Problem 38 - University Calculus: Early Transcendentals

Question

In Exercises \(37-40,\) find the average value of \(F(x, y, z)\) over the given region. \(F(x, y, z)=x+y-z\) over the rectangular solid in the first octant bounded by the coordinate planes and the planes \(x=1, y=1,\) and \(z=2\).

Step-by-Step Solution

Answer

The average value of the function over the given region is 0.

Step 1: Define the Region and the Function

First, identify the function and the region over which the average value is being calculated. The function given is \( F(x, y, z) = x + y - z \). The region is a rectangular solid in the first octant bounded by the planes \( x = 0 \), \( y = 0 \), \( z = 0 \), \( x = 1 \), \( y = 1 \), and \( z = 2 \). This defines a box with dimensions \( x: [0, 1] \), \( y: [0, 1] \), \( z: [0, 2] \).

Step 2: Write the Formula for Average Value

The average value \( \langle F \rangle \) of a function \( F(x, y, z) \) over a region \( V \) is given by:\[ \langle F \rangle = \frac{1}{V} \int \int \int_{V} F(x, y, z) \, dx \, dy \, dz \]where \( V \) is the volume of the region.

Step 3: Calculate the Volume of the Region

The volume \( V \) of the rectangular solid can be calculated as the product of its dimensions:\[ V = (1 - 0)(1 - 0)(2 - 0) = 2 \]

Step 4: Set Up the Triple Integral

Set up the integral to find the total value of \( F \) over the region. The limits of integration for \( x \), \( y \), and \( z \) are from 0 to 1, 0 to 1, and 0 to 2 respectively:\[ \int_{0}^{1} \int_{0}^{1} \int_{0}^{2} (x + y - z) \, dz \, dy \, dx \]

Step 5: Evaluate the Integral

Perform the integration step by step. Start with the inner integral with respect to \( z \):\[ \int_{0}^{2} (x + y - z) \, dz = [(x+ y)z - \frac{z^2}{2}]_{0}^{2} = (x + y)(2) - \frac{(2)^2}{2} = 2x + 2y - 2 \]Now, integrate the result with respect to \( y \):\[ \int_{0}^{1} (2x + 2y - 2) \, dy = [2xy + y^2 - 2y]_{0}^{1} = 2x(1) + (1)^2 - 2(1) = 2x + 1 - 2 = 2x - 1 \]Finally, integrate with respect to \( x \):\[ \int_{0}^{1} (2x - 1) \, dx = [x^2 - x]_{0}^{1} = (1)^2 - 1 = 1 - 1 = 0 \]

Step 6: Calculate the Average Value

To find the average value, divide the result of the integral by the volume of the region:\[ \langle F \rangle = \frac{1}{2} \times 0 = 0 \]

Key Concepts

Triple Integrals

A triple integral is a way to extend the concept of integration to three-dimensional spaces. This type of integral allows us to calculate quantities like volume, mass, or the total output of a function over a three-dimensional region. Think of it as summing up an infinite number of infinitesimally small values within a solid.

When evaluating a triple integral, you typically integrate one variable at a time, moving from the innermost to the outermost integral. For example, if you have \[\int_{a}^{b} \int_{c}^{d} \int_{e}^{f} F(x, y, z) \, dz \, dy \, dx\]this indicates you integrate with respect to "z" first, then "y", and finally "x".

In the given problem, we used a triple integral to calculate the total value of the function \(F(x, y, z) = x + y - z\) within a specified rectangular solid region. By doing this, we essentially take the sum of the function’s outputs across the entire volume.

• Step-by-step integration ensures each dimension is accounted for sequentially.
• Limits of integration determine the boundaries for each variable.

Understanding triple integrals is key in multivariable calculus as they enable the analysis of more complex systems that are naturally three-dimensional.

Average Value of a Function

Finding the average value of a function in multivariable calculus expands the idea of finding a simple average. Here, rather than summing a list of data points, we consider an entire volume. The average value of a function gives insight into its typical value over a given region.

The formula for calculating the average value of a function \(F(x, y, z)\) over a volume \(V\) is:\[\langle F \rangle = \frac{1}{V} \int \int \int_{V} F(x, y, z) \, dV\]This essentially means you compute the total value of \(F\) via integration and then divide by the volume size. Doing this gives an indication of the function's behavior across the region.

• For large volumes, it helps to determine what the function's output is like on average, smoothing out any extreme or isolated variations.
• It helps in analyzing distributions and density over a given space.

In our example, after calculating the triple integral, we divided by the rectangular solid's volume to get the average value, equalling zero. This indicates that across the entire solid, the function maintains a neutral balance between positive and negative values.

Rectangular Solid

A rectangular solid, often referred to as a cuboid, is a three-dimensional geometric figure bounded by six rectangle-shaped faces. Consider it like a box, where each corner forms a right angle.

In mathematical exercises, a rectangular solid is defined by specifying its dimensions along each of the three axes: width, height, and depth. Finding the volume is straightforward, achieved by multiplying these dimensions together as \(V = \text{length} \times \text{width} \times \text{height}\).

In our exercise, the rectangular solid is confined to the first octant, meaning all the coordinates (x, y, z) are non-negative. The bounding planes define this space as \(x = 1\), \(y = 1\), and \(z = 2\). This sets the region over which we take our triple integral.

• This shape's simplicity makes it ideal for initial multivariable calculus problems, allowing easier setup and understanding of bounds during integration.
• The concept extends to real-world applications such as computing volumes of warehouses or packages.

Knowing how to work with a rectangular solid is crucial, laying the groundwork for tackling more complex regions in advanced problems.