the integral of a function f : R n  R over a (parametrized) path (a path integral),

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

2. c(t) 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) b The line integral of F over c is defined to be F(c(t)) • c(t) dt denoted (One possible interpretation is the work done to move a particle along the path through the vector field.) a F • ds . c In R3 this integral is sometimes written (F1i+F2j+F3k) • (dxi+dyj+dzk) or (F1dx+F2dy+F3dz) or c c b dx dy dz ( F1 — + F2 —+ F3— ) dt . dt dt dt a

3. 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 . (a) Find F • ds . c c(t) = (1 , 2t , 4) (t + t2 , 5t + 3 , – t2– 4t – 3) • (1 , 2t , 4) = F(c(t)) • c(t) = t + t2+ 10t2 + 6t– 4t2 – 16t – 12 = 7t2 – 9t – 12 5 5 (7t2 – 9t – 12) dt = 7t3/3 – 9t2/2 – 12t = 800 / 6 = 400 / 3 1 t = 1

4. (b) Find F • ds . b (2u , 4u3 + 4u , 8u) b(u) = F(b(u)) • b(u) = (u4 + 3u2 + 2 , 5u2 + 8 , – u4– 6u2–8) • (2u , 4u3 + 4u , 8u) = 14u5 + 10u3 – 28u 2 2 (14u5 + 10u3 – 28u) du = 7u6/3 + 5u4/2 – 14u2 = 800 / 6 = 400 / 3 u = 0 0 (c) Do c(t) and b(u) describe the same path? Yes, since c(u2+1) = b(u).

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 d — f(c(t)) = f (c(t)) • c(t) . We then observe that dt b b b f(c(t)) • c(t) dt = F • ds = F(c(t)) • c(t) dt = f(c(t)) = c a a t =a f(c(b)) – f(c(a)) . (See Theorem 3 on page 440.)

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 (x1 , y1 , z1) to (x2 , y2 , z2) will be equal to no matter what path is chosen. f(x2 , y2 , z2) –f(x1 , y1 , z1) (x2 , y2 , z2) (x1 , y1 , z1)

7. 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).

8. c(t) = (1, 2t , 3t2) (e) Find F • ds and F • ds . b(t) = (3 , 15 , 63) c b (t + t2 , t + t3 , – t2–t3) • (1 , 2t , 3t2) = F(c(t)) • c(t) = t + t2+ 2t2 + 2t4 – 3t4 – 3t5 = t + 3t2 – t4 – 3t5 (18t + 2 , 66t + 2 , – 78t– 2) • (3 , 15 , 63) = F(b(t)) • b(t) = 54t + 6 + 990t + 30 – 4914t – 126 = – 3870t – 90 4 10913 – ——— 5 F • ds = (t + 3t2 – t4 – 3t5) dt = c 1 1 F • ds = (– 3870t – 90) dt = – 2025 b 0 (f) Find V • ds and V • ds . V • ds = V • ds = c b c b

9. (f) Find V • ds and V • ds . c b V • ds = V • ds = f(4 , 16 , 64) –f(1 , 1 , 1) = c b (4)2 + (16)2 + (64)2 (1)2 + (1)2 + (1)2 ————————– ——————— = 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 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)? c(t) = (3 cos t , 3 sin t) for 0 t 2 (b) How can we parametrize the path b(t)? 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 .

11. (c) Suppose F(x,y) is a gradient vector field. Find F • ds and F • ds . c b Since F = f for some f(x,y), then we must have F • ds = F • ds = f(3,0) –f(3,0) = 0 c b

12. Look at the first homework problem in Section 7.2 (#2a, page 447): x dy–y dx = c (F1dx+F2dy) Note that the line integral is written in the following form: c This implies that the vector field F = ( ) is integrated over the given path – y , x , 0