Problem 39 - Thomas Calculus

Question

Regions with many holes Green's Theorem holds for a region \(R\) with any finite number of holes as long as the bounding curves are smooth, simple, and closed and we integrate over each component of the boundary in the direction that keeps \(R\) on our immediate left as we go along (see accompanying figure). $$\begin{array}{l}{\text { a. Let } f(x, y)=\ln \left(x^{2}+y^{2}\right) \text { and let } C \text { be the circle }} \\ {x^{2}+y^{2}=a^{2} . \text { Evaluate the flux integral }}\end{array}$$ $$\oint_{C} \nabla f \cdot \mathbf{n} d s$$ b. Let \(K\) be an arbitrary smooth, simple closed curve in the plane that does not pass through \((0,0) .\) Use Green's Theorem to show that $$\oint_{K} \nabla f \cdot \mathbf{n} d s$$ has two possible values, depending on whether \((0,0)\) lies inside \(K\) or outside \(K .\)

Step-by-Step Solution

Answer

Flux around circle \(C\) is \(4\pi\); for \(K\), \(4\pi\) if \((0,0)\) inside, 0 otherwise.

Step 1: Understand the Vector Field

For a function \(f(x, y) = \ln(x^2 + y^2)\), its gradient \(abla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})\) is computed to find the vector field we need. Calculating the partial derivatives, we have: \(\frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2}, \frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2}\). Thus, \(abla f = \left(\frac{2x}{x^2 + y^2}, \frac{2y}{x^2 + y^2}\right)\).

Step 2: Compute Normal Vector

The unit normal vector to a circle \(x^2 + y^2 = a^2\) at any point is given by \(\mathbf{n} = \frac{1}{a}(x, y)\). This is because the gradient \( (x, y) \) of the circle can be normalized to find \(\mathbf{n}\).

Step 3: Set Up Flux Integral Over Circle

The flux integral \(\oint_{C} abla f \cdot \mathbf{n} \, ds\) needs computation. Substitute \(abla f = \left( \frac{2x}{x^2+y^2}, \frac{2y}{x^2+y^2} \right)\) and \(\mathbf{n} = \frac{1}{a}(x, y)\). Then, \(abla f \cdot \mathbf{n} = \frac{2x^2}{a(x^2+y^2)} + \frac{2y^2}{a(x^2+y^2)} = \frac{2(x^2+y^2)}{a(x^2+y^2)} = \frac{2}{a} \). The integral becomes \(\oint_{C} \frac{2}{a} \, ds\).

Step 4: Evaluate Flux Integral Over Circle

Since \(\frac{2}{a}\) is constant along the circle, the integral \(\oint_{C} \frac{2}{a} \, ds = \frac{2}{a} \cdot 2 \pi a = 4 \pi\).

Step 5: Apply Green’s Theorem for Arbitrary Curve

To evaluate the integral \(\oint_{K} abla f \cdot \mathbf{n} \, ds\) over arbitrary curve \(K\), consider whether \((0,0)\) lies inside or outside \(K\). According to Green’s Theorem in vector form, the flux integral over a closed curve with respect to the radial component is essentially contributing \(4\pi\) if \((0,0)\) is inside \(K\), with zero contribution if outside, due to an absence of singularities in that region. The flux present around the origin yields a value of \(4\pi\) if \((0,0)\) is enclosed.

Step 6: Conclusion

\(\oint_{K} abla f \cdot \mathbf{n} \, ds\) is \(4\pi\) if \((0,0)\) is inside \(K\); otherwise, it is 0, as \(abla f\) contributes only when encompassing \((0,0)\).

Key Concepts

Flux Integral

In vector calculus, a flux integral is used to calculate the flow of a vector field through a surface. It quantifies how much of the vector field passes through the surface. In our exercise, we're evaluating the flux integral of the gradient of the function \(f(x, y) = \ln(x^2 + y^2)\) over a closed curve like a circle. The core idea is to integrate the vector field across a curve, which in this case is expressed as \( \oint_{C} abla f \cdot \mathbf{n} \, ds \), where \(abla f\) represents the vector field, and \(\mathbf{n}\) is the unit normal vector to the curve.An essential component of this process is ensuring that the curve is oriented correctly. The orientation must keep the region of interest on the left side as one traverses the curve. This integrates the flux vector's effect encountered in the vector field, ensuring a meaningful interpretation of results, whether inside or outside a given area. When utilizing Green's Theorem, this concept becomes vital, as it relates a flux integral around a closed curve to a double integral over the region it encloses.

Vector Field

The concept of a vector field involves associating a vector to every point in a particular space. For the function \(f(x, y) = \ln(x^2 + y^2)\), this involves computing its gradient, \(abla f\). Here, \(abla f\) gives us the direction and rate of the steepest ascent of the function, essentially mapping out a vector field over the plane.

Calculating the Gradient

The gradient of \(f\) is calculated by taking partial derivatives with respect to \(x\) and \(y\):

• \(\frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2}\)
• \(\frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2}\)

This results in the vector field \(abla f = \left( \frac{2x}{x^2+y^2}, \frac{2y}{x^2+y^2} \right)\). The vector field describes how the function \(f\) changes at any point \((x, y)\), which is critical for understanding the flux integrals. The challenge is interpreting these changes as they relate to the curve's flow direction and the region's properties.

Unit Normal Vector

In mathematics, a unit normal vector is a vector that is perpendicular to a surface or curve and has a magnitude of one. It's crucial when we want to find the component of a vector field that passes perpendicularly through a curve. For a circle of radius \(a\), described by \(x^2 + y^2 = a^2\), its unit normal vector at any given point is \(\mathbf{n} = \frac{1}{a}(x, y)\).

Why Use a Unit Normal Vector?

The unit normal vector helps determine the direction of the field at the boundary in a perpendicular manner. This orthogonal orientation is vital for calculating the flux, as it ensures that the integral measures the vector field's correct effect thorough the boundary. Without the unit normal, calculations could misinterpret the field's direction, leading to incorrect measurement of flow through the surface. By standardizing direction and length, the vector computations are simplified and more accurate.

Smooth Simple Closed Curve

In calculus, a smooth simple closed curve is a continuous loop in a plane that does not intersect itself and is differentiable at all points. Think of it as a non-overlapping loop, similar to a circle or an ellipse, that smoothly bends without any sharp corners.

Why They Matter in Green's Theorem

Using smooth simple closed curves is essential when applying Green's Theorem. The theorem relates a line integral around such a curve to a double integral over the region it encompasses. These curves ensure well-defined integrals, where the behavior and properties of boundaries do not contribute singularities or complex paths that could complicate calculations. When solving exercises like the one provided, ensuring the curves in question meet the criteria for smoothness and simplicity is crucial for applying the theorem adequately. It allows the transformation of complex line integrals into more manageable double integrals, facilitating the computation of interesting properties like area, mass, or flux based on the vector field behavior around and within the curve.