Problem 10 - Thomas Calculus

Question

Evaluate \(\int _ { C } ( x - y + z - 2 ) d s\) where \(C\) is the straight-line segment \(x = t , y = ( 1 - t ) , z = 1 ,\) from \(( 0,1,1 )\) to \(( 1,0,1 )\) .

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Parameterize the Curve

The curve \(C\) is already given in terms of parameter \(t\), where \(x = t\), \(y = 1 - t\), and \(z = 1\). As \(t\) varies from 0 to 1, the point \((x, y, z)\) moves from \((0,1,1)\) to \((1,0,1)\). Hence, we have the parametrization: \(\mathbf{r}(t) = (t, 1-t, 1)\) for \(t\) in \([0, 1]\).

Step 2: Find the Derivative of the Parametrization

We need to find the derivative \(\mathbf{r}'(t)\) of the parametrization \(\mathbf{r}(t) = (t, 1-t, 1)\). Differentiating with respect to \(t\), we get \(\mathbf{r}'(t) = (1, -1, 0)\).

Step 3: Calculate the Magnitude of the Derivative

The magnitude of the derivative \(\mathbf{r}'(t)\) is used to determine the infinitesimal arc length \(ds\). Calculate \(|\mathbf{r}'(t)| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}\). This indicates that \(ds = \sqrt{2} \, dt\).

Step 4: Substitute into the Integral

The integral \(\int_C (x - y + z - 2) ds\) becomes \(\int_0^1 (t - (1-t) + 1 - 2) \sqrt{2} \, dt\). Simplify this to get \((2t - 1)\sqrt{2}\), leading to the integral: \(\int_0^1 (2t - 1) \sqrt{2} \, dt\).

Step 5: Evaluate the Integral

Evaluate the integral \(\sqrt{2} \int_0^1 (2t - 1) \, dt\). The antiderivative of \(2t - 1\) is \(t^2 - t\). Evaluate this from 0 to 1: \((1^2 - 1) - (0^2 - 0) = 0\). Thus, the value of the integral is \(\sqrt{2} \times 0 = 0\).

Key Concepts

Parametrization

Parametrization is a technique that transforms a problem from a complicated form into a simpler one using a parameter, often denoted as \( t \).
In this exercise, parametrization converts a geometric curve into a form that is easier to work with using an equation. The line segment \( C \) is described parametrically by the equations \( x = t \), \( y = 1-t \), and \( z = 1 \), where \( t \) moves from 0 to 1.
By re-defining the position of a point on a curve with respect to \( t \), we form a new description, \( \mathbf{r}(t) = (t, 1-t, 1) \).
This representation helps break down movement into more manageable pieces, leading to a simpler way to evaluate the integral over the curve.

Arc Length

Arc length measures the distance along a curve.
When computing line integrals, it's critical to incorporate arc length to ensure accuracy. The infinitesimal arc length estimated through \( ds \), is crucial for integrating over curves, as it provides the necessary differential piece of a curve’s full length.
To find \( ds \), we first calculate the derivative of the parametrization \( \mathbf{r}(t) \). Deriving \( \mathbf{r}(t) = (t, 1-t, 1) \) gives \( \mathbf{r}'(t) = (1, -1, 0) \).
A key step is finding the magnitude of this derivative, \(|\mathbf{r}'(t)| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}\), showing that each small segment \( ds \) equals \( \sqrt{2} \, dt \). This tiny length element is then used in integral evaluation.

Integral Evaluation

Integral evaluation is the process of calculating the total value of an integral, often signifying summing over a curve.
This process uses the parametrized form and arc length to simplify calculation.
First, substitute the parameters and \( ds = \sqrt{2} \, dt \) into the integral: \( \int_C (x - y + z - 2)\, ds \) becomes \( \int_0^1 (t - (1-t) + 1 - 2)\, \sqrt{2} \, dt \).
By simplifying, it translates to \( (2t - 1)\sqrt{2} \). To solve \( \sqrt{2} \int_0^1 (2t - 1) \, dt \), find the antiderivative, which is \( t^2 - t \).
Evaluating this from 0 to 1 provides \( (1^2 - 1) - (0^2 - 0) = 0 \). Hence, the integral's value is \( \sqrt{2} \times 0 = 0 \), simplifying to \( 0 \). This reflects the symmetry of the curve’s path relative to the integrand.