Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The volume is \(\frac{128\pi}{3} \left( 1 - \frac{\sqrt{2}}{2} \right)\).

Step 1: 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.

Step 2: 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 \(0 \leq \rho \leq 4\).

Step 3: 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:\[V = \int_0^{2\pi} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_0^4 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\]

Step 4: 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:\[V = 2\pi \times \frac{64}{3} \times \left( 1 - \frac{\sqrt{2}}{2} \right) = \frac{128\pi}{3} \left( 1 - \frac{\sqrt{2}}{2} \right)\]

Step 5: 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.

Key Concepts

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.

When performing volume integration in spherical coordinates, it's essential to pay attention to the limits of integration. These limits define the boundaries of the solid and ensure that the integral only covers the desired region. In our exercise, the region is determined by the spherical and conical shapes intersecting in specific ways.

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.

In the context of our problem, converting Cartesian equations to spherical coordinates simplifies calculations significantly. For instance, the equation of a sphere \(x^2 + y^2 + z^2 = 16\) becomes \(\rho = 4\). Similarly, the equation of a cone \(z = \sqrt{x^2 + y^2}\) translates to \(\phi = \frac{\pi}{4}\).

• 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:\[\int_a^b \int_c^d \int_e^f f(\rho, \phi, \theta) \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\]

• 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.

In our problem, this integral process helped determine the volume of a complicated solid defined by intersecting regions, like a sphere and a cone. The setup was especially important, aligning each variable's limits to their physical meanings—you start with \(\rho\), span into \(\phi\), and finally encompass with \(\theta\), covering the entire spatial region needed.