Problem 6 - Thomas Calculus

Question

In Exercises \(5-16\) , use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) $$\mathbf{F}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}$$ a. Cube \(D :\) The cube cut from the first octant by the planes \(x=1, y=1,\) and \(z=1\) b. Cube \(D :\) The cube bounded by the planes \(x=\pm 1\) \(y=\pm 1,\) and \(z=\pm 1\) c. Cylindrical can \(D :\) The region cut from the solid cylinder \(x^{2}+y^{2} \leq 4\) by the planes \(z=0\) and \(z=1\)

Step-by-Step Solution

Answer

Flux is 3 for a, 0 for b, and \(32\pi\) for c.

Step 1: Understand the Divergence Theorem

The Divergence Theorem states that the flux of a vector field \( \mathbf{F} \) across a closed surface \( S \) is equal to the triple integral of the divergence of \( \mathbf{F} \) over the volume \( D \) enclosed by \( S \). Mathematically, this is expressed as \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{D} (abla \cdot \mathbf{F}) \, dV \).

Step 2: Calculate the Divergence of \( \mathbf{F} \)

Given \( \mathbf{F} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \), we compute the divergence using the formula \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) \). This simplifies to \( abla \cdot \mathbf{F} = 2x + 2y + 2z \).

Step 3a: Find the Flux for Cube in First Octant

For the cube \( D \) cut from the first octant with planes \( x = 1, y = 1, z = 1 \), compute \( \iiint_D (2x + 2y + 2z) \, dV \) over \( x, y, z \) in \([0, 1] \). Set up the integral: \( \int_0^1 \int_0^1 \int_0^1 (2x + 2y + 2z) \, dx \, dy \, dz \). Evaluating this triple integral results in a flux of 3.

Step 3b: Find the Flux for Cube with Negative Bounds

For the cube \( D \) with bounds \( x = \pm 1, y = \pm 1, z = \pm 1 \), compute the integral \( \iiint_D (2x + 2y + 2z) \, dV \) over \( x, y, z \) in \([-1, 1] \). The integral becomes \( \int_{-1}^1 \int_{-1}^1 \int_{-1}^1 (2x + 2y + 2z) \, dx \, dy \, dz \). The symmetry causes the net flux to compute to zero.

Step 3c: Find the Flux for Cylindrical Region

For the cylindrical can cut by \( x^2 + y^2 \leq 4 \) and \( z = 0 \) to \( z = 1 \), calculate \( \iiint_D (2x + 2y + 2z) \, dV \). Convert to cylindrical coordinates: \( x = r \cos \theta, y = r \sin \theta, z = z \), where \( r \leq 2 \), \( \theta \in [0, 2\pi] \), and \( z \in [0, 1] \). The integral collapses to \( \int_0^{2\pi} \int_0^2 \int_0^1 (2r \cos\theta + 2r \sin\theta + 2z) \, r \, dz \, dr \, d\theta \), simplifying to a flux of \( 32 \pi \).

Key Concepts

Flux Calculation

Flux is a key concept when dealing with vector fields. It represents the amount of a field passing through a surface. In the context of the Divergence Theorem, flux calculation becomes more meaningful as it relates the flow across a surface to the behavior within a volume.
To calculate flux using the Divergence Theorem, we equate it to the integral over a volume. This simplifies our work from potentially complex surface integrals to easier triple integrals: - The idea is to evaluate the divergence of the given vector field.- Then, integrate this result over the volume enclosed by the surface.
For example, in the exercise provided, you can see how the divergence simplifies our calculations. We only need to find the integral of a simple expression like \( 2x + 2y + 2z \) over the volume \( D \). Depending on how the volume is defined, this can lead to different results. Overall, flux calculation through this method becomes a streamlined process, allowing problems that would have been tedious to solve manually to be tackled swiftly.

Vector Fields

Vector fields represent functions that assign a vector to every point in space. They can depict various physical fields, like the magnetic field around a magnet or the flow field in a fluid. Understanding vector fields is essential when working with the Divergence Theorem, as it pertains to how these fields behave over a volume.
In the exercise, we have a vector field expressed as \( \mathbf{F} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \). Each component function tells us how the field behaves in the respective directions - \( x \), \( y \), and \( z \). Analyzing these expressions helps us understand the source strength or intensity of the fields in different regions.
A vector field can - Vary linearly in each direction, such as \( x^2 \), \( y^2 \), and \( z^2 \) do,- Or even represent more complex dependencies in multiple dimensions.
Once this is clear, calculating divergence, which is the sum of the partial derivatives of each component, becomes straightforward. This step is crucial as it allows the transformation to a volume integral, making flux calculations practical.

Triple Integrals

Triple integrals are mathematical tools used to integrate functions over three-dimensional spaces. They extend the concept of integration from one and two dimensions to three, making them vital for problems involving volumes, such as those tackled using the Divergence Theorem.
In the context of the exercise, triple integrals help in finding the total field effect within a given volume \( D \):- You start by setting up the integral bounds corresponding to the volume's geometry,- Then integrate the divergence over all three coordinates typically \( x, y, z \).
For example, integrating over a cube involves using bounds that correspond to its sides, such as \([0,1]\) or \([-1,1]\). This leads to expressions involving these variables being evaluated over the entirety of the cube's volume. Triple integrals are especially useful in handling complex shapes. - For the cylindrical region in the exercise, cylindrical coordinates make evaluation simpler by aligning the integration bounds with the shape's natural symmetry.
Thus, mastering triple integrals is crucial when working on volumetric calculations, offering clarity and precision in results.