Problem 22 - 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 \(\begin{array}{l}{\text { a. by taking } \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi, \text { and }} \\ {\text { integrating } \mathbf{F} \cdot d \mathbf{r} \text { over the circle. }} \\\ {\text { b. by applying Stokes' Theorem. }}\end{array}\)

Step-by-Step Solution

Answer

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

Step 1: Calculate the Gradient of f

The function given is \( f(x, y, z) = (x^2 + y^2 + z^2)^{-1/2} \). To find \( abla f \), we compute the partial derivatives: \[ \frac{\partial f}{\partial x} = -x(x^2 + y^2 + z^2)^{-3/2}, \quad \frac{\partial f}{\partial y} = -y(x^2 + y^2 + z^2)^{-3/2}, \quad \frac{\partial f}{\partial z} = -z(x^2 + y^2 + z^2)^{-3/2}. \] Therefore, \( abla f = \left( -\frac{x}{(x^2 + y^2 + z^2)^{3/2}}, -\frac{y}{(x^2 + y^2 + z^2)^{3/2}}, -\frac{z}{(x^2 + y^2 + z^2)^{3/2}} \right) \).

Step 2: Parameterize the Circle in the xy-plane

We are given the circle \( x^2 + y^2 = a^2 \). This can be parameterized using the vector function \( \mathbf{r}(t) = (a\cos t)\mathbf{i} + (a\sin t)\mathbf{j} \) for \( 0 \leq t \leq 2\pi \).

Step 3: Compute dr and F · dr

Differentiating \( \mathbf{r}(t) = (a\cos t)\mathbf{i} + (a\sin t)\mathbf{j} \) gives \( d\mathbf{r} = (-a\sin t)\mathbf{i} dt + (a\cos t)\mathbf{j} dt \). Then \( \mathbf{F} \cdot d\mathbf{r} = \left( -\frac{a\cos t}{a^3}, -\frac{a\sin t}{a^3}, 0 \right) \cdot \left( -a\sin t \ dt, a\cos t \ dt, 0 \right) = -a^2\sin t \cos t dt + a^2 \cos t \sin t dt = 0 \).

Step 4: Integrate the Dot Product Over the Circle

The line integral becomes \[ \int_{0}^{2\pi} (\mathbf{F} \cdot d\mathbf{r}) = \int_{0}^{2\pi} 0 \ dt = 0 \]. Thus, the circulation is zero.

Step 5: Verify Results Using Stokes' Theorem

Stokes' Theorem relates the circulation of a vector field around a closed loop to the surface integral of the curl of the field. \( abla \times abla f = \mathbf{0} \) by definition of the gradient, because the curl of a gradient field is always zero. Thus, the integral of the curl over any surface is zero, confirming that the circulation is zero.

Key Concepts

Gradient

The gradient is a vector operation that indicates the direction of the most rapid increase of a function. For a scalar field, like our function \( f(x, y, z) = (x^2 + y^2 + z^2)^{-1/2} \), the gradient, denoted \( abla f \), is a vector field. It is formed by taking the partial derivatives of \( f \) with respect to each variable:

• \( \frac{\partial f}{\partial x} \)
• \( \frac{\partial f}{\partial y} \)
• \( \frac{\partial f}{\partial z} \)

This gives us a vector that points in the direction of the function's greatest rate of increase and whose magnitude is the rate of increase. In the given problem, the gradient was calculated as \( abla f = \left( -\frac{x}{(x^2 + y^2 + z^2)^{3/2}}, -\frac{y}{(x^2 + y^2 + z^2)^{3/2}}, -\frac{z}{(x^2 + y^2 + z^2)^{3/2}} \right) \). This represents the vector field \( \mathbf{F} \). A key takeaway is that gradients are derived from scalar fields, and they point towards the steepest ascent of the field.

Curl

The concept of curl is used to measure the rotation of a vector field. When dealing with a vector field \( \mathbf{F} \), the curl is a new vector field, denoted by \( abla \times \mathbf{F} \). It provides a way to describe the ``twisting" or ``circulating" nature of the field.
For gradient fields, such as \( \mathbf{F} = abla f \), the curl is always zero. This is because gradients result from scalar fields that change smoothly and without rotating.

• Mathematically, for any scalar field \( f \), \( abla \times (abla f) = \mathbf{0} \).

In our specific problem, this property is used to verify the results using Stokes' Theorem. Since the curl of \( abla f \) is zero, the assignment confirmed no circulation around the given path.

Line Integral

The line integral is a method used to integrate functions over a curve. In simple terms, it allows us to sum up the influence of a field along a path.
When a vector field \( \mathbf{F} \) is integrated along a curve \( C \), often parameterized by a vector function like \( \mathbf{r}(t) \), it gives us a measure of the field's influence along that path. The integral can be expressed as:\[ \int_C \mathbf{F} \cdot d\mathbf{r} \]In the exercise, the path is a circle in the \( xy \)-plane, and the field’s influence was computed to be zero, confirming there is no net circulation. This shows that the field does not impart any cumulative effect along the circle.

Vector Field

A vector field assigns a vector to every point in space. It's a fundamental concept in understanding physical phenomena such as magnetic and gravitational fields.
In the exercise, the vector field \( \mathbf{F} = abla f \) was derived from the scalar function \( f \). This means at every point \((x, y, z)\), the vector field gives a direction and magnitude based on the gradient of \( f \).

• This concept is particularly important for visualizing and analyzing the behavior of fields in physics and engineering.
• Vector fields can represent various natural forces and flows, offering insight into the distribution and direction of those forces.

In the problem, understanding how the vector field behaves was crucial for concluding that its circulation around a closed path is zero, in line with Stokes’ Theorem.