Problem 9 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \( \sqrt{2} \).

Step 1: Parameterize the Curve

The curve \(C\) can be parameterized in terms of \(t\) as follows: \(x = t\), \(y = 1-t\), and \(z = 0\). The parameter \(t\) will vary from \(0\) to \(1\).

Step 2: Find the Derivative of the Parameterization

Calculate the derivative of each component of the parameterization with respect to \(t\). We have \(\frac{dx}{dt} = 1\), \(\frac{dy}{dt} = -1\), and \(\frac{dz}{dt} = 0\).

Step 3: Compute the Differential Arc Length

The differential arc length \(ds\) of the curve is calculated as \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt = \sqrt{1^2 + (-1)^2 + 0^2} \, dt = \sqrt{2} \, dt\).

Step 4: Integrate the Function Over the Curve

Substitute \(x = t\) and \(y = 1-t\) into the integral. Therefore, the integral becomes \(\int_{0}^{1} (t + (1-t)) \, \sqrt{2} \, dt = \int_{0}^{1} 1 \, \sqrt{2} \, dt\).

Step 5: Perform the Integration

Integrate the function: \(\int_{0}^{1} 1 \, \sqrt{2} \, dt = \sqrt{2} \cdot (t) \bigg|_{0}^{1} = \sqrt{2} \cdot (1 - 0) = \sqrt{2}\).

Key Concepts

Parameterization

Parameterization is a technique used in mathematics to represent a curve with one or more variables. In line integrals, parameterization allows us to express the coordinates of the points on a curve concerning a single parameter, often denoted by \(t\). This simplifies the process of evaluating integrals over curves.
For instance, in the exercise at hand, the curve \(C\) is a straight-line segment. We have described it using the formulas \(x = t\), \(y = 1 - t\), and \(z = 0\). Here, \(t\) acts as the parameter, varying from \(0\) to \(1\).

• \(t = 0\) corresponds to the point \((0, 1, 0)\).
• \(t = 1\) corresponds to the point \((1, 0, 0)\).

This linear transformation helps us transition smoothly along the curve and is vital for calculating other related quantities, such as arc length.

Arc Length

The arc length of a curve is the distance along the curve between two points. Calculating the arc length of a parameterized curve involves differentiating its components and applying the Pythagorean theorem in the context of calculus.
For our curve \(C\), with parameterization \(x = t\), \(y = 1 - t\), and \(z = 0\), the calculation of arc length utilizes the derivatives of these components:

• \(\frac{dx}{dt} = 1\)
• \(\frac{dy}{dt} = -1\)
• \(\frac{dz}{dt} = 0\)

The differential arc length \(ds\) is determined by the formula:
\[ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt = \sqrt{2} \, dt\]
Thus, it gives us a fundamental element for integration in a line integral, reflecting how the curve 'accumulates' length along its path.

Integral Calculus

Integral calculus is an essential tool for finding the total accumulation of quantities over a given interval, whether they are lengths, areas, volumes, or other types of total sums. Line integrals, which integrate over paths or curves, extend this idea beyond straight segments or simple intervals.
In the context of the exercise, we're tasked with evaluating \(\int_{C}(x+y) \, ds\) over a line where the expression \(x+y\) is integrated over the differential arc length \(ds\). After setting our parameterization and calculating \(ds\), we substitute \(x\) and \(y\) using their parameterized equations.
This results in the integral:
\[\int_{0}^{1} (t + (1-t)) \, \sqrt{2} \, dt = \int_{0}^{1} 1 \, \sqrt{2} \, dt\]
Integral calculus then helps in computing the definite integral over the interval \([0,1]\), yielding \( \sqrt{2} \cdot (1) = \sqrt{2} \).
Understanding line integrals necessitates grasping the power of parameterization and arc length calculations to translate physical curves and spaces into manageable mathematical expressions.