Problem 34 - 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 \(\mathbf{r}_{2}(t)=t \mathbf{i},-a \leq t \leq a\) $$\mathbf{F}=-y^{2} \mathbf{i}+x^{2} \mathbf{j}$$

Step-by-Step Solution

Answer

Both circulation and flux around path C are 0.

Step 1: Understand the Path C

The path \( C \) is a combination of a semicircular arch and a straight line segment. The semicircular arch is parameterized as \( \mathbf{r}_1(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \) for \( 0 \leq t \leq \pi \). The line segment is parameterized as \( \mathbf{r}_2(t) = t \mathbf{i} \) for \( -a \leq t \leq a \).

Step 2: Define the Field \(\mathbf{F}\)

The vector field is given by \( \mathbf{F} = -y^2 \mathbf{i} + x^2 \mathbf{j} \). This means that for any point \((x, y)\), \( \mathbf{F} \) equals \( -y^2 \) in the \( \mathbf{i} \) direction and \( x^2 \) in the \( \mathbf{j} \) direction.

Step 3: Circulation over \(\mathbf{r}_1(t)\)

Calculate the line integral of \( \mathbf{F} \) along the semicircular path \(\mathbf{r}_1(t)\). The tangent vector is \( \mathbf{r}_1'(t) = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} \). The circulation is given by \( \int_{0}^{\pi} \left( -a^2 \sin^2 t + a^2 \cos^2 t\right) dt = \int_{0}^{\pi} a^2 \cos(2t) dt = 0 \).

Step 4: Circulation over \(\mathbf{r}_2(t)\)

Compute the line integral of \( \mathbf{F} \) along the line segment \(\mathbf{r}_2(t)\). Notice \( y = 0 \) for this path, thus \( \mathbf{F} \) is simplified to \( t^2 \mathbf{j} \). The circulation integral becomes \( \int_{-a}^{a} 0 \cdot dt = 0 \).

Step 5: Flux over \(\mathbf{r}_1(t)\)

Use the normal vector to calculate the flux across semicircular \(\mathbf{r}_1(t)\). The outward normal requires rotating \( \mathbf{r}_1'(t) \) by 90 degrees, resulting in \( \mathbf{n} = (-a \cos t \mathbf{i} - a \sin t \mathbf{j}) \). The dot product \( \mathbf{F}\cdot\mathbf{n} \) yields \( a^3 \sin^3 t + a^3 \cos^3 t\), and the flux integral becomes zero after evaluation.

Step 6: Flux over \(\mathbf{r}_2(t)\)

For the segment \(\mathbf{r}_2(t)\), since \(y = 0\) and all flow is horizontal, the normal vector along \(x\)-axis contributes no flux. Therefore, the integral for flux across \(\mathbf{r}_2(t)\) results in zero.

Key Concepts

Line Integrals

Line integrals are a fundamental concept in vector calculus used to compute quantities along a curve. Imagine walking along a path in a vector field like a river, where each step experiences a force or a pull to a particular direction at each point.
This "path integral" helps us calculate important values such as the work done by a force or total flow across a boundary. In our problem, we consider two parts: the semicircular arch and the straight line back.

• For the semicircular path, we determine the force's alignment to the path using the integral: \( \int\mathbf{F} \cdot d\mathbf{r} \), where \(d\mathbf{r}\) is the tangent vector.
• If the field's force aligns perfectly with the path, the integral yields a non-zero value, representing how much the field pushes along the curve.
• However, if the majority pushes in opposition or tangentially cancels out, the integral results in less or even zero.

Overall, for a precise output in our exercise, it was noted that both path segments produced a zero line integral, illustrating that there's no net energy transfer along the path.

Flux Calculation

Flux in vector calculus is akin to measuring how much a fluid flows through a surface. Our exercise asks for the flux over two connected paths—a curve and a straight line—where the vector field \( \mathbf{F} \) exerts a flow-like influence.

• To find the flux through these paths, one must calculate the dot product of the field and the unit normal vector along the path, yielding \( \mathbf{F} \cdot \mathbf{n} \).
• The result reflects how much of the vector field is passing through or is normal to the boundary.
• In the semicircular arc, aligning the field's normals to the adjusted path direction yields an integrated flux of zero.This is due to symmetrical distribution and equal oppositional components of force.
• Flipping to the straight line, the recognized field doesn’t flow perpendicularly, maintaining a horizontal direction, leading the flux integral through it also to zero.

This outcome highlights how field orientation critically affects the flux calculations across open and closed paths.

Circulation in Vector Fields

Circulation measures how much a vector field winds around a closed path and is used to evaluate whether a field is conservative or rotational.
The exercise engages us in computing the circulation around a composite path: a semicircular arch and a line.

• For the semicircular arch, the field's components are analyzed for their overall rotational effect using the circulation formula \( \oint_C \mathbf{F} \cdot d\mathbf{r} \).
• Here, while initially calculated components \( -a^2 \sin^2 t + a^2 \cos^2 t \) seem like potential contributors, when evaluated through integration, they culminate in zero net circulation.
• Shifting to the straight line path, the field net effect, oriented along the direct path and residing solely in one dimension (\( t^2 \mathbf{j} \)), restricts rotational contribution, yielding zero when integrated over symmetrical bounds.

This confirms that the vector field exerts no cumulative rotational influence along our closed system.

Parameterization of Paths

Parameterization of paths is a technique employed to describe curves using a single variable, often simplifying calculations over a region.
In this exercise, two types of geometric paths are parameterized to ease the integration and vector field evaluations.

• The semicircular path, \( \mathbf{r}_1(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \), uses the parameter \( t \), ranging from 0 to \( \pi \). Here, \(a\) specifies the semicircle radius, and trigonometric functions articulate its circular nature.
• In contrast, the line segment, \( \mathbf{r}_2(t) = t \mathbf{i} \), extending from \(-a\) to \(a\), uses a straightforward parameter \( t \), giving a simple linear path along the x-axis.
• This systematic representation turns potentially complex coordinate-based calculations into manageable integrative processes, allowing easy evaluation of effects such as circulation and flux.

Parameterization is a powerful tool in calculus, simplifying how various fields interact with curves in space.