Problem 30 - Thomas Calculus

Question

Flux across a circle Find the flux of the fields $$\mathbf{F}_{1}=2 x \mathbf{i}-3 y \mathbf{j} \quad \text { and } \quad \mathbf{F}_{2}=2 x \mathbf{i}+(x-y) \mathbf{j}$$ across the circle $$\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi$$

Step-by-Step Solution

Answer

Flux for \( \mathbf{F}_1 \) is \(-a^2 \pi\) and for \( \mathbf{F}_2 \) is \(2a^2 \pi\).

Step 1: Parametrize the Circle

The circle is parameterized by the vector function \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \). This represents a circle centered at the origin with radius \( a \).

Step 2: Determine the Tangent Vector

The tangent vector to the circle is found by differentiating \( \mathbf{r}(t) \) with respect to \( t \). This gives \( \mathbf{r}'(t) = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} \).

Step 3: Find the Normal Vector

The outward normal vector to the circle can be determined by rotating the tangent vector \( \mathbf{r}'(t) \) counter-clockwise by 90 degrees. Therefore, \( \mathbf{n}(t) = \frac{\mathbf{r}'(t) \cdot \mathbf{k}}{||\mathbf{r}'(t)||} = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \).

Step 4: Evaluate Flux for \( \mathbf{F}_1 \)

The flux of the field \( \mathbf{F}_1 = 2x \mathbf{i} - 3y \mathbf{j} \) across the circle is given by:\[ \text{Flux} = \int_{0}^{2\pi} \mathbf{F}_1 \cdot \mathbf{n}(t) \, a \, dt \]. Substituting \( x = a \cos t \) and \( y = a \sin t \): \[ \int_{0}^{2\pi} (2a \cos t) \cos t - (3a \sin t) \sin t \; dt \] becomes \[ 2a^2 \int_{0}^{2\pi} \cos^2 t \, dt - 3a^2 \int_{0}^{2\pi} \sin^2 t \, dt \]. Each integral evaluates to \( \pi \), giving \( 2a^2\pi - 3a^2\pi = -a^2\pi \).

Step 5: Evaluate Flux for \( \mathbf{F}_2 \)

The flux of \( \mathbf{F}_2 = 2x \mathbf{i} + (x-y) \mathbf{j} \) is evaluated similarly: \[ \text{Flux} = \int_{0}^{2\pi} (2a \cos t \cos t + a(\cos t - \sin t) \sin t) \, a \, dt \], resulting in \[ 2a^2 \int_{0}^{2\pi} \cos^2 t \, dt + a^2 \int_{0}^{2\pi} (\cos t \sin t - \sin^2 t) \, dt \]. The first integral is \(2a^2\pi\) and the second evaluates to zero, so the result is \(2a^2\pi\).

Key Concepts

vector fields

Vector fields are mathematical constructs where a vector, consisting of both magnitude and direction, is assigned to every point in a region of space. In simpler terms, think of a vector field as an invisible force field like a wind map, where each vector points in the direction the wind is blowing at a specific location.
For example, the vector fields given in the problem— \( \mathbf{F}_1 = 2x\mathbf{i} - 3y\mathbf{j} \) and \( \mathbf{F}_2 = 2x\mathbf{i} + (x-y)\mathbf{j} \)—translate to regions following specific rules:

• \( \mathbf{F}_1 \) references vectors pointing east when the x-value is positive and south gravitational direction when the y-value is positive.
• \( \mathbf{F}_2 \) possesses vectors pulling things eastward and is subjected to change in a complicated pattern depending on both x and y coordinates.

Recognizing the distinct attributes and computations of these fields can elucidate interactions with paths or boundaries, as is the case when calculating flux across a circle.

parametrization

Parametrization is the process of describing a geometric object through equations that depend on a parameter. Consider it a way to make tracing out the shape or path of a curve efficient and mathematical.
In this exercise, the circle is parameterized with \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \). This parametric equation means:

• \( a \cos t \) corresponds to the horizontal distance from the circle's center, thereby defining the circle's width as the parameter \( t \) changes.
• \( a \sin t \) details the vertical distance away from the origin, making it ideal for describing the circle's height.

The range given, \( 0 \leq t \leq 2\pi \), lets the parameter \( t \) trace the entire circle from \( t = 0 \) back to the full round-up at \( t = 2\pi \). This clear mapping is key to further calculating tangents and normals.

tangent vector

A tangent vector at a point on a curve is the vector that points in the direction in which the curve is heading as it still sits at the point itself. It's like predicting the immediate next move of a roller coaster from where you currently stand, moving along the track.
The exercise calculates the tangent vector to the circle as:

• \( \mathbf{r}'(t) = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} \)

The derivative \( \mathbf{r}'(t) \) emerges by differentiating the position vector \( \mathbf{r}(t) \), revealing how the position changes with time.
For instance:

• The \( -a \sin t \) term indicates movement along the i or horizontal axis inverted, reflecting how the circle curves away steadily.
• Likewise, the \( a \cos t \) is the j or vertical directional counterpart, highlighting continuity in upward curvature.

Tangent vectors thus afford insight into the shape's ongoing motion effectively.

normal vector

Normal vectors are perpendicular to the tangent vectors and often lie on the path's surface or curving edge. If you imagine driving and resting perpendicular to a traffic flow on a curving road, the direction you're facing is analogous to a normal vector.
For the circle in question, the normal vector is found as:

• \( \mathbf{n}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \)

Here, the normal vector directly stems from a 90-degree rotation of the corresponding tangent vector. This ensures it stands at right angles to the tangent at any point calculated.
This distinct orientation describes how the path's boundary (circle) interacts with forces (like vector fields) head-on as they approach directly at every point, making it crucial to compute flux effectively.
Whether engaging in physics or engineering, the role of a normal vector proves vital in understanding internal field dynamics as they interface with exterior constraints or shapes.