Problem 40 - Thomas Calculus

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 of \( F(x, y, z) = xyz \) over the region is 1.

Step 1: Understand the Problem

We need to find the average value of the function \( F(x, y, z) = xyz \) over a cube in the first octant. The cube is bounded by the planes: \( x = 0 \), \( x = 2 \), \( y = 0 \), \( y = 2 \), \( z = 0 \), and \( z = 2 \).

Step 2: Determine the Volume of the Region

The volume \( V \) of the cube is the product of its side lengths. Here, each side of the cube equals 2, as it extends from 0 to 2 along the x, y, and z axes. Therefore, the volume is \( V = 2 \times 2 \times 2 = 8 \).

Step 3: Set Up the Triple Integral

To find the average value of \( F(x, y, z) = xyz \), we calculate the triple integral over the region and divide by the volume. The triple integral is set up as: \[\int_{0}^{2} \int_{0}^{2} \int_{0}^{2} xyz \, dx \, dy \, dz.\]

Step 4: Evaluate the Inner Integral

Evaluate the inner integral with respect to \( x \): \[\int_{0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} = 2.\]Now, the integral becomes \[\int_{0}^{2} \int_{0}^{2} 2yz \, dy \, dz.\]

Step 5: Evaluate the Middle Integral

Evaluate the integral with respect to \( y \): \[\int_{0}^{2} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{2} = \frac{2^2}{2} = 2.\]Now, the integral becomes \[\int_{0}^{2} 4z \, dz.\]

Step 6: Evaluate the Outer Integral

Evaluate the integral with respect to \( z \): \[\int_{0}^{2} z \, dz = \left[ \frac{z^2}{2} \right]_{0}^{2} = \frac{2^2}{2} = 2.\]The evaluated integral is 8.

Step 7: Compute the Average Value

Divide the result of the integral by the volume of the cube to find the average value:\[\text{Average value} = \frac{8}{8} = 1.\]

Key Concepts

Understanding Triple Integrals

Triple integrals extend the concept of single and double integrals to functions of three variables. This powerful tool allows us to calculate the volume under a surface in three-dimensional space. Imagine slicing through a 3D object with multiple layers. The triple integral gathers up these slices to give us an aggregate value over a specified region. Calculating the triple integral involves setting up a series of three integrals, each addressing one dimension.

• The innermost integral accounts for variation along the x-axis.
• The middle integral considers the y-axis.
• The outermost integral deals with the z-axis.

To set up a triple integral for a function like \( F(x, y, z) = xyz \) over a cube, we need to define the limits for each variable. For this scenario, all are from 0 to 2. The integral becomes \( \int_0^2 \int_0^2 \int_0^2 xyz \, dx \, dy \, dz \), where each integration follows sequentially to calculate the contribution of each dimension piece by piece.

Navigating the First Octant

The first octant is a specific region in three-dimensional space. It is where all three coordinates

• x
• y
• z

are non-negative, meaning they are either zero or positive. Imagine cutting the space around the origin (0, 0, 0) with the coordinate planes: x-y, y-z, and z-x. These cuts give us eight distinct regions called octants. The first octant is the one located where x, y, and z are all positive, making it the default starting space for many mathematical explorations in 3D.In our exercise, the function \( F(x, y, z) = xyz \) is evaluated in this first octant, specifically within a cube bounded by planes \( x=0 \), \( x=2 \), \( y=0 \), \( y=2 \), \( z=0 \), and \( z=2 \). This means we only consider the segment of the universe where x, y, and z stay within the 0 to 2 range. It helps simplify complex spatial problems by focusing only on this non-negative octant.

Volume Calculation in 3D Space

Volume calculation in 3D involves determining the space enclosed within boundaries. For regular shapes, like cubes or spheres, the calculation can be straightforward. In the case of the cube in the first octant, it extends evenly along the x, y, and z axes. Each side of the cube is of length 2.To find the volume \( V \) of such a cube, we simply multiply the lengths of its sides:\[V = 2 \times 2 \times 2 = 8\]This calculation is crucial before proceeding with integrating functions across this region. It's especially key when determining average values, like in this example. We perform the triple integral over this region to gather the aggregate contribution from each point within the cube. Once we have the result of the integral, we divide by the volume to find an average value. Here, the function \( F(x, y, z) = xyz \) is integrated over the cube's volume, and then the result is divided by 8 to yield the average value inside the cube, which is 1.