Problem 39 - University Calculus: Early Transcendentals

Question

In Exercises \(37-40,\) find the average value of \(F(x, y, z)\) over the given region. \(F(x, y, z)=x^{2}+y^{2}+z^{2}\) 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 of \(F(x, y, z)\) over the cube is 1.

Step 1: Understand the Region of Integration

The cube is bounded by the coordinate planes and the planes \(x=1\), \(y=1\), and \(z=1\). This means the bounds for \(x\), \(y\), and \(z\) are all from 0 to 1, forming a cube.

Step 2: Set Up the Triple Integral

To find the average value of \(F(x, y, z)\), we use the formula for the average value of a function over a region \(R\):\[ \frac{1}{\text{Volume of } R} \int \int \int_R F(x, y, z) \, dV \]The function is \(F(x, y, z) = x^2 + y^2 + z^2\). For a cube \(0 \leq x, y, z \leq 1\), the volume is 1.

Step 3: Compute the Volume of the Region

The cube extends from 0 to 1 in each direction, so its volume is \(1 \times 1 \times 1 = 1\).

Step 4: Formulate the Triple Integral

Write the integral as: \[ \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (x^2 + y^2 + z^2) \, dx \, dy \, dz \] This integral will calculate the volume under the function \(F(x, y, z)\) over the specified region.

Step 5: Evaluate the Integral with Respect to x

Integrate \(x^2\) with respect to \(x\) first: \[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} \]Thus, the integral simplifies to: \[ \int_{0}^{1} \int_{0}^{1} \left(\frac{1}{3} + y^2 + z^2\right) \, dy \, dz \]

Step 6: Evaluate the Integral with Respect to y

Next, integrate with respect to \(y\):\[ \int_{0}^{1} (\frac{1}{3} + y^2) \, dy = [\frac{1}{3}y + \frac{y^3}{3}]_{0}^{1} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]This results in the integral:\[ \int_{0}^{1} \left(\frac{2}{3} + z^2 \right) \, dz \]

Step 7: Evaluate the Integral with Respect to z

Now integrate with respect to \(z\):\[ \int_{0}^{1} (\frac{2}{3} + z^2) \, dz = [\frac{2}{3}z + \frac{z^3}{3}]_{0}^{1} = \frac{2}{3} + \frac{1}{3} = 1 \]

Step 8: Calculate the Average Value

The average value of the function over the region is given by:\[ \text{Average} = \frac{1}{\text{Volume}} \times \text{integral result} = \frac{1}{1} \times 1 = 1 \]

Key Concepts

Triple Integrals

Triple integrals are an extension of the idea of single and double integrals, used to calculate the volume under a surface in three-dimensional space. They allow us to integrate functions of three variables over a specified region. In our exercise, we aim to find the average value of the function \(F(x, y, z) = x^2 + y^2 + z^2\) over a cubical region. This involves setting up and evaluating a triple integral.The process of triple integration involves performing integration over each of the three variables, often in an order that simplifies computation. Starting with the innermost integral, you evaluate towards the outer layers of integration. These integrals are commonly evaluated with bounds defined by the region of interest, in this case, a cube defined by the limits 0 to 1 for each variable. By computing these integrals, we find a cumulative measure over the volume occupied in three-dimensional space.

Average Value of a Function

The concept of the average value of a function is crucial when analyzing how a function behaves over a specific region. To calculate the average value of a function in calculus, especially over a multi-dimensional region, you would essentially sum (integrate) the function's values over the entire region and then divide by the measure of that region, which in three dimensions is the volume.In our specific example, the average value is calculated using:\[\text{Average} = \frac{1}{\text{Volume of } R} \int \int \int_R F(x, y, z) \, dV\]By determining the integral of \(F(x, y, z) = x^2 + y^2 + z^2\) over the unit cube in the first octant and dividing by the volume of the cube, we obtain the average value, which gives us a sense of the function's typical value over the region.

Volume of a Region

The volume of a region in three-dimensional space is integral to understanding and solving problems involving triple integrals. In the context of our exercise, the volume is simply calculated because the region of integration is a unit cube.Since the region is defined as a cube in the first octant bound by \(x = 1\), \(y = 1\), and \(z = 1\), the bounds of integration are each from 0 to 1. The volume of this cube can be calculated easily by multiplying its side lengths: \[1 \times 1 \times 1 = 1\]This straightforward volume is crucial as it serves as the denominator in calculating the average value of the function. It highlights how the integral is proportional to the size of the region over which it is evaluated.

Cube in First Octant

The concept of a cube in the first octant is fundamental when solving multivariable calculus problems, as it defines a specific and familiar region of integration. The first octant is where all three coordinates \(x\), \(y\), and \(z\) are positive, essentially one-eighth of three-dimensional space.In our exercise, the cube lies within this first octant, bounded by the coordinate planes (\(x=0\), \(y=0\), \(z=0\)) and the planes \(x=1\), \(y=1\), and \(z=1\). Estimating volumes or averages across this region enables a uniform treatment of each variable due to the equal bounds (0 to 1), simplifying the integration process. Understanding this spatial configuration aids in visualizing how the function behaves and integrates over the specified region. This uniformity makes the calculations more straightforward, as all integration limits are identical.