Problem 63 - University Calculus: Early Transcendentals

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 of the function is \( \frac{2}{3} \).

Step 1: Identify the Volume of the Region

The region of integration is the cylinder with radius 1, height from \(z = -1\) to \(z = 1\), represented in cylindrical coordinates as \(0 \leq r \leq 1\), \(0 \leq \theta \leq 2\pi\), \(-1 \leq z \leq 1\). The volume \(V\) of this cylinder is given by integrating over these limits: \[ V = \int_{0}^{2\pi} \int_{0}^{1} \int_{-1}^{1} r\, dz\, dr\, d\theta = 2\pi \cdot \int_{0}^{1} r \cdot 2\, dr = 2\pi \cdot \int_{0}^{1} 2r\, dr.\]

Step 2: Calculate the Volume of the Cylinder

Compute the integral to find the volume: \[ V = 2\pi \cdot \int_{0}^{1} 2r\, dr = 2\pi \cdot \left[ r^2 \right]_0^1 = 2\pi \cdot 1 = 2\pi. \] This is the volume of the cylinder.

Step 3: Set Up the Integral for the Average Value

The average value of the function is given by \( \frac{1}{V} \int \int \int_{R} f(r, \theta, z)\, dV \), where \( f(r, \theta, z) = r \). Therefore, the integral we need is: \[ \int_{0}^{2\pi} \int_{0}^{1} \int_{-1}^{1} r \cdot r\, dz\, dr\, d\theta. \] Simplified, it becomes: \[ \int_{0}^{2\pi} \int_{0}^{1} r^2 \cdot 2\, dr\, d\theta. \]

Step 4: Evaluate the Integral

Calculate the integral \( \int_{0}^{2\pi} \int_{0}^{1} 2r^2\, dr\, d\theta \):\[ \int_{0}^{2\pi} 2\left[ \frac{r^3}{3} \right]_0^1 d\theta = \int_{0}^{2\pi} \frac{2}{3} d\theta = \frac{2}{3} \times 2\pi = \frac{4\pi}{3}. \]

Step 5: Calculate the Average Value

The average value \(f_{avg}\) of the function over the region is \( \frac{1}{V} \times \int \int \int_{R} f(r, \theta, z)\, dV \):\[ f_{avg} = \frac{1}{2\pi} \times \frac{4\pi}{3} = \frac{2}{3}. \] Therefore, the average value is \( \frac{2}{3} \).

Key Concepts

Average Value of a Function

The average value of a function provides a way to analyze its behavior over a specified region. This is particularly useful in multivariable calculus where functions depend on more than one variable. In essence, calculating the average value involves integrating the function over the domain and dividing by the domain's measure.

To determine the average value of a function like \(f(r, \theta, z) = r\) over a cylindrical region, the formula used is:

• Average Value \(f_{avg} = \frac{1}{V} \int \int \int_{R} f(r, \theta, z) \, dV\).
• Here, \(V\) represents the volume of the region over which we average the function.
• \(dV\) is the differential volume element in the coordinates we are using.

By applying this formula, one assesses not only individual values but how the function acts across the entire region.

Multivariable Calculus

Multivariable calculus explores functions of several variables---for instance, functions like \(f(x, y, z)\). In solving problems, it's essential to understand how changes in one variable affect others. This understanding can guide us when setting up integrals over complex geometric regions.

In this context, cylindrical coordinates provide a natural fit for regions shaped like cylinders. They use \(r\) (radius), \(\theta\) (angle), and \(z\) (height) to describe space.

• Equations in cylindrical coordinates can simplify integration and calculations in cylindrical regions.
• Volume calculations often involve three integrals, each representing a step through the distinct dimensions of the region.
• This helps in breaking down the process of integration into simpler, more manageable parts.

This dimension-wise thinking is key in managing the complexities of multivariable calculus.

Integration in Polar Coordinates

Polar coordinates offer a useful method to perform integration for functions in circular domains. When dealing with cylindrical regions in 3D, we extend this concept by adding the height dimension \(z\), leading to cylindrical coordinates.

In cylindrical coordinates:

• \(r\) represents the radial distance from the origin.
• \(\theta\) is the angular position in the plane.
• Integrations involve the differential element \(r \cdot dz \cdot dr \cdot d\theta\).

The integration process, like finding the average value of the function \(f(r, \theta, z) = r\), shows how integrals are set up:

• First, identify the bounds for each variable based on the region.
• Perform the integration iteratively from the inside out.

This technique simplifies the handling of complex functions in regions that are not rectangular or even traditional in shape.