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. a. The circle \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\) b. The ellipse \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(4 \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)

Step-by-Step Solution

Answer

Circulation: \( \mathbf{F}_1=0 \), \( \mathbf{F}_2=2\pi \) on the circle; \( \mathbf{F}_1=0 \), \( \mathbf{F}_2=8\pi \) on the ellipse. Flux: \( \mathbf{F}_1=2\pi \), \( \mathbf{F}_2=0 \) for both curves.

Step 1: Understanding Circulation

The circulation of a vector field \( \mathbf{F} \) around a closed curve \( C \) is given by the line integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) is the tangent vector to the curve. We evaluate this integral for both fields \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \) around the given curves.

Step 2: Calculate Circulation for Field \( \mathbf{F}_1 \) on Circle

First, we have \( \mathbf{F}_1 = x \mathbf{i} + y \mathbf{j} \) and the circle \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \). Substituting \( x = \cos t \) and \( y = \sin t \) into \( \mathbf{F}_1 \) gives \( \mathbf{F}_1 = \cos t \mathbf{i} + \sin t \mathbf{j} \). The tangent vector \( d\mathbf{r} = (-\sin t \mathbf{i} + \cos t \mathbf{j}) dt \). Therefore, \( \mathbf{F}_1 \cdot d\mathbf{r} = (\cos t)(-\sin t) + (\sin t)(\cos t) = 0 \). The circulation is \( \oint_0^{2\pi} 0 \ dt = 0 \).

Step 3: Calculate Circulation for Field \( \mathbf{F}_2 \) on Circle

With \( \mathbf{F}_2 = -y \mathbf{i} + x \mathbf{j} = -\sin t \mathbf{i} + \cos t \mathbf{j} \), and \( d\mathbf{r} \) as before, \( \mathbf{F}_2 \cdot d\mathbf{r} = (-\sin t)(-\sin t) + (\cos t)(\cos t) = 1 \). The circulation is \( \oint_0^{2\pi} 1 \ dt = 2\pi \).

Step 4: Calculate Circulation for Field \( \mathbf{F}_1 \) on Ellipse

The ellipse is \( \mathbf{r}(t) = \cos t \mathbf{i} + 4\sin t \mathbf{j} \). Substituting into \( \mathbf{F}_1 = x \mathbf{i} + y \mathbf{j} \), we get \( \mathbf{F}_1 = \cos t \mathbf{i} + 4\sin t \mathbf{j} \). The tangent vector is \( d\mathbf{r} = (-\sin t \mathbf{i} + 4\cos t \mathbf{j}) dt \). With \( \mathbf{F}_1 \cdot d\mathbf{r} = (\cos t)(-\sin t) + 4(\sin t \cos t) = 3\sin t \cos t \). The circulation evaluates to \( 3 \int_0^{2\pi} \sin t \cos t \ dt = 0 \).

Step 5: Calculate Circulation for Field \( \mathbf{F}_2 \) on Ellipse

For \( \mathbf{F}_2 = -y \mathbf{i} + x \mathbf{j} = -4\sin t \mathbf{i} + \cos t \mathbf{j} \), and \( d\mathbf{r} \) as before, \( \mathbf{F}_2 \cdot d\mathbf{r} = (-4\sin t)(-\sin t) + \cos t \cdot 4\cos t = 4 \). The circulation is \( \oint_0^{2\pi} 4 \ dt = 8\pi \).

Step 6: Understanding Flux

The flux of a vector field \( \mathbf{F} \) across a closed curve \( C \) is given by the line integral \( \oint_C \mathbf{F} \cdot \mathbf{n} \ ds \), where \( \mathbf{n} \) is the unit normal vector to the curve. We will use the normal vector perpendicular to \( d\mathbf{r} \) to evaluate the flux.

Step 7: Calculate Flux for Field \( \mathbf{F}_1 \) across Circle

The normal vector to \( d\mathbf{r} = (-\sin t \mathbf{i} + \cos t \mathbf{j}) \) is \( \mathbf{n} = (\cos t \mathbf{i} + \sin t \mathbf{j}) \). Therefore, \( \mathbf{F}_1 \cdot \mathbf{n} = (\cos t)(\cos t) + (\sin t)(\sin t) = 1 \). The flux is \( \oint_0^{2\pi} 1 \ dt = 2\pi \).

Step 8: Calculate Flux for Field \( \mathbf{F}_2 \) across Circle

For the normal vector \( \mathbf{n} \) as before, \( \mathbf{F}_2 \cdot \mathbf{n} = (-\sin t)(\cos t) + (\cos t)(\sin t) = 0 \). The flux is \( \oint_0^{2\pi} 0 \ dt = 0 \).

Step 9: Calculate Flux for Field \( \mathbf{F}_1 \) across Ellipse

For the ellipse, \( \mathbf{n} = \frac{1}{4}(4\cos t \mathbf{i} + \sin t \mathbf{j}) \), normalizing the vector, the dot product is \( \mathbf{F}_1 \cdot \mathbf{n} = (\cos t)(4\cos t) + 4(\sin t)(\frac{1}{4}\sin t) = 1 \). The flux is \( \int_0^{2\pi} 1 \ dt = 2\pi \).

Step 10: Calculate Flux for Field \( \mathbf{F}_2 \) across Ellipse

Similarly, \( \mathbf{F}_2 \cdot \mathbf{n} = (-4\sin t)(4\cos t) + (\cos t)(\frac{1}{4}\sin t) = 0 \). The flux evaluates to \( \oint_0^{2\pi} 0 \ dt = 0 \).

Key Concepts

Vector Field

A vector field in mathematics is essentially a function that assigns a vector to every point in space. Think of it like a map that shows the direction and strength of a certain force at various locations. For instance, in a gravitational field around Earth, each point in space has a vector pointing towards the center of the Earth, representing the force of gravity at that point.

In this exercise, we look at two different vector fields:

• **\(\mathbf{F}_1 = x \mathbf{i} + y \mathbf{j}\):** This field points outward from the origin, as each vector is aligned with the coordinate pair \((x, y)\) it corresponds to.
• **\(\mathbf{F}_2 = -y \mathbf{i} + x \mathbf{j}\):** This field, on the other hand, produces vectors that are always perpendicular to those of \(\mathbf{F}_1\), swirling around the origin.

Understanding the behavior of vector fields is critical when analyzing circulation and flux, as these attributes depend on how the vectors interact with different paths or curves in space.

Line Integral

The concept of a line integral is vital for calculating circulation and flux in vector fields. Line integrals allow us to accumulate a quantity along a curve. For circulation, they help us understand how a vector field moves along a path;
for flux, they describe the field's action across the path.

Mathematically, the line integral of a vector field \(\mathbf{F}\) over a curve \(C\) is expressed as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \). Here, \(d\mathbf{r}\) is the differential element of the curve, resembling tiny vectors that run tangent to the curve.

In simpler terms, imagine walking along a path with a wind blowing in a particular direction. The line integral would provide the total wind experienced during this journey. If the path is closed, forming a loop, the line integral gives the circulation, indicating how strongly the field encourages motion around that loop.

Closed Curve

A closed curve is a path that starts and ends at the same point without any breaks in between. Examples include circles, ellipses, and any loop that perfectly connects back to its starting position. They are crucial in computing circulation and flux because they provide a bounded path for these calculations.

In our exercise, both the circle and the ellipse are closed curves:

• The **circle** is given by the parametric equations \(\mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j}\), with \(0 \leq t \leq 2\pi\) indicating a full revolution.
• The **ellipse**, on the other hand, has the parametric form \(\mathbf{r}(t) = (\cos t) \mathbf{i} + (4 \sin t) \mathbf{j}\), stretching the circle in the \(\mathbf{j}\) direction.

These curves act like closed loops around which we analyze the behavior of the vector fields, determining how much of the vector field is moving along the curve itself (circulation) or across it (flux).
Studying these interactions on closed curves allows us to explore more complex geometric and physical properties of the fields involved.