Chapter 5: Problem 163
a. Find the volume of the solid \(S_{1}\) inside the unit sphere \(x^{2}+y^{2}+z^{2}=1\) and above the plane \(z=0\) b. Find the volume of the solid \(S_{2}\) inside the double cone \((z-1)^{2}=x^{2}+y^{2}\) and above the plane \(z=0\) c. Find the volume of the solid outside the double cone \((z-1)^{2}=x^{2}+y^{2}\) and inside the sphere \(x^{2}+y^{2}+z^{2}=1\)
Short Answer
Step by step solution
Understanding the Geometry and Setting up the Integral for Part a
Evaluating the Integral for Part a
Understanding the Geometry of Part b
Evaluating the Integral for Part b
Understanding the Geometry of Part c
Evaluating the Volume for Part c
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Spherical Coordinates
- \( \rho \) (the radial distance from the origin),
- \( \phi \) (the angle from the positive z-axis),
- \( \theta \) (the angle from the positive x-axis in the xy-plane).
This coordinate system elegantly expresses surfaces and volumes of spheres which simplifies integrals. In the case of a unit sphere centered at the origin, for example, the radius \( \rho \) ranges from 0 to 1. The angle \( \phi \) sweeps from 0 to \( \pi/2 \) for the upper hemisphere, and \( \theta \) rotates from 0 to \( 2\pi \).
One can determine the volume of a section of the sphere using the volume integral in spherical coordinates: \[ V = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. \]
This triple integral incorporates factors that account for volume stretching in spherical geometry.
Cylindrical Coordinates
- \( r \) (the radial distance from the z-axis),
- \( \theta \) (the angle within the xy-plane from the positive x-axis),
- \( z \) (the height along the z-axis).
This coordinate system is especially useful when dealing with problems like computing the volume of cones. The equation \((z - 1)^2 = x^2 + y^2\) describes a double cone centered on the z-axis. Using cylindrical coordinates, \(r^2 = x^2 + y^2 = (z - 1)^2\), helps simplify the volume calculation of such geometric structures by defining the bounds for \( r \) and \( z \).
The volume integral for a cone in cylindrical coordinates is expressed as: \[ V = \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{z-1} r \, dr \, dz \, d\theta. \] This fairly straightforward integration handles radial symmetry which is typical in conical volumes.
Volume of a Sphere
For example, for a unit sphere centered at the origin, the volume can be computed using the integral: \[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta. \]
This setup involves slicing the sphere with planes perpendicular to the z-axis, representing each 'slice' in computer as elements of volume. The integral yields the well-known formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \).
In calculations involving only the upper half (hemisphere) of the sphere, the integration of the limits of \( \phi \) is adjusted, like \[ \phi: 0 \rightarrow \frac{\pi}{2}. \]
which adjusts the resultant volume to half of the complete sphere, reflecting how only part of the sphere is being considered.
Volume of a Cone
A cone's general equation can be represented in cylindrical coordinates based on radius \( r \) and height \( z \). For instance, a double cone equation like \((z-1)^2 = x^2 + y^2\) translates to \( r = z - 1 \) in cylindrical form, making it simple to define integrals.
The volume of a cone in cylindrical coordinates is calculated via the following volume integral: \[ V = \int_{0}^{2\pi} \int_{0}^{h} \int_{0}^{z-1} r \, dr \, dz \, d\theta \]where \( h \) is the maximum height above the base plane. For rotational symmetric structures like cones, cylindrical coordinates help condense complexity into manageability in the integration process, producing, ultimately, the formula \( V = \frac{1}{3} \pi r^2 h \). This expression comes from evaluating the above integral, confirming the conical volume is one-third of the cylinder with the same height and base radius.