Problem 28 - Thomas Calculus

Question

Zero circulation Let \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\) . Show that the clockwise circulation of the field \(\mathbf{F}=\nabla f\) around the circle \(x^{2}+y^{2}=a^{2}\) in the \(x y\) -plane is zero \begin{equation} \begin{array}{l}{\text { a. by taking } \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi, \text { and inte- }} \\ {\text { grating } \mathbf{F} \cdot d \mathbf{r} \text { over the circle. }} \\ {\text { b. by applying Stokes' Theorem. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The clockwise circulation of \( \mathbf{F} \) around the circle is zero.

Step 1: Calculate Gradient to Find F

To find the vector field \( \mathbf{F} \), we need to compute the gradient \( abla f \) of the scalar function \( f(x, y, z) = \left(x^2 + y^2 + z^2\right)^{-1/2} \). This involves finding the partial derivatives with respect to \( x \), \( y \), and \( z \).

Step 2: Define the Curve in Parametric Form

We use the given parametric equation \( \mathbf{r}(t) = (a \cos t)\mathbf{i} + (a \sin t)\mathbf{j} \) for the circle, where \( 0 \leq t \leq 2\pi \).

Step 3: Compute the Differentials

Find \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt \). In our parametric form, differentiate \( \mathbf{r}(t) \) with respect to \( t \): \( d\mathbf{r} = (-a \sin t) \mathbf{i} dt + (a \cos t) \mathbf{j} dt \).

Step 4: Evaluate the Line Integral Directly

Integrate \( \mathbf{F} \cdot d\mathbf{r} \) over the interval from \( 0 \) to \( 2\pi \). Since \( \mathbf{F} = abla f \) and we are integrating over a closed path (circle), by the conservative nature of gradient fields, this integral evaluates to zero.

Step 5: Apply Stokes' Theorem

Stokes' Theorem states that \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} \), where \( S \) is any surface bounded by the path \( C \). Compute \( abla \times \mathbf{F} \) and observe that the curl is zero as \( abla f \) is a gradient of a scalar field, confirming that the circulation is zero as well.

Key Concepts

Gradient Field

In vector calculus, a gradient field is a vector field that can be expressed as the gradient of a scalar function. The gradient of a scalar function, such as \( f(x, y, z) = (x^2 + y^2 + z^2)^{-1/2} \), is computed by finding the partial derivatives with respect to each spatial variable. Each partial derivative forms a component of the vector field:

• \( \frac{\partial f}{\partial x} = -\frac{x}{(x^2 + y^2 + z^2)^{3/2}} \)
• \( \frac{\partial f}{\partial y} = -\frac{y}{(x^2 + y^2 + z^2)^{3/2}} \)
• \( \frac{\partial f}{\partial z} = -\frac{z}{(x^2 + y^2 + z^2)^{3/2}} \)

Together, these form the vector field \( \mathbf{F} = abla f \), which is simply the derivative of the scalar field in space. A key feature of gradient fields is that they are conservative, meaning any line integral over a closed path in such a field must be zero. This is because conservative fields have no "circulation" or "curl." Another important property of gradient fields is that they are irrotational, meaning their curl is zero. This foundational concept will be used when applying Stokes' Theorem later on.

Line Integral

A line integral can be thought of as a generalization of the usual integral, but over a curve in space rather than along a straight line. In the context of vector fields, the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) measures the work done by the vector field \( \mathbf{F} \) along a path or curve \( C \). For our exercise, the curve \( C \) is given by the parametric equations \( \mathbf{r}(t) = (a \cos t)\mathbf{i} + (a \sin t)\mathbf{j} \) for \( 0 \leq t \leq 2\pi \).
To compute this line integral, we find \( d\mathbf{r} \) by differentiating \( \mathbf{r}(t) \) with respect to \( t \):

• \( d\mathbf{r} = (-a \sin t) \mathbf{i} \, dt + (a \cos t) \mathbf{j} \, dt \)

Here, the differential path element \( d\mathbf{r} \) reflects the tangent to the curve as it varies with \( t \).
Since \( \mathbf{F} = abla f \) and acknowledges the conservative nature of gradient fields, integrating over the closed circle yields zero. This demonstrates the zero circulation of the vector field around the curve.

Stokes' Theorem

Stokes' Theorem establishes a fundamental relationship between a vector field's curl and its line integral around a closed loop. It states that:\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} \]where:

• \( C \) is the boundary of the surface \( S \)
• \( abla \times \mathbf{F} \) is the curl of \( \mathbf{F} \)

In this exercise, the surface \( S \) can be any surface whose boundary is the given circle in the \( xy \)-plane.
To apply Stokes' Theorem, compute the curl of the vector field \( \mathbf{F} \), and verify that it is zero because \( \mathbf{F} \) is a gradient field. If the curl is zero, the right-hand side of Stokes' Theorem is also zero, confirming that the integral of \( \mathbf{F} \) over the circle is zero too.
Stokes' Theorem is significant because it shifts the problem from a line integral to a surface integral, and often simplifies verification by using properties of the vector field, such as whether the field is conservative or irrotational.

Parametric Equations

Parametric equations are a convenient way to express curves without explicitly stating \( x \) or \( y \) as a single function of the other. They define both \( x \) and \( y \) in terms of another variable, typically referred to as \( t \). For example, the circle \( x^2 + y^2 = a^2 \) can be parameterized as:

• \( x = a \cos t \)
• \( y = a \sin t \)

with \( 0 \leq t \leq 2\pi \).
This representation captures the entire circle by allowing \( t \) to vary over the interval \( 0 \leq t \leq 2\pi \).
Using parametric equations simplifies many mathematical operations, such as integrating over curves, by replacing the multi-variable calculus approach with a single variable calculus along the parameter \( t \). This is particularly useful in vector calculus, where following the path of the curve step-by-step is important for operations such as line integrals.