Problem 21 - Calculus: Early Transcendental Functions

Question

Evaluate the line integral. \(\int_{C} 4 z d s,\) where \(C\) is the line segment from (1,0,1) to (2,-2,2)

Step-by-Step Solution

Answer

The value of the line integral \(\int_{C} 4 z ds,\) over the line segment from (1,0,1) to (2,-2,2) is 2.

Step 1: Parameterize the line

The first step is to find a parameterization for the curve \(C\). It is a straight line that runs between the points (1,0,1) and (2, -2, 2). We generally use the variable \(t\) for parameterization. So let's start to parameterize the curve \(C\) by linear interpolating each component as a function of \(t\). \((x, y, z) = (1-t)+2t, -2t, (1-t) +2t\)

Step 2: Evaluation of \(ds\)

The element of arc length \(ds\) in three-dimensional space for parametrized curves can be defined using the Pythagorean theorem. For our parameterized line \(ds\) can be represented as \(\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} dt \). As our curve is a line segment, \((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = 1\). Therefore, \(ds=dt\).

Step 3: Substitute parameterization and \(ds\) into integral

Now we can substitute \((x, y, z) = (1-t)+2t, -2t, (1-t) +2t\), and \(ds=dt\) into the integral. This transforms the given line integral over the curve \(C\) in three-dimensional space into a simple integral over the parameter \(t\). The integral becomes: \(\int_{0}^{1} 4((1-t) + 2t) dt\).

Step 4: Evaluate the integral

Evaluating the integrated function yields: \(\int_{0}^{1} 4((1-t) + 2t) dt\) = \([4t - 2t^2]|_{0}^{1}\) = \(4-2\)=2.

Key Concepts

Parameterization of a Curve

Parameterization is a technique used to describe a curve by defining a continuous function that maps a single parameter, usually noted as \(t\), to points on the curve. In the context of the exercise, the parameterization allows us to express the line segment, or curve \(C\), as a function of \(t\). This is particularly useful for converting the complex problem of a line integral along a spatial curve into a manageable one-dimensional integral over the parameter

For instance, to parameterize a line segment between two points \((1,0,1)\) and \((2,-2,2)\), we linearly interpolate each component with respect to \(t\). The parameter \(t\) usually varies between 0 and 1, where \(t=0\) represents the starting point of the curve, and \(t=1\) represents the end point. The line segment's parameterization given in the solution is a function of \(t\) that linearly combines the starting and ending point coordinates, resulting in \((x, y, z) = (1-t)+2t, -2t, (1-t) +2t\).

Arc Length (ds) in Parametrized Curves

The concept of arc length, denoted by \(ds\), is crucial in calculating line integrals, especially for parameterized curves. The arc length element for a curve in three-dimensional space, when the curve is given by a parameterization, is calculated using the derivative of the parameterized function with respect to the parameter \(t\)

For the given parameterization of a straight line, \(ds\) simplifies significantly. As we take the derivative of each component with respect to \(t\) and use the Pythagorean theorem, for this straight line, we find that the sum of the squares of the derivatives \( (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 \) equals 1. Thus, the differential arc length \(ds\) is simply equal to \(dt\). This simplification occurs because the line segment's direction does not change, leading to a direct correspondence between the infinitesimal movements along the curve and the parameter interval.

Evaluating Integrals

Evaluating integrals is the final step in solving line integrals of parameterized curves. Once we’ve parameterized the curve and have an expression for \(ds\), we substitute these into the integral to find a single-variable integral with respect to \(t\).

In our exercise, after substitution, the problem transforms into evaluating the integral \(\int_{0}^{1} 4((1-t) + 2t) dt\). This becomes a basic calculus problem, where we integrate the function with respect to \(t\) and apply the limits of integration, which correspond to the bounds of the parameter.

Definite Integration

Here, we use definite integration, which gives us the net area under the curve of the function within specified limits. By integrating our function from \(t=0\) to \(t=1\), we perform a straightforward computation: \(\int_{0}^{1} 4((1-t) + 2t) dt = [4t - 2t^2]|_{0}^{1} = 4 - 2 = 2\). This result is the value of the line integral over the curve \(C\), showcasing how parameterization and the evaluation of \(ds\) facilitate the solution of complex spatial problems.