Problem 40 - 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 cube in the first octant bounded by the coordinate planes and the planes \(x=2, y=2,\) and \(z=2\).

Step-by-Step Solution

Answer

The average value is 1.

Step 1: Identify the Region of Integration

The problem states that the region of integration is a cube in the first octant. The cube is bounded by the planes \(x = 0\), \(y = 0\), \(z = 0\), \(x = 2\), \(y = 2\), and \(z = 2\). Therefore, the region of integration can be described as \(0 \leq x \leq 2\), \(0 \leq y \leq 2\), and \(0 \leq z \leq 2\).

Step 2: Determine the Integral of the Function Over the Region

To find the average value of \(F(x, y, z) = xyz\) over the cube, we first need the volume integral of the function over the specified region. Set up the triple integral:\[\int_{0}^{2} \int_{0}^{2} \int_{0}^{2} xyzdzdy \, dx\].

Step 3: Evaluate the Inner Integral

Integrate \(xyz\) with respect to \(z\):\[\int_{0}^{2} xyz \; dz = xy \left[\frac{z^2}{2}\right]_0^2 = xy \times \frac{4}{2} = 2xy\].

Step 4: Evaluate the Middle Integral

Substitute the result from the inner integral and integrate with respect to \(y\):\[\int_{0}^{2} 2xy \; dy = 2x \left[\frac{y^2}{2}\right]_0^2 = 2x \times 2 = 4x\].

Step 5: Evaluate the Outer Integral

Substitute the result from the middle integral and integrate with respect to \(x\):\[\int_{0}^{2} 4x \; dx = \left[2x^2\right]_0^2 = 2 \times 4 = 8\].

Step 6: Calculate the Volume of the Region

The volume of the region, a cube with side length 2 in the first octant, is calculated as:\[Volume = (2-0) \times (2-0) \times (2-0) = 8\].

Step 7: Compute the Average Value

The average value of a function over a region is given by the integral of the function divided by the volume of the region. Hence:\[\text{Average value} = \frac{\text{Integral}}{\text{Volume}} = \frac{8}{8} = 1\].

Key Concepts

Triple Integrals

Triple integrals are used to evaluate the volume under a surface in three-dimensional space. They work similarly to single and double integrals, but they include an extra dimension.

In the context of multivariable calculus, triple integrals allow us to integrate functions over three variables—typically represented as \( x, y, \) and \( z \). They are excellent for calculating volumes and averages over 3D regions.

Here's how you set up a triple integral:

• Choose the region over which to integrate. This could be a 3D shape, like a cube or a sphere.
• Establish the limits of integration for each variable. These limits define the boundaries of the region.
• Integrate the function in a step-by-step manner—starting with one variable and moving outward to others.

In the original exercise, the triple integral is used to compute the integral of the function \( F(x, y, z) = xyz \) over a specified cube region.

Average Value of a Function

The average value of a function within a given region is a concept that helps in understanding the function’s mean behavior over space. For multivariable functions, this is calculated as the integral of the function divided by the volume of the region over which it's integrated.

To find the average value of \( F(x, y, z) = xyz \) over a specific region, we first perform a triple integration over that region to find the total sum of the function’s values. Then, we divide this sum by the size of the region, which for 3D problems is its volume. This gives you how the function behaves on average across the entire region.

In the exercise, it is shown that the average value over the described cube is computed to be 1.

Regions of Integration

The region of integration refers to the 3D space over which a triple integral is calculated. Selecting the right region is crucial, as it determines the limits for each variable in your integrals.

In Cartesian coordinates, regions are often defined by planes. For example, in the given exercise, the region is a cube bound by the planes \( x = 0, y = 0, z = 0 \) and \( x = 2, y = 2, z = 2 \). This specifies a cube in the first octant of the coordinate system.

• First octant means: all coordinates \( (x, y, z) \) are positive.
• The cube is specifically defined by the set of inequalities \( 0 \leq x \leq 2, 0 \leq y \leq 2, 0 \leq z \leq 2 \).

Understanding your region's geometric boundaries ensures accurate calculation for triple integrals, which is crucial in finding the average value of functions as in the exercise.

Integration in Cartesian Coordinates

Integration in Cartesian coordinates involves using \( x, y, \) and \( z \) as the primary dimensions. It’s the most straightforward framework for integrating functions over rectangular or cuboidal regions.

For the cube region in this exercise, each dimension is integrated over a defined range between boundaries set by planes (such as \( x=2 \) or \( y=0 \)).

• The process starts by integrating with respect to one variable, treating all others as constants, then moving to the next.
• This approach allows building up complexity stage by stage until the complete integration is solved.

Each step of the integration process addressed by integrating inside-out from the inner integral (\( z \)), through middle (\( y \)), to the outer (\( x \)), following the limits established by the region's boundaries.