Problem 65 - Thomas Calculus

Question

Find the average value of the function \(f(\rho, \phi, \theta)=\rho\) over the solid ball \(\rho \leq 1\)

Step-by-Step Solution

Answer

The average value of the function is \( \frac{3}{4} \).

Step 1: Understand the Problem

We need to find the average value of the function \( f(\rho, \phi, \theta) = \rho \) over a ball described in spherical coordinates by \( \rho \leq 1 \). The region is a solid sphere of radius 1 centered at the origin.

Step 2: Set up the Integral

The average value of a function over a region \( R \) is given by \( \frac{1}{\text{Volume of } R} \int_R f(\rho, \phi, \theta) \, dV \). Here, \( dV \) in spherical coordinates is \( \rho^2 \, ext{sin} \, \phi \, d\rho \, d\phi \, d\theta \).

Step 3: Calculate the Volume of the Sphere

The volume of the sphere \( \rho \leq 1 \) is calculated using the integral \( \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^2 \, ext{sin} \, \phi \, d\rho \, d\phi \, d\theta \) which simplifies to \( \frac{4}{3}\pi \).

Step 4: Set up the Triple Integral

We need to evaluate \( \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho \cdot \rho^2 \, ext{sin} \, \phi \, d\rho \, d\phi \, d\theta \), which is the integral of the function times the differential volume element, over the region \( \rho \leq 1 \).

Step 5: Integrate with Respect to \( \rho \)

Perform the integration of \( \rho^3 \) with respect to \( \rho \): \( \int_0^1 \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_0^1 = \frac{1}{4} \).

Step 6: Integrate with Respect to \( \phi \)

Next, integrate \( \frac{1}{4} \cdot \text{sin} \, \phi \) with respect to \( \phi \): \( \int_0^{\pi} \text{sin} \, \phi \, d\phi = [-\text{cos} \, \phi]_0^{\pi} = 2 \). Thus, this integral is \( \frac{1}{4} \cdot 2 = \frac{1}{2} \).

Step 7: Integrate with Respect to \( \theta \)

Finally, integrate \( \frac{1}{2} \) with respect to \( \theta \): \( \int_0^{2\pi} \, d\theta = 2\pi \). Thus, the integral is \( 2\pi \cdot \frac{1}{2} = \pi \).

Step 8: Calculate the Average Value

The average value is given by \( \frac{1}{\text{Volume of sphere}} \cdot \text{Result of integral} = \frac{1}{\frac{4}{3}\pi} \cdot \pi = \frac{3}{4} \).

Key Concepts

Spherical Coordinates

Spherical coordinates provide a way to represent points in three-dimensional space using three variables:

• \( \rho \)
• \( \phi \)
• \( \theta \)

These variables define a point in terms of its distance from a fixed origin and two angles. \( \rho \) represents the radial distance from the origin, \( \phi \) is the polar angle measured from the positive z-axis, and \( \theta \) is the azimuthal angle in the xy-plane from the positive x-axis. This system is especially useful in dealing with objects that have spherical symmetry, such as spheres or parts of spheres.For instance, in the exercise, the solid ball is defined by \( \rho \leq 1 \). This region includes all points from the origin extending to a maximum radial distance of 1.

Triple Integrals

Triple integrals allow us to calculate the total of a function over a three-dimensional region. In spherical coordinates, these integrals take into account the volume elements unique to this coordinate system.The volume element in spherical coordinates is given by:\[ dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]Using triple integrals, one can integrate scalar fields over volumes. The process involves successive integration with respect to each coordinate, usually in a specific order such as \( \rho \), \( \phi \), and then \( \theta \). Each integral might require substitution based on the limits dictated by the region of integration, and in our case, the limits are dictated by the sphere's parameters.

Volume of a Sphere

The volume of a sphere in spherical coordinates can be determined using a triple integral. The formula used is crucial in determining the average value of a function over the sphere.For a sphere with radius equal to 1, the volume is found by evaluating:\[\int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^2 \, \sin \phi \, d\rho \, d\phi \, d\theta\]This integral simplifies to the well-known formula:\[ \frac{4}{3} \pi \] Understanding this calculation is essential. It allows for the average value of any function over a spherical region to be found by taking the triple integral of the function over the region and dividing by this volume. This gives a comprehensive understanding of how the space is represented mathematically, notably for symmetrical objects like spheres.