Problem 52 - Essential Calculus

Question

A manufacturer drills a hole through the center of a metal sphere of radius \(R\). The hole has a radius \(r\). Find the volume of the resulting ring.

Step-by-Step Solution

Answer

The volume of the resulting ring is given by \( V_ring = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi r^3\).

Step 1: Compute the volume of the original sphere

Use the formula for the volume of a sphere which is given by \( V = \frac{4}{3} \pi R^3\) where R is the radius of the original sphere.

Step 2: Compute the volume of the drilled sphere

The drilled sphere can also be regarded as a smaller sphere taken out of the bigger one. Thus, the volume of the drilled sphere can be calculated by using the same formula as in Step 1, substituting r for R. Hence the volume of the drilled sphere is \( V = \frac{4}{3} \pi r^3\)

Step 3: Calculate the volume of the resulting ring

The volume of the resulting ring is attained by subtracting the volume of the drilled sphere from the original sphere. Mathematically, this can be represent as: \( V_ring = V_{sphere} - V_{drilled} = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi r^3\).

Key Concepts

Sphere

A sphere is a three-dimensional object where every point on its surface is equidistant from its center. This distance is known as the radius. Spheres are perfectly symmetrical and have no edges or vertices.

When calculating the volume of a sphere, we use the formula: \[ V = \frac{4}{3} \pi R^3 \] where \( R \) is the radius of the sphere. This formula helps find the amount of space inside the sphere, a frequently needed calculation in many practical scenarios.

Consider the properties of a sphere:

• A sphere has constant width at all points on its surface.
• Its symmetry means that all diameters are equal, and this fact is useful for understanding volumes derived from or altered by spheres.

Cylindrical Hole

A cylindrical hole can be visualized as a tunnel passing straight through an object, such as a sphere in this case. If you pierce a sphere with a cylindrical hole, you are essentially subtracting space from the sphere.

This hole is characterized by:

• The length of the hole, which is often equal to twice the radius of the sphere if the hole passes through the center.
• Its own radius \( r \) which is crucial for calculating the volume of the material removed.

To find the volume of the material removed by drilling a hole, recalculate it by treating it as a smaller sphere being cut out.
Apply the volume of a sphere formula:\[ V_{drilled} = \frac{4}{3} \pi r^3 \]where \( r \) is the radius of the hole.

This gives a neat way to understand the subtraction necessary to find the leftover volume in more complex geometric problems.

Calculus

Calculus is a branch of mathematics that studies continuous change. It is invaluable when solving problems involving geometric shapes, like our sphere with a hole. Calculus approaches such problems analytically, often breaking them down into simpler parts.

While the original exercise doesn't delve deeply into calculus techniques, understanding calculus concepts can enhance your problem-solving prowess.

Consider how calculus relates to:

• Determining areas and volumes as limits of sums of infinitesimally small parts, a process that aids in calculating complex geometric shapes.
• Using integration to find the volume of an irregular shape, like our sphere with a hole.

In this context, calculus provides tools and foundational principles that allow you to extend simple volume formulas for more complex configurations, offering a deeper understanding of volume and geometry.