is parametrized by a variable t that describes motion along the curve at a variable speed kx0(t)k. Rather than reparameterizing the path in terms of s, another way to directly give the line integral is Line integrals Math 131 Multivariate Calculus Z Z b D Joyce, Spring 2014 f ds = f(x(t)) kx0(t)k dt. x a Preview. We’ll start our study of line integrals. For the next few meetings we’ll study line integrals. Example 1. Evaluate the scalar line integral First, we’ll see just what line integrals are, both Z scalar line integrals and vector line integrals. Sec- (3x + xy + z3) ds ond, we’ll discuss Green’s theorem. Green’s theo- x rem relates a double integral over a plane region to a line integral around the boundary of that plane where x is the path x(t) = (cos 4t, sin 4t, 3t) for region. Finally, we’ll look at conservative vector t ∈ [0, 2π]. fields and curls. Vector line integrals. The vector line integrals Scalar line integrals. So, what is a line integral? we’re going to look at are integrals of vectors of It’s an integral along a curve. An ordinary integral, Z b a vector field F dotted with unit tangent vectors f(s) ds, is the integral along the line segment T for the curves. Since the integrand is actually a a scalar, the dot product of vectors, the value of [a, b]. We’ll generalize this integral by changing the these integrals is also a scalar. line segment [a, b] of the x-axis to a segment of any curve in Rn. When we do that, we’ll take the inte- A vector line integral of a vector-valued function n m n grand f(x) to be a scalar-valued function Rn → R. F : R → R along a path x :[a, b] → R is the (Soon, we’ll look at vector-valued functions, too.) integral The natural parameterization of a curve is by its Z b F(x(t)) · x0(t) dt arclength s that we discussed a while ago. For that a parameterization, a path x : R → Rn in Rn has the 0 Z property that kx (s)k = 1 for all s. It means that a which we’ll also denote F · ds. Note that the ds point is travelling the path at unit speed. At each x point x(s) on the path, the function f has a value, in this notation is a vector, not the scalar ds we Z b just used for the scalar line integrals. f(x(s)). Just as the ordinary integral, f(s) ds, These vector line integrals can be given in terms a sums the values of f along the line segment [a, b] of the unit tangent vector T of a path. Recall that Z b T is defined by on the x-axis, the line integral, f(x(s)) ds, sums a x0(t) dx the values of f along the path x(s) as s varies from T(t) = = . a to b. This line integral is variously denoted kx0(t)k ds Z Z Z b f = f ds = f(x(s)) ds. The vector line integral equals x x a Typically, a path is not given already Z parametrized by its arclength s. Instead, it (F · T) ds, x

1 R and here’s the proof. Then we can rewrite the vector integral x F · ds as

Z b Z F(x(t)) · x0(t) dt F · ds x a b b Z Z x0(t) 0 = F(x(t)) · kx0(t)k dt = F(x(t)) · x (t) dt 0 a a kx (t)k Z Z b = (F · T) ds = (M x0(t) + N y0(t) + P z0(t)) dt x a Z b  dx dy dz  Thus, if we interpret the vector differential ds as = M dt + N dt + P dt a dt dt dt being the product T ds, we can literally define the Z b vector line integral as = M dx + N dy + P dz a Z Z dx F · ds = F · T ds. Here, the differential dx is a notation for dt. x x dt That last integral is usually abbreviated This formulation doesn’t depend on the path, that Z is it doesn’t depend on the speed a moving point M dx + N dy + P dz. goes along the curve. But it does depend on which x end of the curve it starts as. If you traverse the The part of the expression after the integral sign, curve in the other direction, the value of the integral namely, is negated. Thus, the vector line integral depends M dx + N dy + P dz, on the orientation of the curve. is called a differential form. In this course, we’re Example 2. Evaluate the vector line integral taking differential forms as being just notational de- vices, but they can be abstracted in a way to make Z them elements of some abstract algebraic object. (3z, y2, 6z) · ds x Example 3. Evaluate the line integral over the path x(t) = (cos t, sin t, t/3) for t ∈ [0, π]. Z (x2 − y) dx + (x − y2) dy C Differential forms. When the integrand F and where C is the line segment from (1, 1) to (3, 5). the path x are given by their coordinate functions, This will require choosing a parametrization of that vector integrals are written using what are called line segment. differential forms. Let’s stick to R3 for a while. Let the vector field F be given coordinatewise as F = (M,N,P ), that is, Math 131 Home Page at http://math.clarku.edu/~djoyce/ma131/ F(x, y, z) = (M(x, y, z),N(x, y, z),P (x, y, z)) and the path x be given, as usual, as

x(t) = (x(t), y(t), z(t)).

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