Chapter 13: Problem 20
In Problems 15-22, use spherical coordinates to find the indicated quantity. Volume of the solid within the sphere \(x^{2}+y^{2}+z^{2}=16\), outside the cone \(z=\sqrt{x^{2}+y^{2}}\), and above the \(x y\)-plane
Short Answer
Expert verified
The volume is \(\frac{128\pi}{3} \left( 1 - \frac{\sqrt{2}}{2} \right)\).
Step by step solution
01
Understand the Problem in Spherical Coordinates
We are asked to find the volume of a solid where it is within a sphere and outside a cone. The sphere has the equation \(x^2+y^2+z^2=16\), which in spherical coordinates becomes \(\rho=4\) since \(\rho^2=x^2+y^2+z^2\). The cone's equation is \(z=\sqrt{x^2+y^2}\), which translates to \(\phi=\frac{\pi}{4}\) in spherical coordinates, as the relation between cylindrical and spherical coordinates gives it this specific angle.
02
Set the Limits of Integration
In spherical coordinates, a point \((x, y, z)\) is represented as \((\rho, \theta, \phi)\) with relationships \(x=\rho \sin \phi \cos \theta\), \(y=\rho \sin \phi \sin \theta\), \(z=\rho \cos \phi\). Given \(\rho=4\) and \(\phi > \frac{\pi}{4}\), the limits are \(0 \leq \theta \leq 2\pi\), \(\frac{\pi}{4} \leq \phi \leq \frac{\pi}{2}\), and \(0 \leq \rho \leq 4\).
03
Set Up the Integral for Volume
The volume element in spherical coordinates is \(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\). The integral for the volume \(V\) is thus:\[V = \int_0^{2\pi} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_0^4 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\]
04
Evaluate the Integral
Firstly, compute the innermost integral:\[\int_0^4 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^4 = \frac{64}{3}\]Next, compute the integral with respect to \(\phi\):\[\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin \phi \, d\phi = \left[ -\cos \phi \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = 1 - \frac{\sqrt{2}}{2}\]Finally, solve the integral with respect to \(\theta\):\[\int_0^{2\pi} d\theta = 2\pi\]Therefore, the volume is:\[V = 2\pi \times \frac{64}{3} \times \left( 1 - \frac{\sqrt{2}}{2} \right) = \frac{128\pi}{3} \left( 1 - \frac{\sqrt{2}}{2} \right)\]
05
Simplify the Final Expression
The final result for the volume can be expressed as:\[V = \frac{128\pi}{3} \left( 1 - \frac{\sqrt{2}}{2} \right)\]This represents the volume of the solid within the given constraints.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
volume integration
Volume integration is a method used to calculate the volume of a three-dimensional solid. It involves integrating a volume element over a defined region. In spherical coordinates, this volume element is represented as \(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\). This form considers how spherical coordinates are characterized by radius \(\rho\), polar angle \(\phi\), and azimuthal angle \(\theta\).
- The integration process typically involves setting up triple integrals, where each integral corresponds to one of the variables \(\rho\), \(\phi\), or \(\theta\).
- Each integral evaluates just part of the geometric configuration that defines the region, such as a sphere or a cone in the context of our problem.
spherical coordinate system
The spherical coordinate system is a three-dimensional system that is especially well-suited for problems involving spheres or spherical-like shapes. It uses three parameters to define a point in space: \(\rho\), \(\phi\), and \(\theta\).
- \(\rho\) is the radial distance from the origin to the point.
- \(\phi\) is the polar angle, measured from the positive z-axis.
- \(\theta\) is the azimuthal angle, measured in the xy-plane from the positive x-axis.
- The simplicity in using spherical coordinates comes from the direct representation of spherical symmetry in the equations and their respective integration limits.
- This representation not only simplifies calculations but also helps visualize the volume within specific boundaries like that of a conical region or spherical section.
triple integrals
Triple integrals extend the concept of integration to three dimensions. They're used to calculate quantities like volume and mass by integrating a function over a three-dimensional region.
In spherical coordinates, a triple integral has the form:\[\int_a^b \int_c^d \int_e^f f(\rho, \phi, \theta) \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\]
In spherical coordinates, a triple integral has the form:\[\int_a^b \int_c^d \int_e^f f(\rho, \phi, \theta) \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\]
- Here, \(f(\rho, \phi, \theta)\) describes the density or other relevant property of the volume, though it may be 1 if finding just the physical volume.
- The integration proceeds from the innermost to the outermost integral, solving step by step while considering each variable's limits.