Problem 181 - Calculus Volume 3

Question

Evaluate line integral \(\oint_{C} y^{2} d x+x^{2} d y,\) where \(C\) is the boundary of a triangle with vertices \((0,0),(1,1),\) and \((1,0), \quad\) with the counterclockwise orientation.

Step-by-Step Solution

Answer

The value of the line integral is \( \frac{1}{3} \).

Step 1: Understand the Problem

We need to evaluate the line integral \( \oint_{C} y^{2} dx + x^{2} dy \), where \( C \) is the boundary of the triangle with vertices \( (0,0), (1,1), (1,0) \). The counterclockwise orientation means the path follows the edges \( (0,0) \rightarrow (1,0) \rightarrow (1,1) \rightarrow (0,0) \).

Step 2: Parameterize Each Segment

Consider the three segments of the path \( C \): Segment 1 from \( (0,0) \) to \( (1,0) \), Segment 2 from \( (1,0) \) to \( (1,1) \), and Segment 3 from \( (1,1) \) to \( (0,0) \). We will parameterize each segment separately.

Step 3: Segment 1: Parameterize and Integrate

For the segment \( (0,0) \rightarrow (1,0) \), let \( x = t, y = 0 \) where \( t \) ranges from 0 to 1. Here, \( dx = dt \) and \( dy = 0 \). The integral becomes \( \int_{0}^{1} 0 \cdot dt + \int_{0}^{1} t^{2} \cdot 0 \cdot dt = 0. \)

Step 4: Segment 2: Parameterize and Integrate

For the segment \( (1,0) \rightarrow (1,1) \), let \( x = 1, y = t \) where \( t \) ranges from 0 to 1. Here, \( dx = 0 \) and \( dy = dt \). The integral becomes \( \int_{0}^{1} t^{2} \cdot 0 \cdot dt + \int_{0}^{1} 1^{2} \cdot dt = \int_{0}^{1} dt = 1. \)

Step 5: Segment 3: Parameterize and Integrate

For the segment \( (1,1) \rightarrow (0,0) \), let \( x = 1-t, y = 1-t \) where \( t \) ranges from 0 to 1. Here, \( dx = -dt \) and \( dy = -dt \). The integral becomes \( \int_{0}^{1} (1-t)^2 (-dt) + \int_{0}^{1} (1-t)^2 (-dt). \) Simplifying, you get \( -\int_{0}^{1} 2(1-t)^2 dt. \) Integrate this to find \(- \frac{2}{3}. \)

Step 3: Sum the Integrals

Now sum the integrals for all three segments: \( 0 + 1 - \frac{2}{3} = \frac{1}{3}. \) Thus, the value of the line integral is \( \frac{1}{3}. \)

Key Concepts

Parameterization

When working with line integrals, parameterization is a crucial step. It involves expressing each segment of the path as a function of a single parameter, usually denoted by \( t \). Each point on the path can be described by equations of the form \( x = f(t) \) and \( y = g(t) \), where \( t \) ranges between two specific values, often between 0 and 1.

For example, the path from \((0,0)\) to \((1,0)\) can be represented by \(x = t\) and \(y = 0\). Here, \(t\) spans from 0 to 1, which conveys the entire segment as \(t\) moves from start to end. This process simplifies the integral, as it turns a function of two variables into a function of a single parameter.

This way, integrals that seemed initially difficult to evaluate become straightforward as they turn into single-variable integrals. Success in solving line integrals heavily relies on the careful and correct parameterization of the path sections.

Triangle Boundary

The path, or boundary \(C\), in this exercise is the perimeter of a triangle. The vertices at \((0,0)\), \((1,1)\), and \((1,0)\) depict a right triangle. To evaluate the integral over this boundary, the path needs to be divided into individual segments. Each side of the triangle forms one segment of the boundary.

The triangle's boundary is typically understood by its edges:

• From \((0,0)\) to \((1,0)\) forms the base of the triangle.
• Moving from \((1,0)\) to \((1,1)\) creates the vertical leg.
• Finally, closing the triangle is the hypotenuse from \((1,1)\) back to \((0,0)\).

These segments are tackled separately when computing the integral, as parameterized paths allow for simpler computation of the otherwise complex line integral along the entire triangle's boundary.

Counterclockwise Orientation

Orientation is vital when evaluating line integrals. It determines the direction we traverse around the path. In this example, we're dealing with a counterclockwise orientation around the triangle's boundary.

A counterclockwise orientation means the path starts at \((0,0)\), moves to \((1,0)\), then towards \((1,1)\), and finally returns to \((0,0)\), effectively tracing the triangle in a consistently leftward direction.

This has implications for the parametrization and evaluation of the integrals. Specifically, it works to ensure the full traversal of the entire path is accounted for properly in line integrals, reflecting positively or negatively based on direction. Counterclockwise paths are commonly used, often aligning with the right-hand rule in mathematical conventions and ensuring consistency in calculations.

Integral Evaluation

Evaluating the line integral involves calculating the integrals for each segment and adding them up. After parameterization, each segment from the boundary is integrated separately, usually simplified into straightforward forms.

For instance:

• The segment from \((0,0)\) to \((1,0)\) yielded an integral of 0, due to the constant \(y = 0\) along this line.
• The next segment from \((1,0)\) to \((1,1)\) provided a unit integral of 1 due to the constant \(x = 1\) throughout this path.
• The final segment from \((1,1)\) to \((0,0)\) contributed an integral of \(-\frac{2}{3}\), resulting from canceling contributions from both terms.

Summing these calculations, the total line integral equates to \(\frac{1}{3}\).

Precise calculation and the correct orientation ensure accurate results. Understanding each step in this evaluation paves the way for mastering more complex line integral problems.