When a curve r (t) = g (t)i + h (t)j + k (t)k, a £ t £ b, passes through the domain of a function f (x, y, z) in space, then the values of f along the curve are given by the composite function f (g (t), h (t), k (t)). Now, if we integrate this composite with respect to arc length from t = a to t = b, we will find the line integral of f along the curve.

figure 1

Suppose that f (x, y, z) is a function whose domain contains the curve r (t) = g (t)i + h(t)j + k (t)k, a £ t £ b. (See figure 1) If we partition the curve into a finite number of sub-arcs, then the typical sub-arc has length D S_k. Now, in each sub-arc, pick a point (x_k, y _k, z _k) and form the sum

If f is continuous and the functions g, h, and k have continuous first derivatives, then

Now, we need to determine what ds is. First of all, it is the length of a sub-arc. So, if r (t) is smooth for a £ t £ b (i.e. v = dr/dt is continuous and never 0), then we can use the equation

to express ds = | v (t)| dt. Therefore,

EXAMPLE 1:

Evaluate

along the curve r = 2t i + t j + (2 - 2t) k, 0 £ t £ 1.

SOLUTION:

EXAMPLE 2:

Find the line integral of

over the curve r = t i + t j + t k, 1 £ t £ ¥ .

SOLUTION:

Line integrals have the useful property that if a curve C is made by joining a finite number of curves C₁, C₂, C₃, . . . C _n end to end, then

EXAMPLE 3:

Integrate

over the path from (0, 0, 0) to (1, 1, 1) given by C₁: r = t i + t ² j, 0 £ t £ 1, and C₂: r = i + j + t k, 0 £ t £ 1.

SOLUTION:

EXAMPLE 4:

Integrate

over the curve y = x ²/2 from (1, 1/2) to (0, 0).

SOLUTION:

First we must parameterize the curve so we can find r (t). Let x = t, then y = t ²/2 and r (t) = t i + (t ²/2) j, 0 £ t £ 1.

The line integral helps us determine the centers of mass and moments of inertia of coil springs and wires in space. To do this, we will treat coil springs and wires like masses distributed along a smooth curve in space. The distribution will be described by a continuous density function d (x, y, z). From this, we can calculate the mass and center of mass.

EXAMPLE 5:

Find the mass of a wire that lies along the curve r (t) = (t ² - 1) j + 2t k, 0£ t £ 1, if the density is d = (3/2)t.

SOLUTION:

EXAMPLE 6:

Find the center of mass of a thin wire lying along the curve r = t i + 2t j + (2/3) t ^3/2 k, 0 £ t £ 2, if the density is

SOLUTION: