Question

In Exercises \(7-12,\) a vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{c} \vec{F} \cdot d \vec{r} .\) \(\vec{F}=\langle y, x\rangle ; C\) is the top half of the unit circle, beginning at (1,0) and ending at (-1,0) .

Short answer

The integral evaluates to 0.

Step-by-step solution

Understand the Problem

We need to evaluate the line integral \( \int_{C} \vec{F} \cdot d\vec{r} \) for the vector field \( \vec{F} = \langle y, x \rangle \) along the curve \( C \), which is the top half of the unit circle starting at (1,0) and ending at (-1,0). This curve can be parametrized, and we need to apply the concept of line integrals over a vector field.

Parametrize the Curve

The top half of the unit circle can be parametrized using the angle \( t \), such that \( x(t) = \cos t \) and \( y(t) = \sin t \), where \( t \) varies from \( 0 \) to \( \pi \). This gives us the curve \( \vec{r}(t) = (\cos t, \sin t) \) for \( t \in [0, \pi] \).

Determine \( d\vec{r} \)

The differential \( d\vec{r} \) is the derivative of \( \vec{r}(t) = (\cos t, \sin t) \) with respect to \( t \), which is \( d\vec{r} = (-\sin t, \cos t) \, dt \).

Compute \( \vec{F} \cdot d\vec{r} \)

Substitute \( \vec{F}(\vec{r}(t)) = \langle \sin t, \cos t \rangle \) and \( d\vec{r} = (-\sin t, \cos t) \, dt \) into \( \vec{F} \cdot d\vec{r} \). This gives \( \vec{F} \cdot d\vec{r} = (\sin t)(-\sin t) + (\cos t)(\cos t) \, dt = (\cos^2 t - \sin^2 t) \, dt \).

Evaluate the Integral

Integrate \( \int_{0}^{\pi} (\cos^2 t - \sin^2 t) \, dt \) using the identity \( \cos 2t = \cos^2 t - \sin^2 t \). This simplifies the integral to \( \int_{0}^{\pi} \cos 2t \, dt \). The antiderivative of \( \cos 2t \) is \( \frac{1}{2}\sin 2t \). Evaluate \( \frac{1}{2}\sin 2t \) from 0 to \( \pi \) to get \( \frac{1}{2}(\sin 2\pi - \sin 0) = 0 \).

Conclusion

The line integral \( \int_{C} \vec{F} \cdot d\vec{r} \) evaluates to 0.

Key concepts

Vector Field

In mathematics, a vector field assigns a vector to each point in a space. For our exercise, the vector field \( \vec{F} = \langle y, x \rangle \) means that at any point \((x, y)\), the vector is comprised of horizontal and vertical components \(y\) and \(x\), respectively. Understanding vector fields is crucial because it allows us to visualize various phenomena, from fluid flow to electromagnetic fields. By assigning vectors to each point on a plane, we can map the direction and magnitude of the object's motion or force experienced at different locations. In the context of line integrals, vector fields help establish how much of the field is influencing an object along a specific path or curve. This is reflected in the dot product \(\vec{F} \cdot d\vec{r}\), which accounts for both the magnitude of the field at a point and how it aligns with the curve.

Parametrization

Parametrization is the process of defining a curve using a parameter, commonly \(t\). For the top half of the unit circle, we use trigonometric functions because they naturally describe circular paths. We set up the parameter \(t\) such that:

• \(x(t) = \cos t\)
• \(y(t) = \sin t\)

This formulates the curve \(\vec{r}(t) = (\cos t, \sin t)\) and neatly captures each point on the unit circle's top half as \(t\) ranges from \(0\) to \(\pi\). By parametrizing a curve, we translate a smooth path into an equation that varies neatly over a specified interval. This method simplifies calculations on complex paths by reducing spatial information into a more manageable form, crucial for integrating vector fields over curves.

Unit Circle

The unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. It is a fundamental concept in trigonometry and calculus since it relates angles to coordinates and lengths in a simple, intuitive way. On the unit circle, any angle \(t\) corresponds to a point \((\cos t, \sin t)\). This circle allows for easy parametrization of curves because of this natural correlation between angles and positions. In the exercise, the curve \(C\) is defined as the top half of the unit circle, starting from point \((1, 0)\) and ending at \((-1, 0)\). It provides a simple yet powerful way to explore fundamental calculus concepts such as line integrals, illustrating the integral of a vector field over a smooth, symmetric path.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables. They play an essential role in simplifying expressions and solving integrals. In the provided solution, the identity \(\cos^2 t - \sin^2 t = \cos 2t\) is used to simplify the integral from a more complex expression into one that's easier to evaluate. This simplification allows for a straightforward integration process. The operation \(\int_{0}^{\pi} \cos 2t \, dt\) yields an antiderivative \(\frac{1}{2} \sin 2t\), which can be easily evaluated to find the integral's value. Trigonometric identities like this one are indispensable tools that help bridge more complex calculus operations with fundamental trigonometric principles.