Problem 63 - Thomas Calculus

Question

Find the average value of the function \(f(r, \theta, z)=r\) over the region bounded by the cylinder \(r=1\) between the planes \(z=-1\) and \(z=1\)

Step-by-Step Solution

Answer

The average value is \( \frac{2}{3} \).

Step 1: Understand the Geometry

The region is a cylindrical volume in cylindrical coordinates where the cylinder is defined by the equation \( r = 1 \). It extends vertically between the planes \( z = -1 \) and \( z = 1 \). Therefore, this region can be described as \( 0 \leq r \leq 1 \), \( 0 \leq \theta \leq 2\pi \), and \( -1 \leq z \leq 1 \).

Step 2: Set Up the Integral for the Average Value

The average value of the function \( f(r, \theta, z) = r \) over a volume \( V \) is given by the formula \( \frac{1}{\text{Volume}} \int_V f(r, \theta, z) \, dV \). In cylindrical coordinates, the volume element \( dV \) is given by \( r \, dr \, d\theta \, dz \).

Step 3: Calculate the Volume of the Region

To find the volume \( V \), we integrate the volume element over the region: \[V = \int_0^1 \int_0^{2\pi} \int_{-1}^1 r \, dz \, d\theta \, dr\].Evaluating the innermost integral: \[\int_{-1}^1 \, dz = 2.\]Substituting this into the volume integral gives:\[V = \int_0^1 \int_0^{2\pi} 2r \, d\theta \, dr.\]

Step 4: Evaluate the Volume Integral

Continuing from Step 3:\[\int_0^{2\pi} \, d\theta = 2\pi,\]\[V = \int_0^1 2r (2\pi) \, dr = 4\pi \int_0^1 r \, dr.\]Now, integrate with respect to \( r \):\[V = 4\pi \left[ \frac{r^2}{2} \right]_0^1 = 4\pi \times \frac{1}{2} = 2\pi.\]

Step 5: Evaluate the Integral for Average Value

Calculate the integral of \( f(r, \theta, z) \) over the region:\[\int_0^1 \int_0^{2\pi} \int_{-1}^1 r^2 \, dz \, d\theta \, dr.\]Using the integral over \( z \) from Step 3:\[\int_{-1}^1 \, dz = 2,\]Substitute into the function integral:\[\int_0^1 \int_0^{2\pi} 2r^2 \, d\theta \, dr = \int_0^1 2r^2 (2\pi) \, dr = 4\pi \int_0^1 r^2 \, dr.\]

Step 6: Complete the Average Value Calculation

Now, integrate \( r^2 \) with respect to \( r \):\[\int_0^1 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^1 = \frac{1}{3}.\]Thus, the integral for average value is:\[4\pi \cdot \frac{1}{3} = \frac{4\pi}{3}.\]Finally, divide by the volume to get the average value:\[\text{Average Value} = \frac{\frac{4\pi}{3}}{2\pi} = \frac{4}{6} = \frac{2}{3}.\]

Step 7: Conclusion

The average value of the function \( f(r, \theta, z) = r \) over the given region is \( \frac{2}{3} \).

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a powerful way to represent points in a three-dimensional space. They are especially useful when dealing with regions that exhibit cylindrical symmetry, such as cylinders and cones. In this coordinate system, a point is described using three parameters: radial distance \( r \), angular coordinate \( \theta \), and height \( z \).

• \( r \) is the distance from the point to the \( z \)-axis. It is the radius in the plane perpendicular to the \( z \)-axis.
• \( \theta \) represents the angle formed with the positive \( x \)-axis, typically measured in radians between 0 and \( 2\pi \) to complete a full circle.
• \( z \) is the same as the \( z \) in Cartesian coordinates, representing the height above the origin on the \( z \)-axis.

Cylindrical coordinates are particularly advantageous for integrating over volumes with rotational symmetry, making computations more straightforward compared to using Cartesian coordinates. The conversion from cylindrical to Cartesian coordinates is given by \( x = r\cos(\theta) \), \( y = r\sin(\theta) \), and \( z = z \). These equations facilitate easy switching between systems as needed.

Integration in Cylindrical Coordinates

When integrating in cylindrical coordinates, it’s crucial to understand the volume element. The differential volume element \( dV \) for integration in cylindrical coordinates is given by \( r \, dr \, d\theta \, dz \).

This accounts for the radial change (\( dr \)), the angular change (\( d\theta \)), and the height change (\( dz \)). The factor of \( r \) arises because, for a small angular segment \( d\theta \), the arc length in the \( xy \)-plane is \( r \, d\theta \), meaning the area element in the radial and angle direction is \( r \, dr \, d\theta \).

To set up an integral, you must:

• Identify the region of integration
• Determine limits for \( r \), \( \theta \), and \( z \)
• Account for the volume element

Integration involves conducting multiple integrations: typically with respect to \( z \) first, followed by \( \theta \), and \( r \) last. This approach will vary depending on the specific region and function you're working with. The arrangement and boundaries of \( r \), \( \theta \), and \( z \) play a crucial role in defining the limits of each integral.

Volume of a Cylindrical Region

The volume of a cylindrical region is a central component when calculating integrals in problems involving cylindrical coordinates. It is imperative to understand the region being described. For example, if a cylinder is bounded by \( r = 1 \), it means that the radius of this region is 1, forming a cylinder that extends around the \( z \)-axis.

In our specific example, the cylinder extends between the planes \( z = -1 \) and \( z = 1 \). This results in a cylindrical object with a full circle at each cross-section between these planes. To find the volume, we perform the definite integral:\[V = \int_0^1 \int_0^{2\pi} \int_{-1}^1 r \, dz \, d\theta \, dr\]Breaking this down:

• The integral \( \int_{-1}^1 \, dz \) gives the height of the cylinder, resulting in 2.
• The integral \( \int_0^{2\pi} \, d\theta \) gives the full rotation along the \( xy \)-plane, yielding \( 2\pi \).
• Finally, integrating with respect to \( r \), \( \int_0^1 r \, dr \), completes the volume calculation.

The complete result is a volume \( V = 2\pi \), confirming the total space taken up by this cylindrical region. Understanding this volume is essential when computing average values of functions over such a region, as seen in the provided problem context.