (t ) ∥r′ (t)∥ dt ∥r′ (t)∥ dx dy dz = ( P(r(t)) + Q(r(t)) + R(r(t)) ) dt dt dt dt = Pdx + Qdy + Rdz

Definition. F · T ds = F · dr = (P, Q, R) ·

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. Definition. The line integral of F along the curve C is defined to be ∫

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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 the path C. .

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. Example. Find the work done by the force field F(x, y, z) = yi + zj + xk in moving a particle from (0, 0, 0) to (1, 1, 1) along the twisted cubic 2 3 .C : r(t) = (t, t , t ) 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 = F · dr = F · T ds = ydx + zdy + xdz C C ) ∫ 1( ∫ 1C d d d 89 = t2 (t) + t3 (t2 ) + t (t3 ) dt = (t2 + 2t4 + 3t3 ) dt = . dt dt dt 60 0 0 . Example. Find the work done by the force field F(x, y, z) = yi + zj + xk in moving 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 = F · dr = F · T ds = ydx + zdy + xdz C C )C ∫ 1( ∫ 1 ∫ 1 d d d 3 = t (t) + t (t) + t (t) dt = (t + t + t) dt = 3 t dt == . dt dt ∫ dt 2 0 0 0 Remark. The integral C F · T ds depends on the path C, not just the end points. .

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. Example. Find the work done by the force field k(xi + yj + zk) kr F(x, y, z) = 3 = 2 in moving a particle along the straight line r (x + y2 + z2 )3/2 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, then r′ (t) = (x′ (t), y′ (t), z′ (t)) = (0, 0, 1). It follows that work done is given by the line integral ∫

W=

=

C

F · dr =

∫ 3 k ( x ( t ) x′ ( t ) + y ( t ) y′ ( t ) + z ( t ) z′ (t ) 0

∫ k 3 d(42 + t2 )

2

0

(42 + t2 )3/2

(x(t)2 k = 2

[

+ y(t)2

+ z(t)2 )3/2

(42 + t2 )− 2 +1 − 32 + 1 3

]3

[

= −√ 0

∫ 3

dt =

]3

k 16 + t2

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0 (42

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= 0

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kt dt + t2 )3/2

k . 20

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∫

Example. The line integral the . direction of the path C.

C

F · dr will change by a minus sign if we reverse

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. Parameterize C : 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 C

∫

F · dq = −

F · dq =

∫Cb a

∫ b

t=a

C

F · dr. In fact,

( P(q(t)) · x1′ (t) + Q(q(t)) · y1′ (t) + R(q(t)) · z1′ (t) )dt =

[ 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, and .

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∫

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 + tOB = (3t, 4t) for t ∈ [0, 1] be the convex −→ −→ combination √ √of vectors OA and OB. Then ds = x′ (t)2 + y′ (t)2 dt = 32 + 42 dt = 5dt, and hence ∫ ∫ 1 ∫ 1 125 . (x2 + y2 ) ds = ( (3t)2 + (4t)2 ) · 5dt = 125 t2 dt = 3 C 0 0 . ∫ Example. Evaluate the line integral y2 dx + x2 dy. where C is the part of the C

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 y2 dx + x2 dy = C ) [ 5 ]1 ∫ 1 ( ∫ 1 2 dx dy t 2t4 2 2 4 2 y (t) + x (t) ( t · (1) + t · 2t)dt = = . dt = + dt dt 5 4 −1 5 −1 −1 .

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∫

Example. Evaluate the line integral C

xyz ds, where C is the path from

A(0, 0, 0) to D(1, 2, 3) consisting of three line segments joining together, the first one is parallel to the x-axis, the second parallel to the y-axis, the third .parallel to the z-axis. Solution. It follows from the description of the path that it starts first from A(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 the remaining two coordinates are constant, in particular, their derivatives with respect to the ( time-parameter)t are zero. With the description, one has ∫

∫

C

xyzds =

∫ 1 0

t · 0 · 0dt +

∫

∫

− →+ − → + −→ xyz AB BC CD ∫ 2 ∫ 3 0

1 · t · 0dt +

0

ds =

1 · 2 · tdt = 2

∫ 3 0

[ ]3 tdt = t2 = 9. 0

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. Fundamental Theorem for Line Integrals. Given a conservative field F = ∇f , then the line∫integral (or∫ work done of F along C) is ∫ C

F · Tds =

C

F · dr =

.by r(t) (a ≤ t ≤ b).

C

∇f · dr = f (r(b)) − f (r(a)), where C is parameterized

Proof. The proof is just a matter of notation and application of fundamental theorem of calculus. ) ∫ ∫ b ( Recall that ∫ b ∂f dy ∂f dz d ∂f dx ∇f · Tds = · + · + · dt = (f (r(t))) dt = ∂x dt ∂y dt ∂z dt a dt C a [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 of the parametrization of C.

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∫

Example. Evaluate the line integral C

F · 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 ) for .0 ≤ t ≤ ln 2. ∫

Solution I. First x′ (t) = 2e2t , y′ (t) = et , z′ (t) = −e−t . F · T ds = C ∫ ln 2 ( ∫ ln 2 ) e2t · 2e2t + et · et + e−t · (−e−t ) dt = (2e4t + e2t − e−2t ) dt = · · · . 0 0 ( ) 1 2 2 2 Solution II. Note that F(x, y, z) = xi + yj + zk = ∇ (x + y + z ) . Hence, 2 it follows from fundamental theorem)of line integral that ( ∫ ∫ 1 2 F · T ds = ∇ (x + y2 + z2 ) · T ds = 2 C C 1 1 (x(ln(2))2 + y(ln(2))2 + z(ln(2))2 ) − (x(ln(0))2 + y(ln(0))2 + z(ln(0))2 ) = 2 2 1 2×2 ln 2 1 2×2×0 2×ln 2 2×(− ln 2) (e + e2×0 + e2×(−0) ) = +e +e ) − (e 2 2 1 4 1 1 1 69 (2 + 22 + 2 − 1 − 1 − 1) = × (17 + ) = . 2 2 4 8 2 .

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. Example. Let F(x, y, z) =

k(xi + yj + zk) be a vector field defined in (x2 + y2 + z2 )3/2 ∫

R3 \ { (0, 0, 0) }. Determine the work done

C

F · 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. k k Solution. Let f (x, y, z) = − = − √ , then its gradient vector field 2 r x + y2 + z2 k(xi + yj + zk) ∇f (x, y, z) = 2 = F(x, y, z), so it follows from the fundamental (x + y2 + z2 )3/2 theorem of line integral that ∫ ∫ k −k −k − = . W= F · dr = ∇f · dr = f (0, 4, 3) − f (0, 4, 0) = 5 4 20 C C 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 f defined on the domain D of the vector field F? That is the same to find the solution f of the equation (P, Q, R) = (fx , fy , fz ). However, we have a necessary condition: 0 = ∇ × ∇f = ∇ × F. In terms of coordinate functions, we have (Ry − Qz , Pz − Rx , Qx − Py ) = (0, 0, 0) i.e. (Ry , Pz , Qx ) = (Qz , Rx , Py ). .

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. Line Integral Independent of Path . Let F be a vector field defined in a region D. The line integral Definition. ∫ I=

C

F · 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 along .every piecewise smooth curve or path in D from A to B. ∫

In this case, we may write C

F · T ds =

∫ B A

F · T ds, because the value of the

integral depends only on the location of end points A and B of C, not on the particular choice of the path joining them. Remark. In fact, the concept of conservative vector field F depend on F and the domain D as well.

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. Example. Suppose that F(x, y, ) = (−2y, 2x) = −2yi + 2xj. Let C1 be the curve of the upper semi-circle from A(10, 0) to B(−10, 0) with center at (0, 0); and C2 be the straight line from A(10, 0) to B(−10, 0) Determine the work done of F .along 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 ∫ π (−y(t)x′ (t) + x(t)y′ (t)) dt = F · Tds = −y dx + x dy = ∫Cπ1 0

C1

2

0

2

100(sin t + cos 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 0

0 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 C

F · T ds also depends on the curve. Moreover, this

method can be used to disprove the vector field F is conservative.

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. Proposition. Let F be a continuous vector field defined on a region D, prove ∫ that 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 any closed curve with the same starting and terminal point A, then the constant path C′ with A for all t is also a curve with the same starting and terminal point A. ∫ It follows from ∫ the path ∫independence of the line integral of F that F · T ds = C C′ F · T ds = C′ F · 0 ds = 0. Conversely, suppose C1 and C2 are two paths, both of them starts from the same point A, and terminates at point B. Let C = C1 ∪ (−C2 ) be a closed path from 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= ∫

C1

C

F · T ds =

F · T ds =

∫

C2

C1

F · T ds +

−C2

F · T ds =

C1

F · T ds −

C2

F · T ds. Hence

F · T ds. So the line integral of F is independence of path.

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∫

Theorem. The line integral C

F · T ds of the continuous vector field F is

independent of path in the plane or space region D ⇔ F = ∇f for some function f defined on D. . Proof. Suppose that F = ∇f for some continuously differentiable function defined on D. It follows from the fundamental theorem of line integral that ∫ C

F · 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, ∫ (x,y,z) then one can define a potential function f (x, y, z) = (x ,y ,z ) F · T ds. It remains 0

0 0

to show that ∇f = F on D. Here we just verify that fx (x, y, z) = P(x, y, z), and the other two can be done similarly. As D is an open, so one may choose a ball 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 a − → segment 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.

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∫

Theorem. The line integral C

F · T ds of the continuous vector field F is

independent of path in the plane or space region D ⇔ F = ∇f for some function f defined on D. . Proof. By the path-independence of line integral, f (x + h, y, z) − f (x, y, z) = ∫ h 0

(P, Q, R) · (1, 0, 0) dt =

∫ B(x,y,z) A(0,0,0)

F · T ds −

∫ E(x+h,y,z)

∫ h 0

A(0,0,0)

F · T ds =

∫

− → F · T ds BE

=

P(x + t, y, z) dt. It remains to show that

f (x + h, y, z) − f (x, y, z) = P(x, y, z). For this, recall that P is continuous on h h→0 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 f (x + h, y, z) − f (x, y, z) radius less than δ. It follows that − P(x, y, z) h ∫ 1 h = ( P(x + t, y, z) − P(x, y, z) ) dt h 0 lim

≤

1 h

∫ h 0

| P(x + t, y, z) − P(x, y, z) | dt