Problem 57 - Thomas Calculus

Question

Volume of a hemisphere Derive the formula \(V=(2 / 3) \pi R^{3}\) for the volume of a hemisphere of radius \(R\) by comparing its cross-sections with the cross- sections of a solid right circular cylinder of radius \(R\) and height \(R\) from which a solid right circular cone of base radius \(R\) and height \(R\) has been removed, as suggested by the accompanying figure.

Step-by-Step Solution

Answer

The volume of a hemisphere is \( \frac{2}{3}\pi R^3 \).

Step 1: Understand the Problem

We need to derive the volume formula for a hemisphere by using its relationship with a cylinder minus a cone, both having a radius and height of \( R \).

Step 2: Set Up the Geometry

Consider a cylinder with radius \( R \) and height \( R \). The volume of this cylinder is \( V_{\text{cylinder}} = \pi R^2 \times R = \pi R^3 \).

Step 3: Calculate the Volume of the Cone

A cone with the same base radius \( R \) and height \( R \) has a volume \( V_{\text{cone}} = \frac{1}{3} \pi R^2 \times R = \frac{1}{3} \pi R^3 \).

Step 4: Find the Volume of the Shape

The volume of the solid obtained by removing the cone from the cylinder is \( V_{\text{solid}} = V_{\text{cylinder}} - V_{\text{cone}} = \pi R^3 - \frac{1}{3} \pi R^3 = \frac{2}{3} \pi R^3 \).

Step 5: Relate to Hemisphere

Observe that the solid we found is equivalent to two halves of hemispheres stacked on top of each other. Thus, the volume of one hemisphere is \( V_{\text{hemisphere}} = \frac{1}{2} \times V_{\text{solid}} = \frac{1}{2} \times \frac{2}{3} \pi R^3 = \frac{2}{3} \pi R^3 \).

Key Concepts

Cylinder and Cone Volumes

To understand the volume of a hemisphere, we need to comprehend how the volumes of cylinders and cones work. A cylinder is a three-dimensional shape with two parallel circles as its top and bottom. Its volume is given by the formula \( V_{\text{cylinder}} = \pi R^2 \times h \), where \( R \) is the radius of the base, and \( h \) is the height. For our exercise, both these dimensions are equal to \( R \). This simplifies the volume of our cylinder to \( \pi R^3 \).

Cones, on the other hand, are pointed structures with a circular base. The volume of a cone is a third of what a cylinder with the same base and height would be. This gives us \( V_{\text{cone}} = \frac{1}{3} \pi R^2 \times h \), and since the base and the height are also \( R \), the cone’s volume becomes \( \frac{1}{3} \pi R^3 \). Recognizing these relationships is crucial for problem-solving in geometry involving volume calculations.

Geometric Relationships

By examining geometric relationships, we can simplify complex problems. In this case, the problem leverages how a hemisphere can be represented by comparing it to a cylinder and a cone. Imagine a cylinder from which a cone with the same dimensions has been subtracted. What remains is a geometric construct that resembles two hemispheres stacked.

This insight highlights the relationship between these shapes, which simplifies the problem and leads to discovering new formulas. Concepts like symmetry, equivalence, and dimension can thus be applied to break down the problem into manageable parts, as seen in this exercise.

Problem Solving in Geometry

Problem-solving in geometry often involves recognizing patterns and relationships among different shapes. In our exercise, identifying that removing the cone from the cylinder yields a shape that simplifies to two hemispheres stacked is key. This not only helps shorten calculations but also provides an intuitive understanding of spatial arrangements.

Approaching geometric problems with a mindset to uncover hidden symmetries or geometric equivalences can be immensely beneficial. Utilizing known formulas, such as those for a cylinder and a cone, to deduce others (like that of a hemisphere) is a common strategy in problem-solving, facilitating a deeper comprehension of geometry.

Derivation of Formulas

Deriving formulas involves analytical thinking and step-by-step logical reasoning. In this case, the formula for the volume of a hemisphere, \( V = \frac{2}{3} \pi R^3 \), is derived by subtracting the volume of a cone from a cylinder and observing the resulting structure.

Every derivation begins with an understanding of fundamental shapes and their volumes, such as cylinders and cones, and using these to understand more complex shapes. By systematically applying known formulas, we can isolate the characteristics of versatile shapes like hemispheres to determine their unique formulas.

• Identify known quantities and shapes, such as base areas and heights.
• Understand the relationship between these shapes.
• Use logical steps to simplify and eventually isolate the shape of interest.

This logical progression ensures that the resulting formula is not only correct but also understandable, enabling students to learn the art of formula derivation with confidence.