Problem 20 - Calculus Early Transcendentals

Question

Evaluate the line integral along the curve C. $$ \begin{array}{l}{\int_{C}\left(x^{2}-y^{2}\right) d x+x d y} \\ {C: x=t^{2 / 3}, \quad y=t \quad(-1 \leq t \leq 1)}\end{array} $$

Step-by-Step Solution

Answer

The line integral evaluates to \(\frac{1}{3}\).

Step 1: Parameterize the curve C

The given curve is parameterized by \( x = t^{2/3} \) and \( y = t \). This means the curve C can be described using the parameter \( t \) within the interval \(-1 \leq t \leq 1\).

Step 2: Express dx and dy in terms of t

Differentiate the parameterization equations with respect to \( t \): \( \frac{dx}{dt} = \frac{2}{3} t^{-1/3} \) and \( \frac{dy}{dt} = 1 \). This means \( dx = \frac{2}{3} t^{-1/3} dt \) and \( dy = dt \).

Step 3: Substitute into the integral

Substitute \( x = t^{2/3} \), \( y = t \), \( dx = \frac{2}{3} t^{-1/3} dt \), and \( dy = dt \) into the integral: \[ \int_{-1}^{1} \left((t^{4/3} - t^2) \frac{2}{3} t^{-1/3} + t^{2/3} \right) dt \]

Step 4: Simplify the integral expression

Simplify the expression within the integral:\( (t^{4/3} - t^2) \frac{2}{3} t^{-1/3} = \frac{2}{3} (t^{4/3 - 1/3} - t^{2 - 1/3}) = \frac{2}{3} (t - t^{5/3}) \) The integral becomes:\[ \int_{-1}^{1} \left( \frac{2}{3} t - \frac{2}{3} t^{5/3} + t^{2/3} \right) dt \]

Step 5: Evaluate the integral

Calculate each part of the integral separately:1. \( \int_{-1}^{1} \frac{2}{3} t \, dt = \frac{2}{3} \left[ \frac{t^2}{2} \right]_{-1}^{1} = \frac{2}{3} \left(1 - \frac{1}{2} \right) = \frac{1}{3} \)2. \( \int_{-1}^{1} -\frac{2}{3} t^{5/3} \, dt = -\frac{2}{3} \left[ \frac{t^{8/3}}{8/3} \right]_{-1}^{1} = 0 \)3. \( \int_{-1}^{1} t^{2/3} \, dt = \left[ \frac{t^{5/3}}{5/3} \right]_{-1}^{1} = 0 \)Add the results: \( \frac{1}{3} + 0 + 0 = \frac{1}{3} \).

Step 6: Conclusion: Value of the line integral

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

Key Concepts

Parameterization

Parameterization is a technique used in mathematics to describe a curve using a parameter, often denoted as \( t \). Instead of expressing \( x \) and \( y \) directly in terms of each other, we express both as functions of \( t \). This allows us to trace the path of the curve as \( t \) varies over a certain interval. For example, if \( x = t^{2/3} \) and \( y = t \), then \( t \) implicitly describes all the points \( (x, y) \) along the curve as \( t \) changes.

Parameterization is quite useful in line integrals since it turns the problem into one-variable calculus by expressing the variables \( x \) and \( y \) as single-variable functions. This step simplifies the process of integration by enabling easier manipulation and substitution of expressions.

Integral Calculus

Integral calculus is a branch of mathematics focusing on the concept of integration. Integration is the process of finding the integral of a function, which can be thought of as the reverse operation to differentiation. In the context of line integrals, integral calculus provides a means to calculate the accumulated value of a field along a curve.

When working with line integrals, we often need to break down a complex integral into simpler parts. This involves techniques such as:

• Substitution, which can simplify the integral by changing variables;
• Recognizing patterns or standard forms that allow for direct integration;
• Breaking down the line integral into smaller, manageable segments.

These techniques help in evaluating integrals effectively and finding solutions to problems involving curves.

Curve Integration

Curve integration refers to the process of integrating a function over a curve in a plane. Unlike standard integration, which deals with a linear path along the x-axis, curve integration involves accounting for the curvature of the path.

To perform curve integration, the curve is typically parameterized, allowing it to be treated as a function of one variable. Once the curve is expressed through parameterization, the differential elements \( dx \) and \( dy \) are derived. By substituting these parameters into the integral, we simplify the expression and calculate the integral over a defined interval of \( t \).

The final step in curve integration is to evaluate the integral using standard calculus techniques. This gives us the value of the line integral, which represents the measure of the vector field along the path given by the curve.