Problem 67 - Calculus: Early Transcendentals

Question

Find the volume common to two spheres, each with radius \( r \), if the center of each sphere lies on the surface of the other sphere.

Step-by-Step Solution

Answer

Volume of intersection is \( \frac{5\pi r^3}{12} \).

Step 1: Understand the Geometry

Two spheres with the same radius \( r \) intersect such that the center of each sphere is on the surface of the other. This geometry implies that the distance between the centers of the spheres is also \( r \).

Step 2: Identify the Shape of Intersection

The intersection of two spheres is a lens-shaped region, specifically a spherical cap for each sphere. Since the spheres are symmetrical and intersect symmetrically, the volume of intersection will be twice the volume of one spherical cap.

Step 3: Calculate the Spherical Cap Height

For this intersecting geometry, the height \( h \) of the spherical cap is equal to the radius because the distance from the center to the surface across the spherical cap is half of the diameter, which equals \( r \). Thus, \( h = r/2 \).

Step 4: Volume of One Spherical Cap

The formula for the volume \( V_{cap} \) of a spherical cap of height \( h \) inside a sphere of radius \( R \) is given by: \[ V_{cap} = \frac{1}{3} \pi h^2 (3R - h). \] Substituting \( R = r \) and \( h = r/2 \), we find: \[ V_{cap} = \frac{1}{3} \pi \left(\frac{r}{2}\right)^2 \left(3r - \frac{r}{2}\right). \] Simplifying, \[ V_{cap} = \frac{1}{3} \pi \frac{r^2}{4} \cdot \frac{5r}{2} = \frac{5\pi r^3}{24}. \]

Step 5: Volume of the Intersection

Since the intersection region consists of two identical spherical caps, the total volume \( V_{intersection} \) is: \[ V_{intersection} = 2 \cdot \frac{5\pi r^3}{24} = \frac{5\pi r^3}{12}. \]

Key Concepts

Spherical Cap

A spherical cap resembles a "cap" that you would find on a bottle, but it's part of a sphere. Imagine cutting a sphere with a plane; the part that stays with the base forms the spherical cap. When two spheres intersect, like in our problem, the overlapping region forms two spherical caps at each intersection. In this situation, the spherical cap is integral to finding the volume of intersection because it represents the part of the sphere above the flat cutting plane.

Spherical caps are significant when studying intersections of spheres because they describe a portion shared between them. Understanding how to determine the volume of a spherical cap is crucial for solving problems involving spheres. In the two intersecting spheres of our problem, each cap has a height equal to half of the radius of the sphere, making it essential to calculate accurately.

Geometry of Spheres

Understanding the geometry of spheres is essential in visualizing and solving mathematical problems involving them. A sphere is a three-dimensional shape where all points on the surface are equidistant from a center point. In our problem, the two spheres intersect with their centers positioned exactly at the surface of each other, forming an intriguing symmetrical configuration.

This symmetry in geometry implies several things:

• The distance between the centers is equal to the radius of either sphere, making the configuration elegant and predictable.
• The intersecting volume forms an axis-aligned shape, leading to the formation of lens-like figures.
• Calculations can exploit this symmetry for simplification, especially in computing shared areas or volumes.

Understanding the geometric placement is vital because it affects how formulas are applied and how the problem is visualized.

Volume Calculation

Volume calculation is key in obtaining the measure of space occupied by a shape, and understanding volume formulas is crucial in geometry problems. For spherical caps formed by intersecting spheres, the volume can be determined using a specific formula that considers both the height and radius of the sphere.

In our exercise, since we know the radius of each sphere and the height of the caps is half of that radius, we can use the formula for the volume of a single spherical cap:\[V_{cap} = \frac{1}{3} \pi h^2 (3R - h)\]

Substituting the known values effectively simplifies the calculation:

• R = r: The full radius of the sphere is used as the maximum extension.
• h = r/2: The height of each cap permits using exactly half the radius.

Solving gives us the volume of one cap, which can then be doubled to compute the total intersecting volume.

Mathematics Problem Solving

When tackling a mathematics problem, especially those involving complex shapes like spheres, it is essential to take a strategic approach. First, fully understand the problem's constraints and define the geometric relationships. Next, establish what you know and decide how it informs the solution. Break down the problem into manageable parts as seen in the original step-by-step solution.

For instance, knowing that we're dealing with spheres which intersect symmetrically allows us to simplify our task into calculating the volume of smaller parts and then combining them. Take advantage of symmetry and known formulas. Implement steps systematically:

• Identify the shapes formed by the intersection, such as the spherical cap.
• Utilize formulas associated with these shapes to find each component volume.
• Ensure calculations are correct by simplifying the results and making logical checks.

Through clear steps and logical deductions, mathematics problem-solving becomes both systematic and efficient.