Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{F}(\mathbf{r})=\mathbf{r} /\|\mathbf{r}\|^{2}=(x \mathbf{i}+y \mathbf{j}) /\left(x^{2}+y^{2}\right)\). (a) Show that \(\int_{C} \mathbf{F} \cdot \mathbf{n} d s=2 \pi\), where \(C\) is the circle centered at the origin of radius \(a\) and \(\mathbf{n}=(x \mathbf{i}+y \mathbf{j}) / \sqrt{x^{2}+y^{2}}\) is the exterior unit normal to \(C\). (b) Show that div \(\mathbf{F}=0\). (c) Explain why the results of parts (a) and (b) do not contradict the vector form of Green's Theorem. (d) Show that if \(C\) is a smooth simple closed curve then \(\int_{C} \mathbf{F} \cdot \mathbf{n} d s\) equals \(2 \pi\) or 0 accordingly as the origin is inside or outside \(C\).

Step-by-Step Solution

Answer

(a) \( 2\pi \); (b) div \( \mathbf{F} = 0 \); (c) Green's Theorem doesn't apply due to singularity at origin; (d) \( 2\pi \) if origin inside, 0 if outside.

Step 1: Express line integral using parameterization

Parameterize the circle \( C \) using \( \mathbf{r}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} \) for \( t \in [0, 2\pi] \). The derivative \( \mathbf{r}'(t) = -a \sin t \mathbf{i} + a \cos t \mathbf{j} \) for the circular path. The exterior normal vector becomes \( \mathbf{n} = \cos t \mathbf{i} + \sin t \mathbf{j} \).

Step 2: Compute the dot product and integral

Compute \( \mathbf{F}(t) \cdot \mathbf{n}(t) = \frac{a(\cos t)^2 + a(\sin t)^2}{a^2} = \frac{a}{a^2} = \frac{1}{a} \). Integrate over the circle: \( \int_0^{2\pi} \frac{1}{a} a \, dt = \int_0^{2\pi} \, dt = 2\pi \), which confirms the result for part (a).

Step 3: Compute divergence of vector field

The vector field is \( \mathbf{F}(x, y) = \frac{x \mathbf{i} + y \mathbf{j}}{x^2 + y^2} \). Compute the divergence: \( \text{div} \, \mathbf{F} = \frac{\partial}{\partial x} \left( \frac{x}{x^2+y^2} \right) + \frac{\partial}{\partial y} \left( \frac{y}{x^2+y^2} \right) = 0 \), confirming part (b).

Step 4: Interpret Green's Theorem

Green's Theorem states \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \text{div} \, \mathbf{F} \, dA \). Since \( \text{div} \, \mathbf{F} = 0 \), the area integral should be zero, but the circle \( C \) includes the origin, where the field is not defined, making this case exempt from the theorem's standard application.

Step 5: Evaluate integral for a general curve

For a smooth simple closed curve \( C \), if the origin is inside, \( \mathbf{F} \) behaves as on a circle, yielding \( 2\pi \). If the origin is outside, \( \mathbf{F} \) is more regular, giving \( 0 \) due to symmetry and the absence of singular points inside \( C \).

Key Concepts

Line Integrals

Line integrals are a fundamental concept in vector calculus, allowing us to sum values along a curve. Consider a vector field and a curve, and think of the line integral as gathering information from the vector field along this curve.
In context,

• We parameterize the curve using the variables or functions that represent it.
• By doing this, we can express the line integral in terms of these parameters.
• The line integral \(\int_{C} \mathbf{F} \cdot \mathbf{n} \, ds\) becomes simpler to calculate once expressed in terms of our parameterization.

For a circle defined by \( C \), parameterized by \( \mathbf{r}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} \), we focus on how the vector field interactions throughout the path, allowing us to eventually sum these interactions to ascertain the value of the integral along the entire curve, which in many cases, provides insightful information about the field and the region encompassed by the curve.

Divergence Theorem

The Divergence Theorem connects the flow (divergence) of a vector field across a closed surface to the behavior of the field inside the surface. It can be seen as a bridge between surface integrals and volume integrals.

• When the divergence of a vector field is zero, it indicates a sort of 'balance' where there is no net creation or annihilation of the field within the region.
• Calculated as \(\text{div} \, \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\) for three-dimensional spaces.

In the case of a two-dimensional vector field like \(\mathbf{F}(x, y) = \frac{x \mathbf{i} + y \mathbf{j}}{x^2 + y^2}\), calculating its divergence confirms whether it behaves consistently over a region.
Green's Theorem in a similar scenario illustrates how the divergence over an area should result in equivalent boundary behavior, but has edge cases when singularities, like an undefined point at the origin, participate.

Green's Theorem

Green's Theorem links line integrals around a closed curve to double integrals over the region it encloses. It's a pivotal part of vector calculus, showing how circulation and flux within a boundary are interrelated.

• Green’s Theorem states: \(\oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \text{div} \, \mathbf{F} \, dA\).
• This is particularly useful when dealing with regions where the field may have singularities or undefined points.

In scenarios where the field includes the origin, the theorem's traditional utilization gets complicated due to the undefined nature at these points. Still, the theorem helps us understand overall behavior by connecting these seemingly isolated events with broader, more continuous behaviors seen in the field’s action over an area.
Although results like in part (a) and the zerodivergence seen in part (b) might intuitively seem contradictory, Green's Theorem itself does not directly apply due to the inclusion of the origin.

Parameterization of Curves

To effectively evaluate complex integrals, parameterization is essential. Think of it as converting a curve into a form that's easier to handle mathematically. Every point on the curve can then be described as a function of one or more parameters, making calculations, such as line integrals, more straightforward.

• For a circle, we use trigonometric functions to express it parametrically: \( \mathbf{r}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} \).
• This approach simplifies finding derivatives and normal vectors for the curve.
• It provides a seamless way to translate a geometric problem into an algebraic form.

The parameterization process is central to solving integrals over curves and provides us an effective handling of different paths and their interaction with vector fields. It helps create a computational pathway from a geometric intuition, enriching the comprehension and solving of complex vector calculus problems.