Problem 22 - Calculus Early Transcendentals

Question

Evaluate the line integral along the curve C. $$ \begin{array}{l}{\int_{C}(y-x) d x+x^{2} y d y} \\ {C: y^{2}=x^{3} \text { from }(1,-1) \text { to }(1,1)}\end{array} $$

Step-by-Step Solution

Answer

The line integral along C is 0, since contributions cancel out.

Step 1: Parameterize the Curve C

The curve is given by the equation \( y^2 = x^3 \). Solving for \( y \) gives \( y = \pm x^{3/2} \). Since the curve moves from \((1, -1)\) to \((1, 1)\), both paths with \( y = -x^{3/2} \) and \( y = x^{3/2} \) are used. We parameterize \( x \) from \( 1 \) to \( 1 \), which is essentially constant in this functional form when the transition reverses direction from \( (1,-1) \to (1,1) \).

Step 2: Evaluate the Integrals Separately for Each Part of C

Since \( C \) consists of two sections, \( C_1: (1, -1) \to (1, 0) \) and \( C_2: (1, 0) \to (1, 1) \), calculate each integral part separately. For \( C_1 \), where \( y = -x^{3/2} \), the line integral involves substituting with this expression: \( \int_{C_1} ((-x^{3/2} - x) \, dx + x^2(-x^{3/2}) \, dy) \) and similarly for \( C_2 \) with \( y = x^{3/2} \).

Step 3: Integrate for Each Segment

For each segment, compute \( \int ((y - x) \, dx + x^2 y \, dy) \) using the specified parameterization. Integrate directly with respect to \( x \), considering that \( dy = \frac{3}{2}x^{1/2}dx \). For both segments, calculate separately: \( \int ((y - x) ) = \int_{1}^{1} (-2) \, dx \) with both \(-x^{3/2}\) and \(x^{3/2}\). The term \(x^2 y \frac{3}{2}x^{1/2} dx\) evaluates similarly across the same limits.

Step 4: Sum the Results of Both Calculations

Add the results from each segment. Each integral will yield zero since the parameter bounds are symmetric despite the flips in the \( y \)-axis for each segment of \( C \). This symmetry reflects that the same function is traversed in opposite directions, causing contributions to cancel each other.

Key Concepts

Parameterization of Curves

The concept of parameterization involves expressing a curve using a parameter, typically denoted by a variable like \( t \). This process allows for an easier evaluation of line integrals through defined curves. Let’s use the given exercise as an illustration. The curve is initially given by the equation \( y^2 = x^3 \). To parameterize this curve, we solve for \( y \) in terms of \( x \), yielding \( y = \pm x^{3/2} \). This indicates two paths concerning the square root function, offering us two expressions for \( y \):

• \( y = -x^{3/2} \)
• \( y = x^{3/2} \)

Within the original problem, the curve transitions from \((1, -1)\) to \((1, 1)\), implicating the need to consider both parameterizations. Parameterization helps break the integral into sections over which the variable ranges incrementally. By transitioning from \((1, -1)\) to \((1, 0)\) using \( y = -x^{3/2} \) and then \((1, 0)\) to \((1, 1)\) with \( y = x^{3/2} \), we maintain complete coverage of the curve's path in manageable steps.

Path Independence

Path independence is a fundamental property of certain types of integrals known as conservative vector fields. When evaluating line integrals over such fields, the integral’s value depends solely on the start and end points, not the path taken between them. In our exercise, the vector field described by the functions \( (y-x) \) and \( x^2 y \) defines the line integral.For the integral to be path independent, there must be a potential function \( f(x, y) \) such that its gradient equals the vector field: \( abla f = (y-x, x^2 y) \). However, within this exercise, the presence of two separated curves \((y = -x^{3/2})\) and \((y = x^{3/2})\) each give distinct paths which hint that path independence may not apply directly. Regardless, because the path from \((1, -1)\) to \((1, 1)\) covers identical distance albeit with direction changes, the symmetry discussed in the problem solution ensures the net integral result would be zero—reflective of balance rather than path independence.

Symmetry in Integration

Symmetry in integration arises when the evaluated integral reverses over a symmetrical path or when cancellations occur due to equivalent, opposite contributions. This is vividly showcased in the given exercise.The given curve, \( C \), is addressed in two symmetrical segments:

• First segment from \((1, -1)\) to \((1, 0)\): follows \( y = -x^{3/2} \)
• Second segment from \((1, 0)\) to \((1, 1)\): switches to \( y = x^{3/2} \)

Both segments' symmetric nature means the respective line integrals mirror each other numerically, yet they oppose in directional traversal along the \( y \)-axis. As a result, even if integrated separately, each part’s motion in opposite directions causes their evaluated results to cancel out each other's contribution entirely. Therefore, the solution reflects how
symmetry plays a pivotal role in causing the final integral value to equal zero. Understanding symmetry helps one anticipate such outcomes in similarly structured problems whereby integral regions counterbalance through spatial or functional symmetry.