Problem 19 - Calculus Early Transcendentals

Question

Use the Divergence Theorem to find the flux of F across the surface ? with outward orientation. $$ \begin{array}{l}{\mathbf{F}(x, y, z)=x^{3} \mathbf{i}+x^{2} y \mathbf{j}+x y \mathbf{k} ; \sigma \text { is the surface of the solid }} \\ {\text { bounded by } z=4-x^{2}, y+z=5, z=0, \text { and } y=0}\end{array} $$

Step-by-Step Solution

Answer

The flux across the surface is approximately 87.77.

Step 1: Understand the Problem

We need to find the flux of the vector field \( \mathbf{F} = x^3 \mathbf{i} + x^2 y \mathbf{j} + xy \mathbf{k} \) across the surface \( \sigma \). This surface is bounded by the surfaces \( z = 4 - x^2 \), \( y + z = 5 \), \( z = 0 \), and \( y = 0 \). It forms a closed surface, allowing us to use the Divergence Theorem.

Step 2: Apply the Divergence Theorem

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

Step 3: Calculate the Divergence of F

The divergence of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). For \( \mathbf{F} = x^3 \mathbf{i} + x^2 y \mathbf{j} + xy \mathbf{k} \), we have:\[ abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(x^2 y) + \frac{\partial}{\partial z}(xy) = 3x^2 + x^2 + 0 = 4x^2 \].

Step 4: Set Up the Volume Integral

The volume \( V \) is bounded by the surfaces \( z = 4 - x^2 \), \( y + z = 5 \), \( z = 0 \), and \( y = 0 \). Within \( y = 0 \) to \( y = 5 - z \), the region would vary.It suggests bounds for \( y \) from 0 to \( 5 - z \) and \( z \) from 0 to \( 4 - x^2 \), where the \(x\) bounds are within \( -2 \leq x \leq 2 \), as the parabola \( 4-x^2 \) stops at these x-values.

Step 5: Evaluate the Triple Integral

The integral for the flux is \[ \iiint_{V} 4x^2 \, dV = \int_{-2}^{2} \int_{0}^{4-x^2} \int_{0}^{5-z} 4x^2 \, dy \, dz \, dx \].Integrate with respect to \( y \) first:\[ \int_{0}^{5-z} 4x^2 \, dy = 4x^2(5-z) \].Next, integrate with respect to \( z \):\[ \int_{0}^{4-x^2} 4x^2(5-z) \, dz = \int_{0}^{4-x^2} (20x^2 - 4x^2z) \, dz \]\[ = 20x^2z - 2x^2z^2 \bigg|_0^{4-x^2} \]\[ = 20x^2(4-x^2) - 2x^2(4-x^2)^2 \]Finally, integrate with respect to \( x \):\[ \int_{-2}^{2} [80x^2 - 20x^4 - 32x^2 + 8x^4 + 2x^6] \, dx \].

Step 6: Simplify and Compute

Combine terms in the integral and calculate the definite integral:\[ \int_{-2}^{2} [48x^2 - 12x^4 + 2x^6] \, dx \].Calculating, we:\[ \int_{-2}^{2} 48x^2 \, dx = 48\left(\frac{8}{3}\right) = 128 \],\[ \int_{-2}^{2} -12x^4 \, dx = -12\left(\frac{32}{5}\right) = -76.8 \],\[ \int_{-2}^{2} 2x^6 \, dx = 2\left(\frac{128}{7}\right) = 36.57 \].Add the results: flux = 128 - 76.8 + 36.57.

Key Concepts

Flux Calculation

Flux represents the quantity of a vector field passing through a surface. When you're given a vector field \( \mathbf{F} \) and a surface \( \sigma \), the flux calculation quantifies how much of \( \mathbf{F} \) penetrates \( \sigma \).

The mathematically inclined way of saying this is: it measures the flow of the vector field through a surface. The Divergence Theorem is a helpful tool here. It lets us move from working on a complex curved surface to an easier volume integral calculation.

Here's the basic idea: the flux \( \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \) over a closed surface is equal to the triple integral of \( abla \cdot \mathbf{F} \) over the volume \( V \) it encloses. In simple terms, it converts a hard-to-solve surface integral into a generally easier volume integral over a region \( V \).

• The surface \( \sigma \) acts like a closed boundary of a region in space.
• The vector field \( \mathbf{F} \) describes what's "flowing." It could be air, water, or energy fields.
• The outward orientation ensures that the vector field \( \mathbf{F} \) is computed in the direction away from the surface, making sure that flux is properly measured.

The Divergence Theorem is a powerful tool. It simplifies some difficult calculations by bringing us into the realm of volumes rather than surfaces.

Vector Fields

A vector field is like a map indicating many vectors at various spatial points. Each point in space has a vector, depicting the direction and magnitude of the flow at that point.

In essence, vector fields visually express how quantities change and move through space. Consider a weather map showing the direction and strength of wind blowing across a region.

• A vector \( \mathbf{F}(x, y, z) = x^{3} \mathbf{i} + x^{2} y \mathbf{j} + xy \mathbf{k} \) is a classic example. It depicts the vector field with components (\( x^{3} \), \( x^{2}y \), \( xy \)) for the different axes.
• The \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors in the x, y, and z directions, respectively.
• Each term (\( x^3 \mathbf{i}, x^2y \mathbf{j}, xy \mathbf{k} \)) signifies how the flow behaves in each spatial direction.

This is a higher-level way to see how systems perform over space and can provide predictions on movement and interaction, crucial in fields like physics, engineering, and meteorology.

Triple Integrals

Triple integrals are essentially three successive integrations. They allow you to calculate a quantity over a three-dimensional region.Think of triple integrals as advanced tools to "add" up pieces over a volume. Consider this analogy: if "double integrals" like a family meeting where each member shares twice, "triple integrals" mean you're bringing in an extended family, each sharing even more widespread insights.

• They are expressed as \( \iiint_V f(x, y, z) \, dV \), quantifying the whole over the volume \( V \).
• This involves a sequence of integrals over the coordinates: first x, then y, and finally z, with the function (e.g., a density function) dictating contributions at each point within the region \( V \).
• Borders in each coordinate are essential. Like in our step-by-step solution, each variable has its own limits, bounding the volume of interest.

When using the Divergence Theorem, the conversion from a surface integral to a volume integral involves a triple integral, which makes solving complex geometric problems more approachable. They are foundational in multi-variable calculus, helping calculate quantities spanning across three-dimensional spaces.