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Chapter 16: Problem 23

Write a triple integral in spherical coordinates giving the volume of a sphere of radius \(K\) centered at the origin. Use the order \(d \theta\) d \(\rho d \phi\)

Short Answer

Expert verified
The volume of the sphere is \(\frac{4\pi K^3}{3}\).

Step by step solution

01

Understand Spherical Coordinates

In spherical coordinates, a point in space is represented by \((\rho, \theta, \phi)\), where \(\rho\) is the radial distance from the origin, \(\theta\) is the azimuthal angle in the \(xy\)-plane from the positive \(x\)-axis, and \(\phi\) is the polar angle from the positive \(z\)-axis. For a sphere of radius \(K\), \(\rho\) extends from 0 to \(K\).
02

Set the Integration Limits

To describe the volume of the sphere, we set the integration limits as follows: - For \(\rho\), it ranges from 0 to \(K\). - For \(\theta\), which measures the angle in the circle of the \(xy\)-plane, it should go from 0 to \(2\pi\). - For \(\phi\), which measures the angle from the positive \(z\)-axis down to the \(xy\)-plane, it goes from 0 to \(\pi\).
03

Establish the Integrand Using Jacobian

In spherical coordinates, the Jacobian (or volume element) is given by \(\rho^2 \sin \phi\). Therefore, the triple integral to calculate the volume is \[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{K} \rho^2 \sin \phi\, d\rho\, d\phi\, d\theta. \]
04

Perform Integration with respect to \(\rho\)

Integrate \(\rho^2\) with respect to \(\rho\) first: \[ \int_{0}^{K} \rho^2 \sin \phi \, d\rho = \left[ \frac{\rho^3}{3} \sin \phi \right]_{0}^{K} = \frac{K^3}{3} \sin \phi. \]
05

Integrate with respect to \(\phi\)

Now integrate the result with respect to \(\phi\): \[ \int_{0}^{\pi} \frac{K^3}{3} \sin \phi \, d\phi = \frac{K^3}{3} \left[ -\cos \phi \right]_{0}^{\pi} = \frac{K^3}{3} \left( 2 \right) = \frac{2K^3}{3}. \]
06

Integrate with respect to \(\theta\)

Finally, integrate the result with respect to \(\theta\): \[ \int_{0}^{2\pi} \frac{2K^3}{3} \, d\theta = \frac{2K^3}{3} [\theta]_{0}^{2\pi} = \frac{2K^3}{3} (2\pi) = \frac{4\pi K^3}{3}. \]

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Key Concepts

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

Spherical Coordinates
Spherical coordinates provide a convenient way to define points in three-dimensional space with respect to a sphere centered at the origin. In this system, each point is described by three parameters:
  • \(\rho\) (rho): The radial distance from the origin. This essentially measures how far the point is from the center of the sphere.
  • \(\theta\) (theta): The azimuthal angle, which is the angle between the point's projection onto the \(xy\)-plane and the positive x-axis. It defines the direction of the point horizontally.
  • \(\phi\) (phi): The polar angle, indicating the angle from the positive z-axis to the point. It shows the elevation angle from the top of the sphere downwards.
This coordinate system is especially useful when dealing with objects that have symmetry around a point, like spheres, because it simplifies the equations used to describe their properties.
Volume of a Sphere
The volume of a sphere can be elegantly calculated using spherical coordinates. A sphere of radius \(K\) can be thought of as all points \((\rho, \theta, \phi)\) such that \(0 \leq \rho \leq K\).
The formula for the volume of a sphere is derived from integrating in spherical coordinates:\[ \frac{4}{3} \pi K^3 \] This formula arises naturally from the integration, when we account for all possible values of \(\theta\), \(\phi\), and \(\rho\) that fill up the sphere.
The triple integral involves integrating \(\rho^2 \sin \phi\) (the Jacobian) over specified limits, which collectively sweep through the entire volume of the sphere.
Jacobian
The Jacobian is a crucial part of understanding how to perform integration in different coordinate systems, like spherical coordinates. In this context, it's a factor that adjusts the volume element from Cartesian coordinates to spherical coordinates.
Specifically, in spherical coordinates, the Jacobian is \(\rho^2 \sin \phi\). This multiplication factor arises from the transformation of the differential volume element:
  • \(dV = dx \, dy \, dz\) becomes \(dV = \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi\).
This accounts for how the radial and angular changes scale the volume element in three dimensions. It's vital when setting up integrals to find volumes or other properties using spherical coordinates.
Integration Limits
Setting up integration limits correctly is key to obtaining the desired result from a triple integral. For a sphere:
  • \(\rho\) goes from 0 to \(K\): This covers all radial distances from the center to the sphere's surface.
  • \(\theta\) goes from 0 to \(2\pi\): This ensures that the integration sweeps through all azimuthal directions, making a full circle around the z-axis in the \(xy\)-plane.
  • \(\phi\) goes from 0 to \(\pi\): This covers all polar angles from the top of the sphere (positive z-axis) to its bottom, encompassing the downward direction to the \(xy\)-plane.
By integrating within these limits, every point inside the sphere's volume is accounted for, ensuring the integral evaluates to the sphere's full volume.

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Most popular questions from this chapter

The density, \(\delta,\) of the cylinder \(x^{2}+y^{2} \leq 4,0 \leq z \leq 3\) varies with the distance, \(r,\) from the \(z\) -axis: $$\delta=1+r \operatorname{gm} / \mathrm{cm}^{3}$$ Find the mass of the cylinder if \(x, y, z\) are in \(\mathrm{cm} .\)

A sphere has density at each point proportional to the square of the distance of the point from the \(z\) -axis. The density is \(2 \mathrm{gm} / \mathrm{cm}^{3}\) at a distance of \(2 \mathrm{cm}\) from the axis. What is the mass of the sphere if it is centered at the origin and has radius \(3 \mathrm{cm} ?\)

Let \(p(x, y)\) be a joint density function for \(x\) and \(y .\) Are the following statements true or false? \(\int_{a}^{b} p(x, y) d x\) is the probability that \(a \leq x \leq b\)

The sphere of radius 2 centered at the origin is sliced horizontally at \(z=1 .\) What is the volume of the cap above the plane \(z=1 ?\)

Which of the following integrals give the volume of the unit sphere? (a) \(\int_{0}^{2 \pi} \int_{0}^{2 \pi} \int_{0}^{1} 1 d \rho d \theta d \phi\) (b) \(\int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{1} 1 d \rho d \theta d \phi\) (c) \(\int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{1} \rho^{2} \sin \phi d \rho d \theta d \phi\) (d) \(\int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{1} \rho^{2} \sin \phi d \rho d \phi d \theta\) (e) \(\int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{1} \rho d \rho d \phi d \theta\)
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