Problem 26 - Thomas Calculus

Question

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) for the vector field \(\mathbf{F}=y \mathbf{i}-x \mathbf{j}\) counterclockwise along the unit circle \(x^{2}+y^{2}=1\) from \((1,0)\) to \((0,1)\) .

Step-by-Step Solution

Answer

\(-\frac{\pi}{2}\)

Step 1: Understand the problem and convert to polar coordinates

The vector field is \( \mathbf{F} = y \mathbf{i} - x \mathbf{j} \). We must evaluate the line integral of this vector field along the unit circle \( x^2 + y^2 = 1 \) from \((1,0)\) to \((0,1)\). The unit circle suggests a conversion to polar coordinates is convenient, where \( x = \cos \theta \) and \( y = \sin \theta \).

Step 2: Parameterize the path

For a counterclockwise path around the unit circle, we can parameterize \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \) with \( t \) ranging from \( 0 \) (at \( (1, 0) \)) to \( \frac{\pi}{2} \) (at \( (0, 1) \)).

Step 3: Express the differential vector

Calculate \( d\mathbf{r} \), the differential of the vector path: \( d\mathbf{r} = \frac{d}{dt}(\cos t \mathbf{i} + \sin t \mathbf{j}) \ dt = (- \sin t \mathbf{i} + \cos t \mathbf{j}) \ dt \).

Step 4: Substitute into the line integral

The line integral becomes \( \int_{0}^{\frac{\pi}{2}} \mathbf{F}(t) \cdot d\mathbf{r}(t) = \int_{0}^{\frac{\pi}{2}} (\sin t \mathbf{i} - \cos t \mathbf{j}) \cdot (-\sin t \mathbf{i} + \cos t \mathbf{j}) \ dt \).

Step 5: Compute the dot product

Calculate the dot product: \( (\sin t \mathbf{i} - \cos t \mathbf{j}) \cdot (-\sin t \mathbf{i} + \cos t \mathbf{j}) = -(\sin^2 t) - (\cos^2 t) = -1 \).

Step 6: Evaluate the integral

The integral simplifies to \( \int_{0}^{\frac{\pi}{2}} -1 \ dt = -t \Big|_{0}^{\frac{\pi}{2}} = - \frac{\pi}{2} + 0 = -\frac{\pi}{2} \).

Key Concepts

Polar Coordinates

Polar coordinates provide a handy way to represent points on a plane using two numbers: the angle \( \theta \) and the radius \( r \). Here, the unit circle, described by the equation \( x^2 + y^2 = 1 \), is effectively parameterized using polar coordinates because every point on the circle can be described as

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

For the unit circle, \( r \) is constant and equals 1. This simplifies to \( x = \cos \theta \) and \( y = \sin \theta \). By using this conversion, problems that involve circular shapes become much simpler to evaluate. You can think of it as translating from a square grid (Cartesian coordinates) to a circular grid (polar coordinates), thus fitting the mathematical problem better to its geometric context.

Vector Field

A vector field assigns a vector to each point in the plane or space. In this problem, the vector field is given as \( \mathbf{F} = y \mathbf{i} - x \mathbf{j} \).

• The \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors along the x-axis and y-axis respectively.
• The function of the field changes depending on the position.

For instance, at a point \( (x, y) \), the vector is determined by switching and negating the coordinates due to the field's definition. This kind of vector field can be represented in contexts such as magnetic or fluid flow problems, where the directions and magnitudes of the vectors represent the direction and rate of flow.

Parameterization

Parameterization is a method used to define a curve by expressing the coordinates of the points on the curve as functions of a single variable, often denoted as \( t \). For the unit circle, parameterization helps represent the circular path through the functions

• \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \)

This bypasses the need for complicated geometric equations, converting them into simpler trigonometric functions. Here, \( t \) is a variable that starts from \( 0 \) when the path is at \((1, 0)\) and ends at \( \frac{\pi}{2} \) when the path is at \((0, 1)\).Parameterizing the path is essential, as it determines the direction and speed at which the line integral is evaluated, giving a smooth framework for integrating the vector field over the curve.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (such as vectors) and returns a single number. In a vector field operation, it is essential in line integrals. The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y \)

This expression helps in finding how one vector projects onto another, especially in calculating integrals of vector fields. In the given exercise, the dot product is used to integrate the vector field along the path. For vector \( \mathbf{F}(t) = (\sin t, -\cos t) \) and differential path \( d\mathbf{r}(t) = (-\sin t, \cos t) dt \), the dot product simplifies to:

• \( (\sin t \mathbf{i} - \cos t \mathbf{j}) \cdot (-\sin t \mathbf{i} + \cos t \mathbf{j}) = -1 \)

This simplification is key to evaluating the line integral over the path, transforming a vector operation into a manageable scalar integral.