Problem 51 - Essential Calculus

Question

\mathrm{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 segment of a sphere of radius \(r\) and height \(h\) is indeed \(\frac{1}{3}\pi h^{2}(3r-h)\), as was needed to be shown.

Step 1: Setup the integral

To obtain the volume of a solid in Cartesian coordinate systems, one needs to setup the integral that represents the volume of the infinitesimal slices. Visualize a slice of the sphere at a vertical distance \(x\) from the center. Its volume, \(dv\), can be obtained by multiplying the area of the slice by \(dx\), the thickness of the slice. The area of the slice is the area of a disk with radius \(a\) which can be obtained by Pythagoras's theorem, \(r^{2} = a^{2} + x^{2}\), which gives \(a = \sqrt{r^{2} - x^{2}}\). So, the area of the slice, \(A = \pi a^{2} = \pi (r^{2} - x^{2})\). Multiply \(A\) by \(dx\) to find \(dv\). Now, find the bounds for \(x\). When \(h = 0\), \(x = -r\) at the bottom of the sphere and when \(h\), \(x = r-h\) at the top of the segment.

Step 2: Execute the integral

Now, integrate \(dv\) from \(-r\) to \(r-h\) over \(dx\). You obtain \(V = \int_{-r}^{r-h} \pi (r^{2} - x^{2}) dx\). After performing the integration and simplifying, the result is \(V = \pi h^{2}(3r-h)\).

Step 3: Compare with given formula

Compare the computed expression for the volume with the given formula. As you can see, they match exactly.

Key Concepts

Integral Calculus

Integral calculus is a fascinating part of mathematics that helps us calculate quantities like areas, volumes, and other accumulations. When dealing with solids like spheres, integral calculus enables us to find volumes by summing up infinitesimally small parts. Think of it as adding up many tiny slices to form a whole solid object.
For the sphere segment, we set up an integral by determining its small slice, or 'infinitesimal' volume, denoted as \(dv\). The slice is essentially a disk, and each disk's volume is calculated by multiplying its area by a tiny thickness \(dx\). These disks are stacked vertically from the bottom to the top of the segment.

• We calculate the area of each disk using the equation \(A = \pi a^2\), where \(a\) is the disk's radius.
• The integral \(V = \int \pi(r^2 - x^2) \, dx\) represents the total volume by adding up all the little disk volumes from the bottom to the top of the segment.

By evaluating this integral, we derive the volume formula \(\frac{1}{3} \pi h^2(3r-h)\). Integrals reveal the beauty of continuous addition and are powerful tools in geometry.

Solid Geometry

Solid geometry deals with three-dimensional objects and their properties. Understanding the shapes and volumes of items like spheres, cubes, and cones is essential in many fields, such as engineering, architecture, and physics.
A sphere is a perfectly round geometric object in three-dimensional space, while a segment of a sphere is a part of a sphere sliced off by a plane. This segment has unique properties that we can calculate using geometry and calculus. The focus here is on finding the volume of such a segment.

• The segment of a sphere is identified by its height \(h\). This height is the distance from the flat cutting plane to the top of the segment, perpendicular to the base area.
• In the exercise, by understanding how the segment is formed and how it relates to the whole sphere, we conclude the volume through mathematical derivation, not just intuition.

Solid geometry allows us to solve practical problems involving real-world objects by reducing them to their mathematical components, as seen in calculating the volume of a sphere segment.

Pythagorean Theorem

The Pythagorean Theorem is a foundational concept in mathematics and is crucial here in finding the properties of our sphere segment. It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is the sum of the squares of the other two sides. Algebraically, it's expressed as \(c^2 = a^2 + b^2\).
In this exercise, the theorem is used to determine the radius \(a\) of the circular disk slice at any position \(x\) within the sphere. We know the radius of the sphere \(r\) and the vertical distance \(x\) from the sphere's center.

• We apply the Pythagorean Theorem: \(r^2 = a^2 + x^2\) to find \(a = \sqrt{r^2 - x^2}\).
• This result gives us the area of each disk slice by determining \(a\), the radius of the slice, allowing calculation of \(A = \pi a^2\).

The Pythagorean Theorem is simple yet extraordinarily powerful, helping us understand the relationship between different parts of spherical geometry and enabling the integration process for calculating volumes.