Problem 29 - Thomas Calculus

Question

Circulation and flux Find the circulation and flux of the fields $$\mathbf{F}_{1}=x \mathbf{i}+y \mathbf{j} \quad \text { and } \quad \mathbf{F}_{2}=-y \mathbf{i}+x \mathbf{j}$$ around and across each of the following curves. \begin{equation} \begin{array}{l}{\text { a. The circle } \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi} \\ {\text { b. The ellipse } \mathbf{r}(t)=(\cos t) \mathbf{i}+(4 \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi}\end{array} \end{equation}

Step-by-Step Solution

Answer

Circle: Circulation 0 for \(\mathbf{F}_1\), \(2\pi\) for \(\mathbf{F}_2\). Flux \(2\pi\) for \(\mathbf{F}_1\), 0 for \(\mathbf{F}_2\). Ellipse: Circulation 0 for both, or \(2\pi\) for \(\mathbf{F}_2\); flux 0.

Step 1: Understanding the Problem

Given the vector fields \(\mathbf{F}_1 = x \mathbf{i} + y \mathbf{j}\) and \(\mathbf{F}_2 = -y \mathbf{i} + x \mathbf{j}\), we need to find their circulation and flux around and across two specific curves: a circle and an ellipse. Circulation is given by the line integral \(\oint_C \mathbf{F} \cdot d\mathbf{r}\), and flux is the surface integral \(\oint_C \mathbf{F} \cdot \mathbf{n} \, ds\), where \(\mathbf{n}\) is the unit normal and \(ds\) is the arc length element.

Step 2: Parameterizing the Circle

The given parameterization for the circle is \(\mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j}\) for \(0 \leq t \leq 2\pi\). The derivative \(d\mathbf{r}/dt = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j}\), which is used as \(d\mathbf{r}\). The unit tangent vector is in the direction of \(d\mathbf{r}/dt\). For flux, the unit normal is the tangent vector rotated by 90 degrees.

Step 3: Calculating Circulation for Circle

For \(\mathbf{F}_1 = x \mathbf{i} + y \mathbf{j}\), the circulation \(\oint_C \mathbf{F}_1 \cdot d\mathbf{r}\) becomes \(\int_0^{2\pi} ((\cos t)(-\sin t) + (\sin t)(\cos t)) \, dt = 0\). For \(\mathbf{F}_2 = -y \mathbf{i} + x \mathbf{j}\), this results in \(\int_0^{2\pi} ((-\sin t)(-\sin t) + (\cos t)(\cos t)) \, dt = \int_0^{2\pi} (1) \, dt = 2\pi\).

Step 4: Calculating Flux for Circle

Using the parameterization, the unit normal vector is \((\cos t) \mathbf{i} + (\sin t) \mathbf{j}\). For \(\mathbf{F}_1\), the flux is \(\oint_C \mathbf{F}_1 \cdot \mathbf{n} \, ds = \int_0^{2\pi} ((\cos t)(\cos t) + (\sin t)(\sin t)) \, dt = \int_0^{2\pi} (1) \, dt = 2\pi\). For \(\mathbf{F}_2\), the vector normal is orthogonal to this field, yielding \(0\).

Step 5: Parameterizing the Ellipse

The given parameterization for the ellipse is \(\mathbf{r}(t) = (\cos t) \mathbf{i} + (4 \sin t) \mathbf{j}\) for \(0 \leq t \leq 2\pi\). Then \(d\mathbf{r}/dt = (-\sin t) \mathbf{i} + (4 \cos t) \mathbf{j}\), used as \(d\mathbf{r}\). Adjust the unit normal as needed for the flux calculation.

Step 6: Calculating Circulation for Ellipse

For \(\mathbf{F}_1\), this becomes \(\int_0^{2\pi} ((\cos t)(-\sin t) + (4 \sin t)(4 \cos t)) \, dt = 16 \int_0^{2\pi} \sin t \cos t \, dt = 0\) due to symmetrical integration bounds. For \(\mathbf{F}_2\), similarly, the integral simplifies and evaluates to \(2\pi\).

Step 7: Calculating Flux for Ellipse

Using the parameterization, the normal vector is adjusted: \(\mathbf{n}(t) = (4 \cos t) \mathbf{i} + (-\frac{1}{4} \sin t) \mathbf{j}\). Evaluating \(\oint_C \mathbf{F}_1 \cdot \mathbf{n} \, ds = 0\) and \(\mathbf{F}_2\) is orthogonal to \(\mathbf{n}(t)\), it simplifies to \(0\).

Key Concepts

Circulation

Circulation refers to the total "flow" of a vector field around a closed path or curve. It's like if you have a wind field and want to know how much the wind is circulating around your path. The precise mathematical tool used to measure this is the line integral. Specifically, circulation is calculated using the line integral of the vector field along the curve:

• For a vector field \( \mathbf{F} \), the circulation around a curve \( C \) is given by \( \oint_C \mathbf{F} \cdot d\mathbf{r} \).

Where \( d\mathbf{r} \) is the differential element of the curve. The dot product \( \mathbf{F} \cdot d\mathbf{r} \) measures how much \( \mathbf{F} \) is aligned with the tangent to the curve.
In simpler terms, if the vector field \( \mathbf{F} \) flows in the same direction as the curve, circulation accumulates positively, otherwise negatively. For instance, in our original exercise, for field \( \mathbf{F}_1 \) around a circle, the circulation vanished (became zero) because the field didn't align or circulate around the path. But for \( \mathbf{F}_2 \), the completed integral came out as \( 2\pi \), indicating a perfect alignment along the way.

Flux

Flux measures how much of the vector field passes through or "escapes" across a surface or curve. Imagine pouring water through a hoop: the flux would quantify how much water goes through that hoop. For vector fields, flux is determined by the surface integral. For our purposes, when working across a curve, it's tied to the integration of the normal component:

• For a vector field \( \mathbf{F} \), flux across a curve \( C \) is \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds \).

Here, \( \mathbf{n} \) is the unit normal vector, which is perpendicular to the curve, and \( ds \) is the differential arc length element. The product \( \mathbf{F} \cdot \mathbf{n} \) tells us how much \( \mathbf{F} \) flows through or cuts through the surface of interest.
For example, in our exercise, for the circle scenario with \( \mathbf{F}_1 \), the flux came out as \( 2\pi \), showing total flow across the whole circle. In contrast, for \( \mathbf{F}_2 \), the flux was zero, revealing orthogonality to the normal, which meant no net flow was passing through the circle.

Line Integral

The line integral is a powerful tool in vector calculus used to integrate functions along a curve or path. It's comparable to summing up small pieces along the path. In our context, it helps measure either circulation or flux. There are different kinds of line integrals depending on the application:

• When using \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), it calculates how a vector field interacts with the path of integration, typically used for circulation.
• Another variant involves integrating scalar functions along curves.

For a vector field along a parameterized path \( \mathbf{r}(t) \) with parameter \( t \):

• The integral becomes \( \int \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt \).

This evaluates the component of \( \mathbf{F} \) along the tangential direction of the path.
In the exercise, line integrals were employed to determine the circulation for various paths like the circle and ellipse. Calculations considered both vector fields \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \). They helped us quantify the extent of their circulation about those paths.

Surface Integral

Surface integrals extend the concept of integration into two dimensions over a surface, often used for assessing flux. Similar to line integrals, but broadly applied to surfaces instead of curves. These integrals are essential for calculating how vector fields interact with surfaces:

• Given a surface \( S \), one might compute \( \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) is the unit normal to the surface and \( dS \) is a small area element.

These integrals sum up how the field penetrates, flows, or remains tangent to minute surface areas, across the entire surface.
In scenarios from our exercise, although dealing with curves, calculations for flux mimic logic pertinent to surface integrals since unit normal vectors were key. Understanding the surface integral concept is crucial for higher-dimensional flux problems and reveals the beautiful link between analysis along lines and across surfaces in vector calculus.