Problem 40 - Thomas Calculus

Question

In Exercises \(37-40,\) find the average value of \(F(x, y, z)\) over the given region. $$ \begin{array}{l}{F(x, y, z)=x y z \text { over the cube in the first octant bounded by the }} \\ {\text { coordinate planes and the planes } x=2, y=2, \text { and } z=2}\end{array} $$

Step-by-Step Solution

Answer

The average value of \(F(x, y, z) = xyz\) over the cube is 1.

Step 1: Determine the Volume of the Cube

Since the cube is in the first octant and bounded by the coordinate planes and the planes \(x=2\), \(y=2\), and \(z=2\), the cube has side lengths of 2. Therefore, the volume \(V\) of the cube is \(V = 2\times 2\times 2 = 8\).

Step 2: Set Up the Triple Integral

To find the average value of \(F(x, y, z) = xyz\) over the cube, set up the triple integral for the region \(0 \leq x, y, z \leq 2\). The integral is \[ \int_0^2 \int_0^2 \int_0^2 xyz \, dx \, dy \, dz \].

Step 3: Evaluate the Inner Integral

Evaluate \( \int_0^2 xyz \, dx \). Treat \(y\) and \(z\) as constants, so the integral becomes \[ \int_0^2 xyz \, dx = yz \left[ \frac{x^2}{2} \right]_0^2 = yz \cdot 2 = 2yz \].

Step 4: Evaluate the Middle Integral

Next, integrate with respect to \(y\): \(\int_0^2 2yz \, dy\). The integral becomes \[ 2z \left[ \frac{y^2}{2} \right]_0^2 = 2z \cdot 2 = 4z \].

Step 5: Evaluate the Outer Integral

Finally, integrate with respect to \(z\): \(\int_0^2 4z \, dz\). The integral becomes \[ \left[ 2z^2 \right]_0^2 = 2 \cdot 4 = 8 \].

Step 6: Calculate Average Value

The average value of \(F(x, y, z) = xyz\) over the cube is given by \(\frac{1}{V} \int \int \int xyz \, dV = \frac{8}{8} = 1\). Thus, the average value is 1.

Key Concepts

Triple Integral

In calculus, a triple integral is used to calculate a volume integral or a hyperspace volume. It's a multivariable calculus concept extending the idea of an integral into three-dimensional space. You work with three variables, usually represented as \(x\), \(y\), and \(z\). When you want to calculate a triple integral, you integrate a function over a three-dimensional space.

For example, you might want to find the mass of a solid object where the density of the object is given by a function \(F(x, y, z)\). The overall process involves three separate integrations, each one considering one variable at a time while keeping the others constant. The standard notation is:
\[ int \, int \, int \, F(x, y, z) \, dx \, dy \, dz.\]

• The first integral calculates over \(x\).
• The second integral does the calculation over \(y\).
• The third, and final, integrates over \(z\).

This step-by-step integration approach helps you confidently work through problems in three-dimensional space.

First Octant

The concept of the first octant comes from dividing three-dimensional space, or \(3D\) coordinate plane, into eight regions based on the axes intersection. Imagine the whole space is divided as if by a glass x, y, and z plane. Each intersection forms what's known as an 'octant'. The first octant is where \(x\), \(y\), and \(z\) are all positive.

In more straightforward terms, for a point to exist in the first octant, it needs to have coordinates such that \((x, y, z) > 0\). This information is crucial when dealing with volume or area calculations, like a cube bounded by certain planes, where the considered space is simply in the positive side of the coordinate axes.
- Note that in only positive values, computational challenges become simpler especially when using integration. - The first octant often represents real-world problems, such as the volume of containers or land plots.

Coordinate Planes

Coordinate planes are three-dimensional planes used as a reference for measuring points in space using a coordinate system. In mathematics, the three primary coordinate planes are:

• The \(xy\)-plane, where \(z = 0\).
• The \(xz\)-plane, where \(y = 0\).
• The \(yz\)-plane, where \(x = 0\).

Each plane acts as a boundary, partitioning space and helping define regions like a cube or other geometric shapes in the context of multivariable calculus. For instance, when you're integrating over a space like a cube in the first octant, these planes serve as the limits for your triple integrals.

These planes help you quickly understand and visualize problems in three dimensions, offering a reference to see how a shape or solution fits into space.

Volume of a Cube

The volume of a cube gives a measure of the space enclosed within its six equal square faces. It's a straightforward geometric calculation essential in many multivariable calculus problems. This characteristic makes cubes a popular choice for exercises involving triple integrals in three-dimensional space.

The formula to calculate the volume is:
\[V = s^3,\]where \(s\) is the side length of the cube. For instance, if each side measures 2 units, the cube’s volume is \(2 \times 2 \times 2 = 8\) cubic units.

• Utilize this concept when setting up integrals over a defined region, especially when regular shapes are involved.
• A cube's geometric simplicity helps in applying boundary conditions efficiently because of its symmetry and equal side lengths.

Understanding how to determine a cube's volume extends directly to calculating more complex figures, setting a foundation for working with three-dimensional integrals and other advancement problems.