Problem 204 - Calculus Volume 3

Question

[T] Find the outward flux of vector field \(\mathbf{F}=x y^{2} \mathbf{i}+x^{2} y \mathbf{j}\) across the boundary of annulus \(R=\left\\{(x, y): 1 \leq x^{2}+y^{2} \leq 4\right\\}=\\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq 2 \pi\\}\) using a computer algebra system.

Step-by-Step Solution

Answer

The outward flux is \(\frac{15\pi}{2}\).

Step 1: Understand the problem

We need to calculate the outward flux of the vector field \(\mathbf{F}=x y^{2} \mathbf{i}+x^{2} y \mathbf{j}\) across the boundary of the annulus \(R\). The flux is calculated by integrating the divergence of the field over the region.

Step 2: Calculate the divergence of the vector field

The divergence of a vector field \(\mathbf{F}=P\mathbf{i}+Q\mathbf{j}\) is given by \(abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\). In our case, \(P=xy^2\) and \(Q=x^2y\). So, we calculate: \[abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(xy^2) + \frac{\partial}{\partial y}(x^2y)=y^2+x^2.\]

Step 3: Set up the double integral in polar coordinates

The annulus is described in polar coordinates as \(1 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\). The differential area element in polar coordinates is \(r\, dr\, d\theta\). So, the flux integral becomes: \[abla \cdot \mathbf{F} = \int_0^{2\pi} \int_1^2 (y^2 + x^2) r\, dr\, d\theta.\] In polar coordinates, \(y=r\sin(\theta)\) and \(x=r\cos(\theta)\). Thus, \(y^2 + x^2 = r^2(\sin^2 \theta + \cos^2 \theta) = r^2\).

Step 4: Evaluate the double integral

Substitute \(y^2 + x^2 = r^2\) into the integral: \[abla \cdot \mathbf{F} = \int_0^{2\pi} \int_1^2 r^3\, dr\, d\theta.\] First, compute the inner integral with respect to \(r\): \[\int_1^2 r^3\, dr = \left[\frac{r^4}{4}\right]_1^2 = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}.\] Next, compute the outer integral with respect to \(\theta\): \[abla \cdot \mathbf{F} = \int_0^{2\pi} \frac{15}{4}\, d\theta = \frac{15}{4} [\theta]_0^{2\pi} = \frac{15}{4} \times 2\pi = \frac{15\pi}{2}.\]

Step 5: Conclusion

Therefore, the outward flux of the vector field across the boundary of the annulus \(R\) is \(\frac{15\pi}{2}\).

Key Concepts

Flux Integral

Flux integrals are a fundamental concept in vector calculus, used to measure the flow of a vector field through a surface. When we talk about the *outward flux* through the boundary of a region, we essentially mean how much of the vector field is passing out of the surface that defines the boundary.
To calculate the flux integral across a surface, we need to evaluate an integral over that surface. The integral takes into account the field vector and the vector normal to the surface. In simpler terms, it can be thought of as summing up all the little contributions of the vector field across each piece of the surface.
The formula typically used for a flux integral is:

• \[\iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \]

where \( \mathbf{F} \) is the vector field and \( \mathbf{n} \) is the unit normal vector to the surface \( S \). For closed surfaces, it tells us the net "flow" coming out or into the volume enclosed by the surface.

Divergence Theorem

The Divergence Theorem, also known as Gauss's Theorem, provides a powerful link between flux integrals and volume integrals. It states that the total flux of a vector field \( \mathbf{F} \) out of a closed surface \( S \) is equal to the integral of the divergence of \( \mathbf{F} \) over the volume \( V \) enclosed by \( S \).
This theorem simplifies the computation of flux integrals by transforming them into a volume integral:

• \[\iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{V} (abla \cdot \mathbf{F}) \, dV\]

By finding the divergence \( abla \cdot \mathbf{F} \), which represents how much the vector field spreads out from a point, we can accurately compute the flux through the boundary of the region \( R \). In our example, the divergence \( y^2 + x^2 \) is calculated, yielding the necessary function to integrate over the specified region.

Polar Coordinates

Polar coordinates provide a natural way to describe locations within a plane using the distance from a reference point and an angle from a reference direction. This is particularly useful for circular or annular regions, such as the annulus in our problem.
Unlike Cartesian coordinates \( (x, y) \), polar coordinates are represented as \( (r, \theta) \), where:

• \( r \) is the radial distance from the origin.
• \( \theta \) is the angular displacement from a reference direction, often the positive x-axis.

In polar coordinates, the transformation equations are:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

The differential area element changes to \( r \, dr \, d\theta \). In our example, the annulus is efficiently described by \( 1 \leq r \leq 2 \) and \( 0 \leq \theta \leq 2\pi \), and effectively used for setting up the integration.

Vector Field Analysis

Vector fields are functions that assign a vector to each point in a subset of space, which can represent quantities such as force, velocity, or flow rates. Analyzing these fields helps us understand how these vectors behave and interact across regions.
In the problem, the vector field \( \mathbf{F}=x y^{2} \mathbf{i}+x^{2} y \mathbf{j} \) describes a field with components that depend on the coordinates \( x \) and \( y \). Breaking down the vector field into its components \( P = xy^2 \) and \( Q = x^2y \) allows us to later calculate the divergence and analyze how the field spreads at each point.
By examining the field within the annular region, we assess behaviors like circulation and outward flow. Understanding these behaviors is crucial for applications in physics and engineering, such as fluid dynamics and electromagnetism, where such fields occur frequently.