91Ó°ÊÓ

A line integral allows us to integrate functions along a curve. It is a powerful concept in vector calculus, especially useful for calculating work done by a force field along a path or for determining flux across a curve. In this context, the expression \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) represents a line integral of the vector field \( \mathbf{F} = y \mathbf{i} - x \mathbf{j} \) along a path \( C \). The curve \( C \) is given by the unit circle. Here, the line integral is evaluated by summing up the product of the vector field and a differential path element over the path.
Applying Green's theorem in this context transforms the problem of calculating a line integral around a circle into a simpler calculation of a double integral over the area contained by the circle. This is particularly helpful when the path is closed and simple, allowing us to focus on just the region bounded by the circle.

A vector field is a map that assigns a vector to every point in space. In two dimensions, this can be visualized as a collection of arrows, each pointing in a specific direction with a certain magnitude. In our exercise, the vector field is described by \( \mathbf{F} = y \mathbf{i} - x \mathbf{j} \). Each point \((x, y)\) in the plane is associated with a vector whose components depend on \(x\) and \(y\).

• The \( y \mathbf{i} \) component suggests the vector's horizontal part will vary with the \( y \)-coordinate.
• The \( -x \mathbf{j} \) component implies the vector's vertical part changes negatively with the \( x \)-coordinate.

Understanding this allows us to determine how the vector field behaves over the unit circle, influencing the line integral's evaluation.

A unit circle is a circle centered at the origin \( (0, 0) \) with a radius of 1. Its equation is given by \( x^2 + y^2 = 1 \). The unit circle is a fundamental concept in trigonometry and calculus, and in this exercise, it is the path denoted as \( C \).
By traveling counterclockwise along the unit circle starting from \( (1,0) \) to \( (0,1) \), we traverse one quarter of the circle. This part of the circle is an arc and represents one-fourth of the entire circle. Hence, when the line integral around the entire circle yields a result such as \( -2\pi \), we need to adjust this by the portion of the circle traversed. Thus, the result is divided by 4, reflecting just the quarter circle traveled, giving the final result of \( -\frac{\pi}{2} \).
Understanding the unit circle and the path is crucial because it informs us how much of the vector field's effect is accumulated in the line integral.