Problem 33 - Thomas Calculus

Question

In Exercises \(31-34,\) find the circulation and flux of the field \(\mathbf{F}\) around and across the closed semicircular path that consists of the semicircular arch \(\mathbf{r}_{1}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq \pi,\) followed by the line segment \(\mathbf{r}_{2}(t)=t \mathbf{i},-a \leq t \leq a\) $$\mathbf{F}=-y \mathbf{i}+x \mathbf{j}$$

Step-by-Step Solution

Answer

Circulation: \( a^2 \pi \), Flux: 0.

Step 1: Understanding the Path and Field

The path consists of a semicircular arch and a line segment. The semicircular path is described by \( \mathbf{r}_{1}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j} \) for \( 0 \leq t \leq \pi \), and the line segment is described by \( \mathbf{r}_{2}(t)=t \mathbf{i} \) for \( -a \leq t \leq a \). The vector field given is \( \mathbf{F} = -y \mathbf{i} + x \mathbf{j} \).

Step 2: Calculate the Circulation Along Semicircular Path

Calculate circulation by evaluating the line integral of \( \mathbf{F} \cdot d\mathbf{r} \). For the semicircle, use \( \frac{d\mathbf{r}_{1}}{dt} = (-a\sin t)\mathbf{i} + (a\cos t)\mathbf{j} \). The integral becomes \( \int_{0}^{\pi} ((a \sin t) a \sin t + (a \cos t) a \cos t) \, dt = \int_{0}^{\pi} a^2 \, dt = a^2 \pi \).

Step 3: Calculate the Circulation Along Line Segment

Evaluate \( \mathbf{F} \cdot d\mathbf{r} \) along the line segment. Here, \( \frac{d\mathbf{r}_{2}}{dt} = \mathbf{i} \), and \( \mathbf{F} = -t\mathbf{i} \). Thus, the integral is \( \int_{-a}^{a} (-t) \, dt = -\left[ \frac{t^2}{2} \right]_{-a}^{a} = 0 \).

Step 4: Combine Circulation Results

The circulation over the closed path is the sum of the integrals over the semicircle and the line: \( a^2 \pi + 0 = a^2 \pi \).

Step 5: Calculate the Flux Across Semicircular Path

Flux through the semicircle surface is calculated using the divergence theorem. Compute \( abla \cdot \mathbf{F} = 0 + 0 = 0 \), so flux through the entire curve is zero.

Step 6: Verify Flux across Whole Region

The flux across the closed curve is zero, as the divergence of \( \mathbf{F} \) is zero, confirming flux across the entire closed region is zero.

Key Concepts

Line Integrals

When working with vector calculus, understanding line integrals is crucial. A line integral is a type of calculus that helps to determine various physical quantities, such as mass or charge, along a curve. It’s also used in the context of force fields, helping us understand work done by or against a force field along a path.

In the context of the given exercise, we used line integrals to determine the circulation of the vector field around the semicircular path and the straight line segment. You compute a line integral by integrating the dot product of the vector field and the vector differential along the path.

The formula used is \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \), where \( C \) denotes the path and \( d\mathbf{r} \) represents a small vector along the path. For the semicircular section, substitutions and simplifications are made using the path’s parametric equations to solve the integral.

Circulation

Circulation in vector calculus refers to the line integral of a vector field around a closed curve, helping us measure the tendency of the field to circulate around the path. It’s particularly useful in physics to understand the behavior of liquids and gases in a flow, as well as electrical currents.

In our exercise, circulation was evaluated over a semicircular arch and straight line path, utilizing the line integral of \( \mathbf{F} \cdot d\mathbf{r} \). For the semicircular path, we achieved a result of \( a^2 \pi \), showing where the force "works with" or "against" the direction of motion. The line segment part of the path returned zero contribution, indicating balanced forces along this segment.

This cumulative effect on the closed path ultimately provided the final circulation value, revealing insights into the field's behavior along the whole path.

Flux

Flux measures how much of a vector field passes through a given surface. It's like counting the number of arrows passing through a net, giving insight into properties like the flow rate of a fluid or air passing through an area. The flux through a closed surface can often be zero if the vector field doesn't diverge through the surface.

According to the exercise solution, the flux across the semicircular path was calculated using divergence. Since the divergence of the vector field turned out to be zero, the flux through the entire region was confirmed as zero. This zero flux outcome indicates the net amount of the field exiting the boundary of the semicircular surface was exactly canceled by the amount entering.

Understanding flux in various scenarios is essential for dealing with problems related to electromagnetism, fluid dynamics, and more.

Divergence Theorem

The divergence theorem is a fundamental theorem linking the flux of a vector field through a closed surface to the divergence of the field over the volume enclosed. It essentially allows us to convert a complex flux calculation over a closed surface into a simple volume integral of the divergence.

In the given problem, the theorem swiftly helped in verifying that the surface flux of the vector field is zero. The theorem shows that when divergence is zero across an area, the flux is naturally zero as well, without needing extensive surface integrals.

By applying the divergence theorem, we bypass labor-intensive calculations directly linking surface and volume integrals. It's a powerful tool simplifying many complex examinations in vector calculus.