Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use spherical coordinates to find the indicated quantity. Find the volume of the solid inside both of the spheres \(\rho=2 \sqrt{2} \cos \phi\) and \(\rho=2\).

Step-by-Step Solution

Answer

The volume is \( \frac{\pi}{3} \).

Step 1: Express Spheres in Spherical Coordinates

The given equations are both in spherical coordinates. The first sphere is given by \( \rho = 2 \sqrt{2} \cos \phi \) and the second sphere by \( \rho = 2 \). In spherical coordinates, \( \rho \) represents the radial distance, \( \phi \) is the polar angle, and \( \theta \) is the azimuthal angle.

Step 2: Determine the Region of Intersection

The region of interest is the common volume of both spheres. For the first sphere, solve \( \rho = 2 \sqrt{2} \cos \phi \) to get \( \rho \leq 2 \cos \phi \). Hence, we are considering \( \rho \) values from \( 0 \) to \( 2 \cos \phi \) for \( \phi \) ranging from \( 0 \) to when the two spheres intersect.

Step 3: Find Limits of Integration for \( \phi \)

To find where the spheres intersect, set \( 2 \sqrt{2} \cos \phi = 2 \). Solving gives \( \cos \phi = \frac{1}{\sqrt{2}} \), which implies \( \phi = \frac{\pi}{4} \). Thus, \( \phi \) ranges from \( 0 \) to \( \frac{\pi}{4} \).

Step 4: Set Limits for Integration Variables

The integration will be set up as follows: \( \rho \) will vary from \( 0 \) to \( 2 \cos \phi \), \( \phi \) from \( 0 \) to \( \frac{\pi}{4} \), and \( \theta \) from \( 0 \) to \( 2\pi \).

Step 5: Write the Integral for the Volume

The volume \( V \) is calculated using the integral in spherical coordinates: \[ V = \int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_0^{2\cos\phi} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]

Step 6: Compute the Integral

First, integrate with respect to \( \rho \): \[ \int_0^{2\cos\phi} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^{2\cos\phi} = \frac{8(\cos\phi)^3}{3} \]. Next, integrate with respect to \( \phi \): \[ \int_0^{\frac{\pi}{4}} \frac{8(\cos\phi)^3}{3} \sin \phi \, d\phi. \] Using the substitution \( u = \cos \phi \), this simplifies to \[ \frac{8}{3} \int_0^{\frac{1}{\sqrt{2}}} u^3 \, du = \frac{8}{3} \left[ \frac{u^4}{4} \right]_0^{\frac{1}{\sqrt{2}}} = \frac{8(1/4)(1/4)}{3 \cdot 4} = \frac{1}{6} . \] Finally, integrate with respect to \( \theta \): \[ \int_0^{2\pi} \frac{1}{6} \, d\theta = \frac{1}{6} [\theta]_0^{2\pi} = \frac{2\pi}{6} = \frac{\pi}{3}. \]

Step 7: Conclusion

The final volume of the solid inside both spheres after evaluating the integral is \( \frac{\pi}{3} \).

Key Concepts

Volume Calculation

To compute the volume of a solid region using spherical coordinates, we need to establish the limits for each spherical variable: \( \rho \), \( \phi \), and \( \theta \).
Spherical coordinates are particularly useful for problems involving symmetry around a point, as they simplify integration over such regions.
In the exercise provided, determining the volume of the solid lying within two spheres requires understanding how the spheres intersect and using those intersection points to set our limits of integration.

• \( \rho \) represents the distance from the origin to a point inside the solid.
• \( \phi \) is the angle from the positive z-axis downwards.
• \( \theta \) denotes the rotational angle around the z-axis.

Once these limits are understood, we define the volume integral in spherical coordinates by multiplying by \( \rho^2 \sin \phi \) (accounting for the Jacobian of the transformation) and integrating over each interval. Each step reduces down the complexity until we find the numerical volume, as in the example where the final result is \( \frac{\pi}{3} \).

Integration in Spherical Coordinates

Integration in spherical coordinates follows a distinct path that accounts for the geometry. When expressing the differential volume element, you use \( \rho^2 \sin \phi \), which accounts for the three-dimensional nature of spherical coordinates.
The integration process typically involves three steps for each variable:

• First, integrate with respect to \( \rho \). This step handles the radial component, where \( \rho \) extends to the boundary defined by the sphere or the intersection points.
• Next, integrate with respect to \( \phi \), the angle from the z-axis, reflecting vertical symmetry in the solid. Often, a change of variables using trigonometric identities aids in simplifying this part.
• Finally, integrate over \( \theta \), capturing the full rotational symmetry around the z-axis, usually ranging from \(0\) to \(2\pi\).

Each integration carefully reduces a dimension, progressing closer to the final volume measure. The practice ensures calculation precision despite the geometrical complexity, demonstrated by reaching the volume result \( \frac{\pi}{3} \) in the exercise.

Intersection of Spheres

Two spheres can intersect in various ways, creating different regions of common volume. Understanding the method to find where these intersections occur is crucial for volume calculations.
To find the intersection of two spheres described in spherical coordinates, set the equations equal to each other. This will help identify the critical angle \( \phi \) where they meet.

• Analyze each equation to determine the range of \( \rho \) for which the inequality holds true. For instance, \( \rho = 2\sqrt{2} \cos \phi \) illustrates that \( \rho \) might vary from \(0\) to \(2\cos \phi \).
• Determine \( \phi \) interval by solving \( \rho \) values at equal conditions intersect, such as the angle \( \phi = \frac{\pi}{4} \) in this example.
• Visualize these regions to understand how they contribute to one solid volume and how the integration bounds are set.

This understanding enables one to effectively set up the bounds of integration, allowing for precise calculation of the shared volume, as shown in this solution process.