Problem 39 - Thomas Calculus

Question

Find the average value of \(F(x, y, z)\) over the given region. region. \(F(x, y, z)=x^{2}+y^{2}+z^{2}\) over the cube in the first octant bounded by the coordinate planes and the planes \(x=1, y=1,\) and \(z=1\)

Step-by-Step Solution

Answer

The average value is 1.

Step 1: Identify the region

The region of integration is a cube in the first octant where all coordinates are positive: bounded by the planes from 0 to 1 for each axis. This means the cube ranges from 0 to 1 for each of the coordinates \(x\), \(y\), and \(z\).

Step 2: Determine the limits of integration

For this problem, since the cube is bounded from 0 to 1 on each axis, the limits of integration are \(0 \leq x \leq 1\), \(0 \leq y \leq 1\), and \(0 \leq z \leq 1\).

Step 3: Set up the volume integral

To find the average value of \(F(x, y, z) = x^2 + y^2 + z^2\) over the cube, we need to integrate the function over the entire volume of the cube. The integral is set up as follows:\[\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (x^2 + y^2 + z^2) \, dz \, dy \, dx\].

Step 4: Calculate the integral

Perform the integration step-by-step:First, integrate with respect to \(z\):\[\int_{0}^{1} (x^2 + y^2 + z^2) \, dz = x^2z + y^2z + \frac{z^3}{3}\bigg|_0^1 = x^2 + y^2 + \frac{1}{3}\].Next, integrate the result with respect to \(y\):\[\int_{0}^{1} (x^2 + y^2 + \frac{1}{3}) \, dy = x^2y + \frac{y^3}{3} + \frac{y}{3} \bigg|_0^1 = x^2 + \frac{1}{3} + \frac{1}{3}\].Finally, integrate with respect to \(x\):\[\int_{0}^{1} (x^2 + \frac{2}{3}) \, dx = \frac{x^3}{3} + \frac{2x}{3} \bigg|_0^1 = \frac{1}{3} + \frac{2}{3} = 1\].

Step 5: Find the volume of the region

The volume of the cube is calculated by multiplying the lengths of its sides. Since the cube extends from 0 to 1 for each dimension (\(x, y, z\)), its volume is \(1 \times 1 \times 1 = 1\).

Step 6: Calculate the average value

The average value \(\bar{F}\) of the function over the region is given by the volume average formula:\[\bar{F} = \frac{1}{\text{{Volume of Region}}} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (x^2 + y^2 + z^2) \, dz \, dy \, dx\].Since the integral is equal to 1 and the volume is 1, \(\bar{F} = \frac{1}{1} = 1\).

Key Concepts

Multivariable Calculus

Multivariable calculus is an extension of calculus where functions have more than one variable. In this specific exercise, we're dealing with a function of three variables, denoted as \( F(x, y, z) = x^2 + y^2 + z^2 \). Understanding functions in multiple dimensions requires us to think not just in terms of lines or curves, but surfaces or volumes. This broader perspective includes concepts like gradients, partial derivatives, and integration over surfaces or volumes.

When solving problems in multivariable calculus, it is common to evaluate the behavior of the function over a specified region, rather than just at a single point or along a single axis. Calculating the average of a function over a region, as done in the exercise, provides insight into the general behavior of the function throughout that space. Multivariable calculus isn't limited to theoretical exercises; it applies to real-world scenarios such as physics fields, fluid dynamics, and any discipline requiring three-dimensional modeling.

Triple Integration

Triple integration is a process within calculus used when dealing with functions of three variables. It's essentially integrating three times: once for each variable. This technique is crucial when we need to calculate a total quantity over a three-dimensional space or volume. In the given exercise, triple integration is used to find the average value of the function \( F(x, y, z) = x^2 + y^2 + z^2 \) within a geometric space known as a cube.

To perform a triple integral, one must consider integration limits, which define the region of space you're examining. Each integral reduces the region one dimension at a time. It's like peeling away dimensions layer by layer until you're left with a scalar quantity. The integration steps are executed in a particular order (e.g., integration with respect to \(z\), then \(y\), and finally \(x\) in this problem). Choosing the correct order can simplify the problem.

• 1st Integration: Along the \(z\)-axis, yielding partially evaluated expressions.
• 2nd Integration: Along the \(y\)-axis, refining these expressions further.
• 3rd Integration: Along the \(x\)-axis, culminating in a single value representing the integral over the entire volume.

Mastery of these techniques allows you to solve complex problems involving physical quantities distributed over 3D spaces.

Region of Integration

The region of integration is the specific three-dimensional space over which you perform the integration of a function. In our example, the region of integration is a cube defined in the first octant of a three-dimensional coordinate system. The first octant is where all the coordinate values (\(x\), \(y\), and \(z\)) are positive.

The boundaries of this cube are set by planes at \(x=0\), \(y=0\), \(z=0\), and the planes \(x=1\), \(y=1\), \(z=1\). Consequently, each of these planes creates a natural limit of integration, ranging from 0 to 1 for all three dimensions.

• The cube's size and location are crucial to solving the problem, as it affects calculations for volume and influences the outcome of the integral.
• Determining these limits accurately ensures the integration properly covers the intended space.
• In applied sciences, defining a region accurately can mean the difference between correct and erroneous results.

Understanding how to define and use regions of integration is fundamental to multivariable calculus and ensures accurate representation of data within a given spatial framework.