Chapter 16: Problem 22
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 a. by taking \(\mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi,\) and integrating \(\mathbf{F} \cdot d \mathbf{r}\) over the circle. b. by applying Stokes' Theorem.
Short Answer
Step by step solution
Calculate Gradient \(\nabla f\)
Parameterize the Circle
Differentiate the Parameterization
Compute \(\mathbf{F} \cdot d\mathbf{r}\)
Integrate \(\mathbf{F} \cdot d\mathbf{r}\) Over the Circle
Apply Stokes' Theorem
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
- The gradient vector \(abla f\) has components representing how much \(f\) changes as you move along the \(x\), \(y\), and \(z\) axes.
- A higher magnitude of the gradient indicates a steeper slope or a more rapid change in that direction.
- If the gradient is zero, the function is at a local maximum, local minimum, or a saddle point at that specific location.
Stokes' Theorem
- The left side of the equation is a line integral over the closed curve \(C\), which computes the circulation of the vector field \(\mathbf{F}\) along that curve.
- The right side involves taking the curl of \(\mathbf{F}\), integrated over some surface \(S\) that \(C\) bounds.
- If the vector field is the gradient of a scalar field \(\mathbf{F} = abla f\), then \(abla \times \mathbf{F} = \mathbf{0}\) because the curl of a gradient is always zero. This simplifies the calculation greatly.
Parameterization of Curves
- Here, \(a\) represents the radius of the circle.
- \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors along the \(x\) and \(y\) axes, respectively.
- The parameter \(t\) runs from 0 to \(2\pi\), ensuring a full loop around the circle.