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Chapter 15: Problem 49

Find the volume of the portion of the solid sphere \(\rho \leq a\) that lies between the cones \(\phi=\pi / 3\) and \(\phi=2 \pi / 3\) $$\phi=2 \pi / 3$$

Short Answer

Expert verified
The volume is \(\frac{2\pi a^3}{3}\).

Step by step solution

01

Understand the Problem

We are tasked with finding the volume of a spherical sector of a sphere defined in spherical coordinates by two conical boundaries. The sphere has radius \(a\) and the sector lies between the angles \(\phi = \pi / 3\) and \(\phi = 2 \pi / 3\).
02

Set Up the Integral for Volume

The volume of a region in spherical coordinates \((\rho, \theta, \phi)\) is determined by the integral \(\int \int \int \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\). The bounds are \(0 \leq \rho \leq a\), \(\pi/3 \leq \phi \leq 2\pi/3\), and \(0 \leq \theta < 2\pi\).
03

Compute the Integral over \(\rho\)

Integrate \(\rho^2\) with respect to \(\rho\) from 0 to \(a\): \[ \int_0^a \rho^2 \, d\rho = \left. \frac{\rho^3}{3} \right|_0^a = \frac{a^3}{3}. \]
04

Compute the Integral over \(\phi\)

Substitute the result of the \(\rho\)-integral into the remaining integral and integrate \(\sin \phi\) with respect to \(\phi\) from \(\pi/3\) to \(2\pi/3\): \[ \int_{\pi/3}^{2\pi/3} \sin \phi \, d\phi = -\cos \phi \bigg|_{\pi/3}^{2\pi/3} = -\left(\cos(2\pi/3) - \cos(\pi/3)\right). \]
05

Compute Specific Cosine Values

Calculate the values of the cosine function: \[ \cos \frac{\pi}{3} = \frac{1}{2}, \quad \cos \frac{2\pi}{3} = -\frac{1}{2}. \]Substitute back: \[ -(-\frac{1}{2} - \frac{1}{2}) = 1. \]
06

Solve the Integral over \(\theta\)

Since the integral is symmetric about \(\theta\), the bounds 0 to \(2\pi\) integrate simply as\[ \int_0^{2\pi} d\theta = 2\pi. \]
07

Combine Results

Combine all the evaluated parts of the integral:\[ \text{Total Volume} = \left(\frac{a^3}{3}\right)(1)(2\pi) = \frac{2\pi a^3}{3}. \]
08

Final Result

The volume of the spherical sector between the given cones is \(\frac{2\pi a^3}{3}\).

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Key Concepts

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

Volume of a Sphere
Finding the volume of a sphere is an application of calculus that involves integration. The general formula for the volume of a sphere of radius \(a\) is \(\frac{4}{3}\pi a^3\). This formula arises from integrating the sphere's surface in spherical coordinates. In our exercise, we are not looking for the whole sphere's volume but only a part of it, known as a "spherical sector". This means we'll have to adjust our formula by considering certain boundaries.
  • The total volume considers all space within a given radius \(a\).
  • In spherical coordinates, a sphere's full volume is achieved by integrating over the entire range of angles and radial distances.
For a spherical sector, you'll only integrate over specific angles, reducing the full sphere's volume to a targeted region.
Conical Boundaries
When dealing with spherical sections, conical boundaries play a significant role. These boundaries are defined by angles in spherical coordinates. Think of a cone extending from the center of the sphere outward. In this exercise, we have two cones defined by the angles \(\phi = \frac{\pi}{3}\) and \(\phi = \frac{2\pi}{3}\).
  • These angles lead to the creation of a wedge-like sector of the sphere.
  • The region between these cones determines where our volume calculation should stop and start.
  • These boundaries make it possible to slice the sphere and limit the integration to a specific part.
Understanding these conical boundaries is essential for accurately calculating volumes in spherical sectors.
Triple Integral
A triple integral extends the idea of integral calculus into three dimensions and is crucial in determining volumes in spherical coordinates. In our case, we use a triple integral to cover three variables鈥擻(\rho\), \(\theta\), and \(\phi\), which are the radial distance, azimuthal angle, and polar angle, respectively.
  • To set up the integral, determine the bounds for each variable.
  • The function inside the integral, \(\rho^2 \sin \phi\), accounts for the spherical volume element.
  • Each integral component (in \(\rho\), \(\phi\), and \(\theta\)) can be computed separately, then multiplied to get the final volume.
The triple integral allows us to elegantly handle complex volume calculations that vary in three different dimensions.
Spherical Sector
A spherical sector is a portion of a sphere that lies between two conical surfaces extending from the sphere's center. Visualize it like slicing a wedge out of a circular cake, where the wedge extends from the center to the surface.
  • The sector's boundaries are determined by specific angles \(\phi\).
  • This sector is just a small slice of the whole sphere, defined by specific radial and angular constraints.
  • Understanding the constraints and boundaries is key to setting up the correct limits for integration.
In our example, the sector between \(\phi = \frac{\pi}{3}\) and \(\phi = \frac{2\pi}{3}\) gives us a manageable portion of the sphere to analyze and calculate.

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Most popular questions from this chapter

(a) find the spherical coordinate limits for the integral that calculates the volume of the given solid and then (b) evaluate the integral. The solid between the sphere \(\rho=\cos \phi\) and the hemisphere \(\rho=2, z \geq 0\)

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In polar coordinates, the average value of a function over a region \(R\) (Section 15.3 ) is given by $$\frac{1}{\operatorname{Area}(R)} \iint_{R} f(r, \theta) r d r d \theta$$ Find the average height of the hemispherical surface \(z=\sqrt{a^{2}-x^{2}-y^{2}}\) above the disk \(x^{2}+y^{2} \leq a^{2}\) in the \(x y\) -plane.
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