Problem 8 - Calculus: Early Transcendental Functions

Question

Use the Divergence Theorem to compute \(\iint_{\partial O} \mathbf{F} \cdot \mathbf{n} d S\). \(Q\) is the rectangular box \(0 \leq x \leq 2,1 \leq y \leq 2,-1 \leq z \leq 2\) \(\mathbf{F}=\left\langle y^{3}-2 x, e^{x z}, 4 z\right\rangle\)

Step-by-Step Solution

Answer

The result of \( \iint_{\partial O} \mathbf{F} \cdot \mathbf{n} d S \), calculated using the Divergence Theorem, is 12.

Step 1: Find the divergence of vector field

First, calculate the divergence of the vector field \( \mathbf{F} \), denoted \( \nabla \cdot \mathbf{F} \). Apply the definition of divergence: \( \nabla \cdot \mathbf{F} = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z} \), where \( F_{1}, F_{2}, F_{3} \) are the component functions of \( \mathbf{F} \). This gives us: \( \nabla \cdot \mathbf{F} = \frac{\partial (y^{3}-2x)}{\partial x} + \frac{\partial e^{xz}}{\partial y} + \frac{\partial 4z}{\partial z} = -2 + 0 + 4 = 2 \).

Step 2: Set up the triple integral

The Divergence Theorem tells us that \( \iiint_{Q} (\nabla \cdot \mathbf{F}) dV = \iint_{\partial O} \mathbf{F} \cdot \mathbf{n} dS \). Now, compute the triple integral \( \iiint_{Q} (\nabla \cdot \mathbf{F}) dV \). Replace \( \nabla \cdot \mathbf{F} \) with 2, which was obtained in Step 1. The integral becomes: \( \iiint_{Q} 2 dV = 2\iiint_{Q} dV \).

Step 3: Copute the volume integral

Evaluate the triple integral \( 2\iiint_{Q} dV \). This is essentially twice the volume of the box Q. Compute the volume of the box using the formula for the volume of a rectangular box \( V = length \times width \times height \). The volume of Q is \( (2 - 0) \times (2 - 1) \times (2 - (-1)) = 2 \times 1 \times 3 = 6 \). Therefore, the triple integral, which is two times the volume of Q, equals: \( 2 \times 6 = 12 \).

Key Concepts

Vector Field

A vector field is a mathematical construct where each point in a space is associated with a vector. Think of it like a map of little arrows pointing in various directions throughout a region. These arrows can represent different quantities like velocity, force, or acceleration at each point in the field.
In the exercise, we are given a specific vector field \( \mathbf{F} = \left\langle y^{3}-2x, e^{xz}, 4z \right\rangle \). This defines how vectors behave at every point within the given space, specifically within the rectangular box \( Q \).

• The first component \( y^{3} - 2x \) tells us about the influence of both \( y \) and \( x \) on the vector field along the x-axis.
• The second component \( e^{xz} \) indicates how the product of \( x \) and \( z \) governs movement along the y-axis.
• The third component \( 4z \) directly relates to the z-axis, indicating that the z component scales linearly with \( z \).

Understanding the behavior of a vector field helps us explore advanced concepts like the divergence theorem, which translates behaviors on surfaces to those happening in an entire volume.

Triple Integral

A triple integral allows us to sum up values over a three-dimensional region. It’s the three-dimensional counterpart of single and double integrals, incorporating three variables as limits to cover a volume. When tackling this, you substitute each function or field within the integral across the specified dimensions.
In the context of the Divergence Theorem, the triple integral is paramount. It relates the divergence of a vector field over a volume directly to the flux through the surface bounding that volume.
In the exercise, we used a triple integral to evaluate \( \iiint_{Q} (abla \cdot \mathbf{F}) \, dV \). This was carried out:

• First, by determining \( abla \cdot \mathbf{F} \), which was shown to be a constant value, 2.
• Then multiplying this divergence constant by the volume of the rectangular box, hence simplifying the computation of the integral.

This approach simplifies complex computations of vector fields within a defined volume, integrating across all points.

Volume of a Rectangular Box

Understanding the volume of a rectangular box is essential for setting up triple integrals. The formula used to calculate this is straightforward: \[ V = \text{length} \times \text{width} \times \text{height} \] This allows you to determine the space the box occupies within a three-dimensional region.
For the rectangular box defined by the boundaries \( 0 \leq x \leq 2 \), \( 1 \leq y \leq 2 \), and \(-1 \leq z \leq 2 \), the dimensions are:

• Length: \( 2 - 0 = 2 \)
• Width: \( 2 - 1 = 1 \)
• Height: \( 2 - (-1) = 3 \)

So, the volume is \( 2 \times 1 \times 3 = 6 \). In the Divergence Theorem application, you ultimately use this to compute the flux over the surface by doubling the box's volume, recognizing how it scales with the divergence constant. Understanding this concept is pivotal not only in geometry but also in vector calculus when evaluating integrals over volumes.