Question

What's wrong? Consider the radial field \(\mathbf{F}=\frac{\langle x, y\rangle}{x^{2}+y^{2}}\) a. Verify that the divergence of \(\mathbf{F}\) is zero, which suggests that the double integral in the flux form of Green's Theorem is zero. b. Use a line integral to verify that the outward flux across the unit circle of the vector field is \(2 \pi\) c. Explain why the results of parts (a) and (b) do not agree.

Short answer

**Question:** Verify the divergence of the radial field \(\mathbf{F} = \frac{\langle x, y \rangle}{x^2 + y^2}\), calculate the outward flux across the unit circle using a line integral, and explain why the results from both calculations do not agree. **Answer:** The divergence of the radial field \(\mathbf{F}\) is zero, but the outward flux across the unit circle, calculated using a line integral, is \(2\pi\). The discrepancy between the two results occurs because Green's Theorem assumes that the vector field \(\mathbf{F}\) is continuously differentiable on a simply connected domain. However, this particular radial field contains a singularity at the origin, which violates the assumptions of Green's Theorem, leading to the contradiction between the two results.

Step-by-step solution

Find the two-dimensional divergence of \(\mathbf{F}\)

The given vector field is \(\mathbf{F} = \frac{\langle x, y \rangle}{x^2 + y^2}\). Calculate the partial derivatives concerning \(x\) and \(y\): $$ \frac{\partial}{\partial x}\left(\frac{x}{x^2 + y^2}\right) \text{ and } \frac{\partial}{\partial y}\left(\frac{y}{x^2 + y^2}\right) $$

Confirm if the divergence is zero

Calculate the divergence of \(\mathbf{F}\) by adding the partial derivatives found in Step 1: $$ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}\left(\frac{x}{x^2 + y^2}\right) + \frac{\partial}{\partial y}\left(\frac{y}{x^2 + y^2}\right) = \frac{y^2 - x^2}{(x^2 + y^2)^2} + \frac{x^2 - y^2}{(x^2 + y^2)^2} = 0. $$ The divergence is indeed zero.

Apply Green's Theorem to show that the double integral's result is zero

Green's Theorem states that: $$ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D (\nabla \cdot \mathbf{F}) dA $$ Since \(\nabla \cdot \mathbf{F} = 0\), the right-hand side of Green's Theorem becomes zero: $$ \iint_D 0 dA = 0 $$

Set up and compute the line integral concerning Green's Theorem for outward flux

The unit circle can be parameterized as \(\mathbf{r}(t) = \langle\cos(t), \sin(t)\rangle\), where \(0 \leq t \leq 2\pi\). The derivative of the parameterization is \(\mathbf{r}'(t) = \langle-\sin(t), \cos(t)\rangle\). The line integral concerning Green's Theorem for the outward flux is: $$ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt $$ Plugging in the parameterization of the unit circle and its derivative into the line integral, we get: $$ \int_0^{2\pi} \frac{\langle\cos(t), \sin(t)\rangle}{\cos^2(t) + \sin^2(t)} \cdot \langle-\sin(t), \cos(t)\rangle dt = \int_0^{2\pi} (-\sin(t)\cos(t) + \sin(t)\cos(t)) dt = 2\pi $$ Thus, the outward flux across the unit circle is \(2\pi\).

Explain the contradiction between parts (a) and (b)

Even though the divergence of \(\mathbf{F}\) is zero, which suggests that the double integral in the flux form of Green's Theorem should be zero, the result of the line integral shows that the outward flux across the unit circle is \(2\pi\). This contradiction occurs because Green's Theorem assumes that the vector field \(\mathbf{F}\) is continuously differentiable on a simply connected domain, which is not the case for this particular radial field, as it contains a singularity at the origin (when \(x^2 + y^2 = 0\)). Due to this singularity, the assumptions of Green's Theorem do not apply, and hence the result of part (a) does not agree with the result of part (b).

Key concepts

Divergence

Divergence is a fundamental concept in vector calculus, describing how much a vector field spreads out or converges inward at a particular point. In simpler terms, it measures the "outflow" density of a vector field from an infinitesimal volume. For a two-dimensional vector field like \( \mathbf{F} = \frac{\langle x, y \rangle}{x^2 + y^2} \), the divergence is calculated by taking the sum of the partial derivatives of its components with respect to their respective directions:

• The partial derivative with respect to \( x \).
• The partial derivative with respect to \( y \).

This yields:\[abla \cdot \mathbf{F} = \frac{\partial}{\partial x}\left(\frac{x}{x^2 + y^2}\right) + \frac{\partial}{\partial y}\left(\frac{y}{x^2 + y^2}\right) = 0,\]indicating that the divergence of \( \mathbf{F} \) is zero everywhere, except at the origin where the function is undefined. This suggests that, generally, the field is neither diverging nor converging anywhere outside the singularity. However, Green’s Theorem, which relates the line integral around a closed curve to a double integral over the region it encloses, cannot be directly applied when singularities are present within the region.

Line Integral

Line integrals are an essential tool for calculating the "accumulated effect" of a vector field along a path or a curve. For the vector field \( \mathbf{F} \) defined earlier, to compute the line integral around a path such as the unit circle, we parameterize the path. The parameterization of the unit circle is \( \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \) with \( 0 \leq t \leq 2\pi \).
Next, we take the derivative, \( \mathbf{r}'(t) = \langle -\sin(t), \cos(t) \rangle \). The line integral then becomes:\[\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt=2\pi.\]The result indicates a net outward flux of \( 2\pi \) across the unit circle. This outcome may seem contradictory given the zero divergence in the earlier divergence calculation, but the presence of the singular point at the origin changes the scenario. The singularity influences the integral in a way that traditional formulas would not predict.

Vector Field Singularities

Understanding vector field singularities is crucial for working with Green's Theorem and similar concepts. A singularity in a vector field is a point where the field does not behave consistently, often not being defined, or having infinite behavior.
For the field \( \mathbf{F} = \frac{\langle x, y \rangle}{x^2 + y^2} \), the singularity appears at the origin \( (0,0) \), since the denominator becomes zero, making the field undefined there. This singular point disrupts the assumptions required for using Green's Theorem, which assumes a continuous and well-behaved field over the entire region of interest.

• The singularity causes inconsistencies when calculating areas and fluxes that include the singular point.
• Green's Theorem, which relies on an underlying assumption of differentiability, doesn't hold as there is a breakdown at this point.

When working with vector fields containing singularities, it's vital to recognize these interruptions and adjust calculations or apply different approaches that account for these points. Understanding and identifying these singularities empower the ability to anticipate how they might affect overall analysis.