§7.2 Line integrals of a vector field In this section, we consider the problem of integrating a vector field along a path. Motivation: (Work done by force F ) Recall Work = Force · Displacement Suppose a particle moves along the path c(t), a ≤ t ≤ b. The particle is experienc- ing the force F (x, y, z ) at the position ( x, y, z ). Along this path, since the direction may be changing, we divide this path into small pieces. The force on each small piece can be written approximately by F (c(t)) · c′(t)∆ t. The total work of this particle along c(t) is b F (c(t)) · c′(t)dt. Za Definition: The line integral of F along the path c(t), a ≤ t ≤ b, is defined to be b F · ds = F (c(t)) · c′(t)dt. Zc Za 5 Example 3. Let F = hx, y, z i and c(t) = (sin( t), cos( t), t ), 0 ≤ t ≤ 2π. Find s c F · d . R 6 §Differential form Let the path c(t) = ( x(t), y (t), z (t)). We write F (x, y, z ) = ( P (x, y, z ), Q (x, y, z ), R (x, y, z )) and also write ds as the differential form ds = ( dx, dy, dz ), then we can rewrite F · ds = P dx + Qdy + Rdz. Zc Zc Moreover, we can derive the following identities: F · ds = P dx + Qdy + Rdz Zc Zc b dx dy dz = P + Q + R dt. Za dt dt dt [Explanation:] 7 Example 4. Evaluate the line integral x2dx + xydy + dz, Zc where c(t) = ( t, t 2, 1) , 0 ≤ t ≤ 1. Method 1: Method 2: 8 §Line integrals of Gradient Fields We call a vector field F is a gradient vector field if F = ∇f, for some scalar valued function f, that is, the vector field F can be expressed as ∂f ∂f ∂f F = , , . ∂x ∂y ∂z Fact. If c(t), a ≤ t ≤ b is a path, and f is a scalar valued function, then ∇f · ds = f(c(b)) − f(c(a)) . (1) Zc Remark: • Recall in Cal. 1, we have known “Fundamental theorem of Calculus (FTC)”: b G′(x)dx = G(b) − G(a). Za Thus, (1) can be viewed as a generalization of FTC for more variables. • If we can recognize the integrand as a gradient, then evaluation of the integral becomes much “easier” by using the generalization of FTC (1). See Example below. 9 2 2 Example 5. Let F (x, y ) = (2 xye x , e x ). Let c(t) = (cos( t), t 2 + 1) , 0 ≤ t ≤ 1. s Find c F · d . R 10 Example 6. Let F (x, y, z ) = (cos( y)ex, − sin( y)ex, π ). Let c(t) = (1 , t, t 2), s 0 ≤ t ≤ 1. Find a function f such that F = ∇f and evaluate c F · d . R 11 §Reparametrization Let c(t) be the path defined on a ≤ t ≤ b. Define the opposite path to c(t) by c−(t) = c(a + b − t). Thus, 1. Line integrals c F · ds (can be interpreted as Work done by force F ): R F · ds = − F · ds Zc− Zc ′ 2. Path integrals c fds = f(c(t)) kc (t)kdt (can be interpreted as Total mass of wire f): R R fds = fds Zc− Zc 12.