Problem 53 - Calculus of a Single Variable

Question

Volume of a Segment of a Sphere Let a sphere of radius \(r\) be cut by a plane, thereby forming a segment of height \(h .\) Show that the volume of this segment is $$\frac{1}{3} \pi h^{2}(3 r-h)$$

Step-by-Step Solution

Answer

The volume of a spherical segment of height \( h \) cut from a sphere of radius \( r \) can indeed be given as \( \frac{1}{3} \pi h^{2}(3 r-h) \), as this is the result obtained after simplifying the difference between the volume of the larger sphere and the combined volumes of the smaller sphere and cone.

Step 1: Establish the Volume of Sphere and Cone

Let's start by establishing the formulas we will use. The volume of a sphere of radius \( r \) is \( \frac{4}{3}\pi r^{3} \) and the volume of a cone of base radius \( r \) and height \( h \) is \( \frac{1}{3}\pi r^{2}h \)

Step 2: Obtain the Sphere Segment and Cone

In our problem, the sphere segment is obtained by removing a smaller sphere of radius \( r-h \) and cone of height \( h \) and radius \( r - h \) from the larger sphere. Therefore, we need to calculate the volume of the smaller sphere and cone. The volume of the smaller sphere is \( \frac{4}{3}\pi (r-h)^{3} \) and the volume of the cone is \( \frac{1}{3}\pi (r-h)^{2}h \)

Step 3: Calculate the Volume of Spherical Segment

The volume of the segment is the difference between the volume of the sphere and the combined volumes of the smaller sphere and cone. Thus, the volume \( V \) of the segment is equal to: \( V = \frac{4}{3}\pi r^{3} - \left(\frac{4}{3}\pi (r-h)^{3} + \frac{1}{3}\pi (r-h)^{2}h\right) \)

Step 4: Simplify

Through simplification, this can be rewritten as \( V = \frac{1}{3} \pi h^{2}(3 r-h) \) thus proving the given formula.

Key Concepts

Sphere Volume Formula

To understand the volume of a spherical segment, it's essential first to grasp the concept of the sphere volume formula. A sphere is a perfectly round object in a three-dimensional space, like a basketball or a planet. The volume of a sphere is the amount of space it occupies, which can be calculated using the formula \( \frac{4}{3} \pi r^3 \). Here, "\( r \)" represents the radius of the sphere - the distance from its center to any point on its surface.

This formula arises from integral calculus, where the sphere is divided into infinitesimally small disks, each with a tiny thickness along its radius. Summing up the volumes of these disks yields the total volume of the sphere. In our exercise, the volume of a single sphere is crucial in understanding how the spherical segment volume is derived.

Cone Volume Formula

The cone volume formula is another pivotal concept in deducing the volume of a spherical segment. A cone is a three-dimensional shape that tapers smoothly from a flat base to a point called the apex. The volume of a cone can be calculated using the formula \( \frac{1}{3} \pi r^2 h \), where "\( r \)" is the base radius, and "\( h \)" is the height of the cone.

In our exercise, we look at a scenario involving a cone with a base radius \( r - h \) and height \( h \). Subtracting the volume of this cone, alongside the smaller sphere, from the original sphere helps us determine the volume of the spherical segment. Familiarity with this formula allows us to manage and simplify our calculations effectively.

Spherical Segment

A spherical segment is a part of a sphere created by slicing through it with a plane. This can be imagined as slicing an orange in half and examining the resulting dome-shaped part. The key to understanding a spherical segment is recognizing it as a combination of parts from a sphere and a cone.

When a sphere is cut, it's beneficial to calculate the remaining segment's volume by subtracting the unwanted parts, such as the smaller sphere and the cone. This exercise's objective is to establish the segment's volume using given measurements, ensuring the subtraction of the smaller components aligns with the predicted formula \( \frac{1}{3} \pi h^2 (3r - h) \).

Through understanding the geometry at play and how each shape interacts, students can better grasp the complexities of spherical segments.

Simplification of Volumes

Simplifying complex volume calculations involves breaking down the problem into more manageable parts. This approach is evident in our exercise, where initially, we calculated the total volume of a sphere. Then we subtracted the volumes of smaller components to find the volume of the spherical segment.

The trick lies in methodically simplifying each term, as shown in our step-by-step solution. Starting with the large sphere's volume \( \frac{4}{3} \pi r^3 \), subtracting out the smaller sphere \( \frac{4}{3} \pi (r - h)^3 \), and the cone \( \frac{1}{3} \pi (r-h)^2 h \). Each simplification step brings us closer to the final, elegant formula \( \frac{1}{3} \pi h^2(3r-h) \).

• Identify each component of the sphere segment.
• Use known formulas to calculate their volumes.
• Subtract these volumes carefully.
• Refine the results with algebraic simplification.

This process not only proves the formula but reinforces our understanding of how these geometric shapes function together.