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Chapter 15: Problem 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 \(z=2\)

Short Answer

Expert verified
The average value is 1.

Step by step solution

01

Understand the Problem

We need to find the average value of the function \(F(x, y, z) = x y z\) over a specified region. The region is a cube in the first octant with boundaries defined by \(x = 2\), \(y = 2\), and \(z = 2\), and the coordinate planes.
02

Set Up the Integral

The average value of a function \(F\) over a volume \(V\) is given by \( \frac{1}{V} \int \int \int_V F(x, y, z) \, dV \). For the given cube, the limits of integration for \(x\), \(y\), and \(z\) each go from 0 to 2.
03

Calculate Volume of the Region

The region is a cube with side length 2. Its volume is \(V = 2 \times 2 \times 2 = 8\).
04

Integrate the Function

We need to evaluate the triple integral \( \int_0^2 \int_0^2 \int_0^2 x y z \, dz \, dy \, dx \). First, integrate with respect to \(z\):\[ \int_0^2 x y z \, dz = \left[ \frac{x y z^2}{2} \right]_0^2 = 2 x y \].
05

Continue Integration Over y

Integrate the resulting expression \(2xy\) with respect to \(y\):\[ \int_0^2 2 x y \, dy = \left[ x y^2 \right]_0^2 = 4 x \].
06

Finish Integration Over x

Now, integrate \(4x\) with respect to \(x\):\[ \int_0^2 4 x \, dx = \left[ 2 x^2 \right]_0^2 = 8 \].
07

Compute Average Value

The average value of the function over the region is given by \(\frac{1}{8} \times 8 = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triple Integration
Triple integration is a powerful mathematical tool used to find values over three-dimensional spaces. It is an extension of single and double integrals to the third dimension. The idea is simple:
  • For a function of three variables like \(F(x, y, z)\), the triple integral allows you to calculate the total accumulation of the function's values over a specified volume in space, such as a cube or sphere.
  • You are integrating one variable at a time within the limits that define the volume.
  • The notation \( \int \int \int_V F(x, y, z) \, dV \) is used, where \(V\) represents the volume of integration.
Triple integration can be thought of as summing up tiny bits of three-dimensional space. In our specific case, you're finding the integral of \(F(x, y, z) = xyz\) over a cube in the first octant bounded by \(x=2\), \(y=2\), and \(z=2\). This cube is represented by integration limits from 0 to 2 for each variable, covering all dimensions of the cube.
Volume of a Region
Finding the volume of a region in multivariable calculus is crucial for applications involving triple integration and average values. The volume is essentially the "size" of the region you're interested in.
  • Just as in real life where you calculate the volume of a soda can or box, in calculus, you calculate the volume of mathematical solids like cubes and spheres.
  • To calculate the volume of a regular-shaped solid, multiply the lengths of its sides. In our example, the cube has side lengths 2, so its volume \(V\) is \(2 \times 2 \times 2 = 8\).
It鈥檚 essential to understand that this volume is important because it鈥檚 used to calculate the average value of a function over that region. The formula for finding the average value of a function integrates the function over the volume and divides by the volume to spread the total value "evenly" over that space.
Average Value of a Function
The average value of a function is an important concept in multivariable calculus used to determine the mean output of a function over a specific region.
  • In simple terms, it tells us what 'average' value we can expect by evaluating a function across all points in a particular region.
  • The formula used to find this is:\[\text{Average Value} = \frac{1}{V} \int \int \int_V F(x, y, z) \, dV\]In this formula, \(V\) is the volume of the region over which the average is calculated.
For our example, \(F(x, y, z)=xyz\), you calculate the triple integral over the cube's volume \(V = 8\), and divide by this volume. The result here turns out to be 1, indicating an even distribution of values for this particular function and region. Essentially, it means that if you imagine this function as a solid shape, its average height is 1.

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Most popular questions from this chapter

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$\int_{1}^{e} \int_{0}^{\ln x} x y d y d x$$

Gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral. $$\int_{-\pi / 3}^{\pi / 3} \int_{0}^{\sec t} 3 \cos t d u d t \quad \text { (the } t u \text { -plane) }$$

Sketch the region of integration, reverse the order of integration, and evaluate the integral. Find the volume of the solid cut from the square column \(|x|+|y| \leq 1\) by the planes \(z=0\) and \(3 x+z=3\)

Evaluate the spherical coordinate integrals. $$\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{2 \sin \phi} \rho^{2} \sin \phi d \rho d \phi d \theta$$

Evaluate the spherical coordinate integrals. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{\sec \phi}(\rho \cos \phi) \rho^{2} \sin \phi d \rho d \phi d \theta$$
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