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A region is space is described. Set up the triple integrals that find the volume of this region using rectangular, cylindrical and spherical coordinates, then comment on which of the three appears easiest to evaluate. The region enclosed by the unit sphere, \(x^{2}+y^{2}+z^{2}=1\).

Short Answer

Expert verified
The spherical coordinates integral is easiest to evaluate due to symmetry.

Step by step solution

01

Understanding the Problem

We are asked to find the volume of the unit sphere defined by the equation \( x^2 + y^2 + z^2 = 1 \) using three coordinate systems: rectangular, cylindrical, and spherical. We need to set up triple integrals for each coordinate system and decide which one seems the easiest to evaluate.
02

Set Up Integral in Rectangular Coordinates

In rectangular coordinates (x, y, z), the region of integration is bounded by the unit sphere equation \( x^2 + y^2 + z^2 \leq 1 \). The volume integral in rectangular coordinates is:\[\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} \, dz \, dy \, dx\]
03

Set Up Integral in Cylindrical Coordinates

In cylindrical coordinates (r, θ, z), the unit sphere transforms into the equation \( r^2 + z^2 = 1 \). The limits for \( r \) range from 0 to 1, for \( \theta \) from 0 to \( 2\pi \), and for \( z \) from \(-\sqrt{1-r^2}\) to \(\sqrt{1-r^2}\). The volume integral in cylindrical coordinates is:\[\int_{0}^{2\pi} \int_{0}^{1} \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} \, r \, dz \, dr \, d\theta\]
04

Set Up Integral in Spherical Coordinates

In spherical coordinates (ρ, φ, θ), the radius \( \rho \) of the sphere is 1. The limits for \( \rho \) are from 0 to 1, for \( \phi \) from 0 to \( \pi \), and for \( \theta \) from 0 to \( 2\pi \). The volume integral in spherical coordinates is:\[\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \, \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]
05

Evaluate and Compare the Integrals

The spherical coordinate system provides the most straightforward integral for evaluating the unit sphere's volume since it directly takes advantage of the sphere's symmetry. The other coordinate systems involve more complex boundaries and integrands.

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Key Concepts

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

Rectangular Coordinates
When discussing triple integrals, rectangular coordinates are one of the most familiar systems. Here points in space are represented as \(x, y, z\). For the unit sphere, the boundary is given by \(x^2 + y^2 + z^2 = 1\).
In this coordinate system, we must integrate over a three-dimensional region confined by these boundaries. The expression for the volume integral of the unit sphere is complex, reflecting the need to adjust limits based on squares and square roots:
  • Outer integral with limits \(-1 ext{ to } 1\) for \(x\)
  • Middle integral for \(y\) limits \(-\sqrt{1-x^2} ext{ to } \sqrt{1-x^2}\)
  • Innermost integral for \(z\) limits \(-\sqrt{1-x^2-y^2} ext{ to } \sqrt{1-x^2-y^2}\)
While rectangular coordinates are often used, they may not be the most efficient system for spheres due to the complex nature of the integrals required.
Cylindrical Coordinates
Cylindrical coordinates provide an efficient way to transform certain problems into more manageable equations, especially when symmetry around an axis is present. In this system, we use \(r, \theta, z\), where \(r\) is the radial distance, \(\theta\) the angle around the z-axis, and \(z\) the vertical position.
For a unit sphere with equation \(x^2 + y^2 + z^2 = 1\), this transforms to \(r^2 + z^2 = 1\). The integration limits become:
  • \(r\) from \(0 ext{ to } 1\)
  • \(\theta\) from \(0 ext{ to } 2\pi\)
  • \(z\) from \(-\sqrt{1-r^2} ext{ to } \sqrt{1-r^2}\)
Including the integral factor \(r\) makes calculations simpler compared to rectangular coordinates. It exploits rotational symmetry, reducing complexity when solving for circular or cylindrical regions.
Spherical Coordinates
Spherical coordinates are ideal for problems involving spheres, as they efficiently use the symmetry of the problem. In this system, points are represented by \(\rho, \phi, \theta\), where \(\rho\) is the radial distance from the origin, \(\phi\) is the polar angle from the z-axis, and \(\theta\) is the azimuthal angle in the x-y plane.
For the unit sphere, we have the limits:
  • \(\rho\) from 0 to 1
  • \(\phi\) from 0 to \(\pi\)
  • \(\theta\) from 0 to \(2\pi\)
This results in a spherical volume integral:\[\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\]The simplicity of these limits leverages the sphere's inherent symmetry, making evaluation straightforward. It's particularly convenient as it naturally defines the entirety of the sphere.
Unit Sphere
A unit sphere is a sphere centered at the origin with a radius of 1. Its equation is given by \(x^2 + y^2 + z^2 = 1\).
The term 'unit' is used because of its radius being exactly one unit. The symmetry in its structure is beneficial when computing integrals in various coordinate systems.
  • Commonly applied in physics and engineering for problems involving spherical symmetry, such as gravitational fields or light from a point source.
  • Allows for straightforward conversion to spherical coordinates, optimizing calculations.
Understanding the geometric and mathematical properties of a unit sphere lays the groundwork for calculating more complex objects' volumes that may also involve spherical symmetry.
Volume Calculation
Calculating the volume of a three-dimensional region involves setting up and solving a triple integral. The choice of coordinate system can drastically affect the simplicity of the integration process.
For the unit sphere:
  • Rectangular coordinates require a multilayered integral due to complex boundaries.
  • Cylindrical coordinates simplify the integral due to radial symmetry, yet still involve square roots.
  • Spherical coordinates allow for the simplest form of the integral due to symmetry, facilitating straightforward calculation.
Choosing the appropriate system is crucial. In this case, spherical coordinates are advantageous for spheres, as they align with the object's natural geometry and reduce integration complexity.

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