In addition to the multiple integral of a function f:RnR over a region in Rn, there are many different types of integrals which can be defined, each of which has its own interpretation and applications. These include

the integral of a function f:RnR over a (parametrized) path (a path integral),

the integral of a vector field F:RnRn over a (parametrized) path (a line integral),

the integral of a function f:RnR over a (parametrized) surface, and

the integral of a vector field F:RnRn over a (parametrized) surface (a surface integral).

We shall now consider integral of a vector field F:RnRn over a (parametrized) path (a line integral).

The line integral of F over c is defined to be F(c(t)) • c (t) dt denoted

b

a

Suppose c(t) for a t b describes a smooth (or “piecewise smooth”) path, the tangent velocity vector c (t) 0 for any t, and F is a vector field.

c(t)c (t)

(One possible interpretation is the work done to move a particle along the path through the vector field.) c

In R3 this integral is sometimes written

(F1dx+F2dy+F3dz) or

c c

(F1i+F2j+F3k) • (dxi+dyj+dzk) or

F • ds .

a

b dx dy dz( F1 — + F2 — + F3 — ) dt . dt dt dt

Example Suppose F = (x + y)i + (x + z)j – (y + z)k ,

c(t) = (t , t2 , 4t + 3) for 1 t 5 , and

b(u) = (u2 + 1 , u4 + 2u2 + 1 , 4u2 + 7) for 0 u 2 .

c (t) = (1 , 2t , 4)

F(c(t)) • c (t) = (t + t2 , 5t + 3 , – t2 – 4t – 3) • (1 , 2t , 4) =

t + t2 + 10t2 + 6t – 4t2 – 16t – 12 = 7t2 – 9t – 12

c

F • ds .

5

1

(7t2 – 9t – 12) dt = 7t3/3 – 9t2/2 – 12t =

t = 1

5

800 / 6 = 400 / 3

(a) Find

(c) Do c(t) and b(u) describe the same path?

Yes, since c(u2+1) = b(u).

b (u) = (2u , 4u3 + 4u , 8u)

F(b(u)) • b (u) =

(u4 + 3u2 + 2 , 5u2 + 8 , – u4 – 6u2 – 8) • (2u , 4u3 + 4u , 8u) =14u5 + 10u3 – 28u

b

F • ds .(b) Find

2

0

(14u5 + 10u3 – 28u) du = 7u6/3 + 5u4/2 – 14u2 =

u = 0

2

800 / 6 = 400 / 3

In general, suppose c(t) for a t b and b(u) = c(h(u)) for c u d describe the same path. By making a change of variables in the line integral of F over c, we can prove that the line integral of F over a path c is the same no matter how the path is parametrized, that is, the line integral is independent of parametrization of the path. (See Theorem 1 on page 437.)

Suppose F = f , that is, F a gradient vector field, and consider the line integral of F over a path c(t) for a t b. From the chain rule, we know that

c

F • ds =

b

a

F(c(t)) • c (t) dt =

d— f(c(t)) = f (c(t)) • c (t) . We then observe thatdt b

a

f(c(t)) • c (t) dt =

f(c(t)) =

b

t = a

(See Theorem 3 on page 440.)f(c(b)) – f(c(a)) .

We see then that the line integral of a gradient field F = f over a path c(t) for a t b depends only on the starting point c(a) and the ending point c(b) of the path.

In other words, the line integral of a gradient field F = f over a path from (x1 , y1 , z1) to (x2 , y2 , z2) will be equal to

no matter what path is chosen.

(x2 , y2 , z2)

(x1 , y1 , z1)

f(x2 , y2 , z2) – f(x1 , y1 , z1)

Example Let F = (x+y)i + (x+z)j – (y+z)k ,

V = xi + yj + zk ,

c(t) = (t , t2 , t3) for 1 t 4 ,

b(t) = (3t + 1 , 15t + 1 , 63t + 1) for 0 t 1 .(a) Is F a gradient vector field? No, since curl F = – 2i 0

(b) Is V a gradient vector field? Yes, since curl V = 0.

Also, V = f where f(x,y,z) =x2 + y2 + z2

————— 2

(c) Do c(t) and b(t) begin at the same point and end at the same point?

Both paths begin at the point (1 , 1 , 1) and end at the point (4 , 16 , 64).

(d) Do c(t) and b(u) describe the same path?

No, they are two different paths from (1 , 1 , 1) to (4 , 16 , 64).

c

F • ds and F • ds .(e) Find

b

c (t) = (1, 2t , 3t2)

F(c(t)) • c (t) = (t + t2 , t + t3 , – t2 – t3) • (1 , 2t , 3t2) =t + t2 + 2t2 + 2t4 – 3t4 – 3t5 = t + 3t2 – t4 – 3t5

b (t) = (3 , 15 , 63)

F(b(t)) • b (t) = (18t + 2 , 66t + 2 , – 78t – 2) • (3 , 15 , 63) =54t + 6 + 990t + 30 – 4914t – 126 = – 3870t – 90

c

F • ds =

b

F • ds =

(t + 3t2 – t4 – 3t5) dt =

4

1

10913– ——— 5

(– 3870t – 90) dt =1

0

– 2025

c

V • ds and V • ds .(f) Find

b c

V • ds = V • ds =

b

c

V • ds and V • ds .(f) Find

b

c

V • ds = V • ds =

b

f(4 , 16 , 64) – f(1 , 1 , 1) =

(4)2 + (16)2 + (64)2 (1)2 + (1)2 + (1)2

———————— – ——————— =2 2

4365—— 2

ExampleLet c(t) be the counterclockwise path in the xy plane along the circle of radius 3 centered at the origin starting and ending at (3,0).Let b(t) be the path in R2 along the rectangle from (3,0) to (0,3) to (–3,0) to (0,–3) to (3,0) .

(a) How can we parametrize the path c(t)?

(b) How can we parametrize the path b(t)?

c(t) = (3 cos t , 3 sin t) for 0 t 2

We can first define each of the line segments of the path separately:

b1(t) = ( 3 – 3t , 3t ) for 0 t 1 ,

b2(t) = (– 3t , 3 – 3t ) for 0 t 1 ,b3(t) = ( – 3 + 3t , – 3t ) for 0 t 1 ,b4(t) = ( 3t , – 3 + 3t ) for 0 t 1 .We then say that b = b1 b2 b3 b4 .

(c) Suppose F(x,y) is a gradient vector field.

Find

c

F • ds and F • ds .

b

Since F = f for some f(x,y), then we must have

c

F • ds = F • ds =

b

f(3,0) – f(3,0) = 0

Look at the first homework problem in Section 7.2 (#2a, page 447):

c

x dy – y dx =

(F1dx+F2dy)

c

Note that the line integral is written in the following form:

This implies that the vector field F = ( ) is integrated over the given path – y , x , 0