1. In addition to the multiple integral of a function f:R n  R over a region in R n, 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:R n  R over a (parametrized) path (a path integral), the integral of a vector field F:R n  R n over a (parametrized) path (a line integral), the integral of a function f:R n  R over a (parametrized) surface, and the integral of a vector field F:R n  R n over a (parametrized) surface (a surface integral). We shall now consider integral of a vector field F:R n  R n over a (parametrized) path (a line integral).

2. 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) c  (t)c  (t) (One possible interpretation is the work done to move a particle along the path through the vector field.) c In R 3 this integral is sometimes written (F 1 dx+F 2 dy+F 3 dz) or cc (F 1 i+F 2 j+F 3 k) (dxi+dyj+dzk) or F ds. a b dx dy dz ( F 1 — + F 2 — + F 3 — ) dt. dt dt dt

3. ExampleSuppose F = (x + y)i + (x + z)j – (y + z)k, c(t) = (t, t 2, 4t + 3) for 1  t  5, and b(u) = (u 2 + 1, u 4 + 2u 2 + 1, 4u 2 + 7) for 0  u  2. c  (t) =(1, 2t, 4) F(c(t)) c  (t) = (t + t 2, 5t + 3, – t 2 – 4t – 3) (1, 2t, 4) = t + t 2 + 10t 2 + 6t – 4t 2 – 16t – 12 = 7t 2 – 9t – 12 c F ds. 5 1 (7t 2 – 9t – 12) dt =7t 3 /3 – 9t 2 /2 – 12t = t = 1 5 800 / 6 = 400 / 3 (a) Find

4. (c) Do c(t) and b(u) describe the same path? Yes, since c(u 2 +1) = b(u). b  (u) = (2u, 4u 3 + 4u, 8u) F(b(u)) b  (u) = (u 4 + 3u 2 + 2, 5u 2 + 8, – u 4 – 6u 2 – 8) (2u, 4u 3 + 4u, 8u) = 14u 5 + 10u 3 – 28u b F ds. (b) Find 2 0 (14u 5 + 10u 3 – 28u) du =7u 6 /3 + 5u 4 /2 – 14u 2 = u = 0 2 800 / 6 = 400 / 3

5. 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 that dt 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)).

6. 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 (x 1, y 1, z 1 ) to (x 2, y 2, z 2 ) will be equal to no matter what path is chosen. (x 2, y 2, z 2 ) (x 1, y 1, z 1 ) f(x 2, y 2, z 2 ) – f(x 1, y 1, z 1 )

7. Example LetF = (x+y)i + (x+z)j – (y+z)k, V = xi + yj + zk, c(t) = (t, t 2, t 3 ) 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) = x 2 + y 2 + z 2 ————— 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).

8. c F ds and F ds. (e) Find b c  (t) =(1, 2t, 3t 2 ) F(c(t)) c  (t) = (t + t 2, t + t 3, – t 2 – t 3 ) (1, 2t, 3t 2 ) = t + t 2 + 2t 2 + 2t 4 – 3t 4 – 3t 5 = t + 3t 2 – t 4 – 3t 5 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 (t + 3t 2 – t 4 – 3t 5 ) dt = 4 1 10913 – ——— 5 (– 3870t – 90) dt = 1 0 – 2025 c V ds and V ds. (f) Find bc V ds = b

9. c V ds and V ds. (f) Find b c V ds = b f(4, 16, 64) – f(1, 1, 1) = (4) 2 + (16) 2 + (64) 2 (1) 2 + (1) 2 + (1) 2 ———————— – ——————— = 2 4365 —— 2

10. Example Let 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 R 2 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: b 1 (t) = ( 3 – 3t, 3t ) for 0  t  1, b 2 (t) = (– 3t, 3 – 3t ) for 0  t  1, b 3 (t) = ( – 3 + 3t, – 3t ) for 0  t  1, b 4 (t) = ( 3t, – 3 + 3t ) for 0  t  1. We then say that b = b 1  b 2  b 3  b 4.

11. (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 = b f(3,0) – f(3,0) = 0

12. Look at the first homework problem in Section 7.2 (#2a, page 447): c x dy – y dx = (F 1 dx+F 2 dy) 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