Problem 143 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

The circulation is \( a^2 \pi \) and the flux is 0.

Step 1: Parameterize the Path

Split the path into two segments: the semicircular arch and the line segment. The semicircular arch is parameterized 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 \( \mathbf{r}_{2}(t) = t \mathbf{i} \) for \( -a \leq t \leq a \).

Step 2: Calculate Circulation along Semicircular Arch

For \( \mathbf{r}_{1}(t) \), find \( \dot{\mathbf{r}}_{1}(t) = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} \).Evaluate the line integral \( \int_{0}^{\pi} \mathbf{F}(\mathbf{r}_{1}(t)) \cdot \dot{\mathbf{r}}_{1}(t) \, dt \).\[ \mathbf{F}(\mathbf{r}_{1}(t)) = -a \sin t \mathbf{i} + a \cos t \mathbf{j}, \]\[ \dot{\mathbf{r}}_{1}(t) = -a \sin t \mathbf{i} + a \cos t \mathbf{j}, \]\[ \int_{0}^{\pi} (-a \sin t \cdot -a \sin t + a \cos t \cdot a \cos t) \, dt = \int_{0}^{\pi} a^2 dt = a^2 \pi.\]

Step 3: Calculate Circulation along Line Segment

For \( \mathbf{r}_{2}(t) \), \( \dot{\mathbf{r}}_{2}(t) = \mathbf{i} \).Evaluate the line integral \( \int_{-a}^{a} \mathbf{F}(\mathbf{r}_{2}(t)) \cdot \dot{\mathbf{r}}_{2}(t) \, dt \).\[ \mathbf{F}(\mathbf{r}_{2}(t)) = -0 \mathbf{i} + t \mathbf{j}, \]\[ \dot{\mathbf{r}}_{2}(t) = \mathbf{i}, \]\[ \int_{-a}^{a} 0 \cdot 1 \, dt = 0. \]

Step 4: Calculate Total Circulation

Add the two components from the previous steps to find the total circulation:\[ a^2 \pi + 0 = a^2 \pi. \]

Step 5: Calculate Flux across Semicircular Arch

Evaluate the surface integral using Green's Theorem, which uses the area within the path. Since the region is a semicircle with radius \( a \), the area is \( \frac{1}{2} \pi a^2 \).The flux is given by \( \int \int (Q_x - P_y) \, dA = 2 \cdot \frac{1}{2} \pi a^2 = \pi a^2. \)

Step 6: Calculate Flux across Line Segment

The line segment has no enclosed area for the vertical flux calculation. Therefore, it does not contribute to the flux. The flux across the line segment is 0.

Step 7: Calculate Total Flux

The flux across the closed path is 0 due to the symmetry of the field. Hence, \( \pi a^2 - \pi a^2 = 0. \).

Key Concepts

Circulation

Circulation in vector calculus measures how much a vector field "circulates" around a given curve. Picture water flowing in a river - circulation would quantify how much that water swirls around a particular path.
For vector fields like \( \mathbf{F} = -y \mathbf{i} + x \mathbf{j} \), we can find circulation by integrating along a parameterized path. Consider a semicircular arch and a straight line segment, as noted in the exercise.

• The circulation over the semicircular part is calculated using line integrals, where we follow the flow of the field over the path's curvature.


• For the semicircular arch, the integral evaluates to \( a^2 \pi \), representing the total swirl around the semicircle.


• For the line segment, the circulation is zero as the field doesn't induce any "spin."

Adding these gives a total circulation of \( a^2 \pi \) for the closed path.

Flux

Flux measures how much of a field passes through a surface. Imagine how sunlight streams through a window; flux calculates the total amount of that light. In vector calculus, it helps us understand the "flow" through surfaces and areas.
The flux of a vector field like \( \mathbf{F} = -y \mathbf{i} + x \mathbf{j} \) across a surface involves evaluating a surface integral. Here, we focus on two path sections:

• For the semicircular path, Green's Theorem aids us by connecting the circulation to flux across the enclosed area. We calculate a positive flux of \( \pi a^2 \).


• For the line segment, there is no area through which the field flows, resulting in zero flux.

Ultimately, while the semicircular path contributes a substantial flux, the line segment doesn't alter the total, leading to a net flux of zero due to symmetrical cancellation of field lines.

Closed Path Integral

A closed path integral involves moving along a closed loop or path and summing a function's values over this loop. It allows us to determine properties like circulation or net change when the path returns to its starting point.
For vector fields, closed path integrals are key for understanding behavior across complete circuits.
In our exercise:

• The semicircular arch and the straight line form the closed path, enabling us to calculate a closed line integral for circulation.


• We integrate along each segment of the path, adding together the contributions from both parts. For this problem, the closed path integral simplifies things by summing these individual integrals, resulting in total circulation \( a^2 \pi \).

This highlights how closed path integrals often simplify complex calculations by bundling together simpler components.

Green's Theorem

Green's Theorem is a powerful tool in vector calculus, linking line integrals around a simple closed curve to a double integral over the plane region bounded by the curve. It is the dimensional equivalent of the fundamental theorem of calculus for integrals involving paths and is a cornerstone for analyzing planar fields.
In this specific exercise:

• Green's Theorem is employed to convert path integrals into area integrals, simplifying the process of computing flux over a semicircular region.


• The theorem states \( \oint_{C} P\,dx + Q\,dy = \iint_{R} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})\,dA \), where \( P = -y \) and \( Q = x \).


• This exercise shows its strength, as it simplifies solving for flux by using the area enclosed by the path, leading to efficient computation of \( \pi a^2 \).

Thus, Green's Theorem transforms what could be a complicated line integral problem into a more tangible area-based calculation.