Problem 16 - Calculus Early Transcendentals

Question

Use the Divergence Theorem to find the flux of F across the surface ? with outward orientation. $$ \begin{array}{l}{\mathbf{F}(x, y, z)=2 x z \mathbf{i}+y z \mathbf{j}+z^{2} \mathbf{k}, \quad \text { where } \sigma \text { is the surface }} \\ {\text { of the solid bounded above by } z=\sqrt{a^{2}-x^{2}-y^{2}} \text { and be- }} \\ {\text { low by the } x y \text { -plane. }}\end{array} $$

Step-by-Step Solution

Answer

The flux of \( \mathbf{F} \) across the surface \( \sigma \) is \( \frac{2\pi}{3} a^5 \).

Step 1: 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 triple integral over the volume \( V \) bounded by \( \sigma \) of the divergence of \( \mathbf{F} \). Mathematically, it is expressed as: \[ \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{V} abla \cdot \mathbf{F} \, dV \] where \( \mathbf{n} \) is the outward normal to the surface. In this problem, \( V \) is the volume bounded by the given surface, and \( \mathbf{n} \) is the normal vector.

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

The divergence of the vector field \( \mathbf{F}(x, y, z) = 2xz \mathbf{i} + yz \mathbf{j} + z^2 \mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x} (2xz) + \frac{\partial}{\partial y} (yz) + \frac{\partial}{\partial z} (z^2) \). Compute each partial derivative: \( \frac{\partial}{\partial x} (2xz) = 2z \), \( \frac{\partial}{\partial y} (yz) = z \), and \( \frac{\partial}{\partial z} (z^2) = 2z \). Therefore, \( abla \cdot \mathbf{F} = 2z + z + 2z = 5z \).

Step 3: Set Up the Triple Integral

Now, set up the triple integral to evaluate \( \iiint_{V} 5z \, dV \). The volume \( V \) is described in cylindrical coordinates by the limits: \( 0 \leq z \leq \sqrt{a^2 - r^2} \), \( 0 \leq r \leq a \), and \( 0 \leq \theta \leq 2\pi \). Convert the integrand to cylindrical coordinates: \( z = z \). So the integral is: \[ \int_{0}^{2\pi} \int_{0}^{a} \int_{0}^{\sqrt{a^2 - r^2}} 5z \, r \, dz \, dr \, d\theta \].

Step 4: Evaluate the Inner Integral

Begin by evaluating the inner integral with respect to \( z \): \( \int_{0}^{\sqrt{a^2 - r^2}} 5z \, rz \, dz = 5r \int_{0}^{\sqrt{a^2 - r^2}} z^2 \, dz \). The integral \( \int z^2 \, dz \) results in \( \frac{z^3}{3} \). Evaluate from \( z=0 \) to \( z=\sqrt{a^2 - r^2} \): \[ \frac{5r}{3} [(\sqrt{a^2 - r^2})^3 - 0] = \frac{5r}{3} (a^2 - r^2)^{3/2} \].

Step 5: Evaluate the Middle Integral

Now, integrate with respect to \( r \) from 0 to \( a \): \( \int_{0}^{a} \frac{5r}{3} (a^2 - r^2)^{3/2} \, dr \). Use a substitution, let \( u = a^2 - r^2 \), \( du = -2r \, dr \). The limits of integration change from \( r = 0 \) to \( r = a \) to \( u = a^2 \) to 0. The integral becomes \[ \frac{-5}{6} \int_{a^2}^{0} u^{3/2} \, du = \frac{5}{6} \int_{0}^{a^2} u^{3/2} \, du \].

Step 6: Evaluate the \( u \)-substitution Integral

The integral \( \int u^{3/2} \, du \) is solved using the power rule: \( \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2} \). Evaluating from 0 to \( a^2 \) gives: \[ \frac{2}{5} (a^2)^{5/2} \]. The result is \( \frac{2}{5} a^5 \). Multiply by \( \frac{5}{6} \) to get \( \frac{10}{30} a^5 = \frac{1}{3} a^5 \).

Step 7: Evaluate the Outer Integral

Finally, integrate with respect to \( \theta \): \( \int_{0}^{2\pi} \frac{1}{3} a^5 \, d\theta = \frac{1}{3} a^5 \int_{0}^{2\pi} \ d\theta = \frac{1}{3} a^5 [\theta]_{0}^{2\pi} = \frac{1}{3} a^5 (2\pi) = \frac{2\pi}{3} a^5 \). This is the total flux across the surface.

Key Concepts

Flux calculation

Flux calculation is a fundamental concept in vector calculus, particularly in relation to the Divergence Theorem. When we talk about flux, we are referring to the amount of a vector field that passes through a surface. Imagine a fluid flowing through a net; the flux tells us how much fluid is passing through that net per unit time.

To calculate the flux of a vector field \( \mathbf{F} \) across a surface \( \sigma \), we need to consider the dot product of the vector field and the outward normal vector \( \mathbf{n} \) of the surface at every point. This is given by

• The formula: \( \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \)
• Where \( \, dS \) represents an infinitesimal area element of the surface \( \sigma \).

This integral is often complex and can be simplified using the Divergence Theorem. The theorem enables a transformation of the surface integral into a volume integral, making computation easier when the surface \( \sigma \) encloses a volume \( V \).

In the given problem, the Divergence Theorem plays a key role in transforming the difficult surface integral into a more manageable triple integral over the volume inside the surface.

Vector fields

Vector fields are essential in understanding various phenomena in physics and engineering. A vector field associates a vector to every point in a space, often representing something like the speed and direction of moving fluid, or the electric or magnetic force in a given space.

In the exercise, the vector field is defined as \( \mathbf{F}(x, y, z) = 2xz \mathbf{i} + yz \mathbf{j} + z^2 \mathbf{k} \). Here,

• The components \( 2xz \), \( yz \), and \( z^2 \) represent the magnitude and direction of the field in the \( x \), \( y \), and \( z \) directions, respectively.
• This can be visualized as a field radiating strength away from or towards points depending on the values of \( x \), \( y \), and \( z \).

Understanding vector fields involves computing their divergence, which describes how much a vector field spreads out or converges at a particular point.

Computing divergence is crucial in applying the Divergence Theorem, which in this exercise involves finding the sum of the rates of change of the components of the vector field.

Triple integrals

Triple integrals are a powerful tool for evaluating many-dimensional integrals over a three-dimensional region, such as when using the Divergence Theorem. They allow for the calculation of quantities over volume, and are expressed in the form

• \( \iiint_{V} \, f(x, y, z) \, dV \)
• Where \( f(x, y, z) \) is the integrand representing a function over the volume \( V \).
• The differential \( dV \) represents the infinitesimal volume element.

In cylindrical coordinates, often used for symmetrical or round regions, \( dV \) is expressed as \( r \, dz \, dr \, d\theta \).

In the exercise, a triple integral is set up to evaluate the divergence of the vector field over the volume enclosed by the surface. With the limits defined for \( z \), \( r \), and \( \theta \), the integral can be evaluated step by step:

• Firstly, integrating over \( z \), then \( r \), and finally \( \theta \).
• This step-by-step integration allows for the precise calculation of volume-based properties such as flux.

Understanding how to set up and evaluate triple integrals is essential in applying these calculations to real-world scenarios efficiently.