Question

Use spherical coordinates to find the volume of the following regions. That part of the ball \(\rho \leq 4\) that lies between the planes \(z=2\) and \(z=2 \sqrt{3}\)

Short answer

The volume of the region in the ball of radius 4, constrained between the planes z = 2 and z = 2√3 is given by the triple integral in spherical coordinates: \(\frac{256\pi(\sqrt{3}-1)}{9}\).

Step-by-step solution

Convert to Spherical Coordinates

In spherical coordinates, a point (x, y, z) is represented by its radius \(\rho\), polar angle \(\theta\), and azimuthal angle \(\phi\). The relationship between the Cartesian coordinates (x, y, z) and the spherical coordinates is the following: \(x=\rho \sin\phi \cos\theta\) \(y=\rho \sin\phi \sin\theta\) \(z=\rho \cos\phi\) In this problem, the given equation \(\rho \leq 4\) already has the radius constraint in spherical coordinates, and the planes are given as \(z=2\) and \(z=2\sqrt{3}\).

Express Planes in Spherical Coordinates

We can use the relationship between Cartesian and spherical coordinates to express the given planes in spherical coordinates. For the plane \(z=2\), we have: \(2 = \rho \cos\phi\) For the plane \(z=2\sqrt{3}\), we have: \(2\sqrt{3} = \rho \cos\phi\) By solving for \(\rho\) and \(\phi\), we can set the proper boundaries for the spherical coordinates. For the first plane: \(\rho = \frac{2}{\cos\phi}\) For the second plane: \(\rho = \frac{2\sqrt{3}}{\cos\phi}\) Thus, the constraints of the region in spherical coordinates are: \(0 \le \rho \le 4\) \(0 \le \theta \le 2\pi\) \(\arccos\left(\frac{2}{\rho}\right) \le \phi \le \arccos\left(\frac{2\sqrt{3}}{\rho}\right)\)

Set Up the Triple Integral

With the integration boundaries established, we can now set up the triple integral. In spherical coordinates, the differential volume element dV is given by \(\rho^2\sin\phi d\rho d\theta d\phi\). Thus, the triple integral for our problem is: \(\int_{0}^{2\pi} \int_{\arccos\left(\frac{2}{\rho}\right)}^{\arccos\left(\frac{2\sqrt{3}}{\rho}\right)} \int_{0}^{4} \rho^2\sin\phi d\rho d\theta d\phi\)

Evaluate the Integral

To find the volume, we can now evaluate the integral: \(\int_{0}^{2\pi} d\theta \int_{\arccos\left(\frac{2}{\rho}\right)}^{\arccos\left(\frac{2\sqrt{3}}{\rho}\right)} \sin\phi d\phi \int_{0}^{4} \rho^2 d\rho\) 1) Perform the innermost integral (rho): \(\frac{\rho^3}{3} \Big|_0^4 = \frac{64}{3}\) 2) Now the middle integral (phi): \((-\cos\phi) \Big|_{\arccos(\frac{2}{\rho})}^{\arccos(\frac{2\sqrt{3}}{\rho})} = \frac{2\sqrt{3}}{\rho} - \frac{2}{\rho}\) 3) Finally, the outermost integral (theta): \(2\pi\left(\frac{2\sqrt{3}}{\rho} - \frac{2}{\rho}\right) \Big|_0^{2\pi} = 4\pi\left(\frac{2\sqrt{3}}{3} - \frac{2}{3}\right)\) Putting it all together, we get the total volume as: \(\frac{64}{3} \times 4\pi\left(\frac{2\sqrt{3}}{3} - \frac{2}{3}\right) = \frac{256\pi(\sqrt{3}-1)}{9}\) Thus, the volume of the region in question is \(\frac{256\pi(\sqrt{3}-1)}{9}\).

Key concepts

Triple Integral

A triple integral is a mathematical tool used to integrate functions of three variables over a three-dimensional region. Think of it as an extension of double integrals used for calculating area, but now applied to volumes. In this specific exercise, we deal with determining the volume of a region in space using the triple integral in spherical coordinates.
When setting up a triple integral for a volume computation, it involves three nested integrals, each corresponding to one of the three variables in your coordinate system. In our case with spherical coordinates:

• \(\rho\) is the radial coordinate representing the distance from the origin.
• \(\theta\) is the azimuthal angle in the xy-plane, measuring rotation from the positive x-axis.
• \(\phi\) is the polar angle from the positive z-axis.

The volume element in spherical coordinates is expressed as \(\rho^2 \sin \phi\ d\rho\ d\theta\ d\phi\), which accounts for the change in volume as you adjust \(\rho\), \(\theta\), and \(\phi\). To find the volume, you integrate this volume element over the specified limits for \(\rho\), \(\theta\), and \(\phi\). The limits are boundaries defining the region of interest. Each nested integral accounts for one variable: \(\rho\), \(\phi\), and \(\theta\).
We start with \(\rho\) and end with \(\theta\), following the conventional order of integration for spherical coordinates.

Volume Calculation

Calculating the volume in spherical coordinates requires both setting up and evaluating the triple integral correctly. The target region is defined by the constraints of the spherical parameters \(\rho\), \(\theta\), and \(\phi\). Here’s how we approach this task:
1. **Identify the Region:** The region we're interested in is a portion of a sphere, with radius \(\rho \leq 4\), between the planes \(z = 2\) and \(z = 2\sqrt{3}\).
2. **Express in Spherical Coordinates:** The equations for the planes are expressed as \(\rho \cos \phi = 2\) and \(\rho \cos \phi = 2\sqrt{3}\). This helps set the limits for \(\phi\).
3. **Set Limits for Integration:** - \(\rho\) ranges from 0 to 4
- \(\theta\) spans the full circle from 0 to \(2\pi\)
- \(\phi\) is contained between \(\arccos\left( \frac{2}{\rho}\right) \) and \(\arccos\left( \frac{2\sqrt{3}}{\rho}\right)\)
4. **Evaluate Triple Integral:** Finally, compute the triple integral to find the volume: \[ V = \int_{0}^{2\pi} \int_{\arccos\left(\frac{2}{\rho}\right)}^{\arccos\left(\frac{2\sqrt{3}}{\rho}\right)} \int_{0}^{4} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta\]
The proper evaluation of these integrals will yield the total volume of the region, showcasing both the elegance and complexity inherent in using spherical coordinates.

Coordinate Transformation

Coordinate transformation involves changing from one set of coordinate systems to another, making it easier to solve integrals involving complex geometries. In this exercise, transforming from Cartesian coordinates (x, y, z) to spherical coordinates (\(\rho, \theta, \phi\)) simplifies the setup of the problem.Spherical coordinates are particularly advantageous when dealing with problems involving spheres, hemispheres, or any symmetry about a central point. Here's how the transformation works:

• \(x = \rho \sin \phi \cos \theta\)
• \(y = \rho \sin \phi \sin \theta\)
• \(z = \rho \cos \phi\)

These transformations are used to convert the integration problem into a form where it is easier to apply the limits of integration for spherical geometries.
In our specific example, the region's geometry (part of a sphere between two planes) leads naturally to limits for \(\rho\), \(\theta\), and \(\phi\) in spherical coordinates. This allows us to express the planes' constraints more efficiently in terms of \(\phi\), simplifying boundary setup. As a result, spherical coordinates turn complex three-dimensional geometry into manageable integrals, harnessing the symmetry of the problem effectively. Understanding how and why to switch between different coordinate systems is a powerful skill in multivariable calculus.