Question

Find the volume of the solid generated by revolving about the \(x\) -axis the region in the first quadrant bounded by the circle \(x^{2}+y^{2}=r^{2}\), the \(x\) -axis, and the line \(x=r-h, 0

Short answer

The volume of the spherical segment is \( \frac{\pi h^2}{3} (3r-h) \).

Step-by-step solution

Understand the Problem

We are asked to find the volume of the solid generated by revolving the region in the first quadrant that is bounded by the circle \(x^{2}+y^{2}=r^{2}\), the \(x\)-axis, and the line \(x=r-h\) about the \(x\)-axis. This involves using the method of disks or washers.

Set Up the Integral

We will use the disk method for this problem. The radius of each disk is given by the \(y\)-coordinate in the circle's equation, so we solve \(x^2 + y^2 = r^2\) for \(y\) to get \(y = \sqrt{r^2 - x^2}\). The volume of each disk is \(\pi y^2\), and we integrate from \(x = r-h\) to \(x = r\) to find the volume of the solid. Thus, the integral setup is \(V = \int_{r-h}^{r} \pi (r^2 - x^2) \, dx\).

Calculate the Integral

Simplify and evaluate the integral: \[ V = \pi \int_{r-h}^{r} (r^2 - x^2) \, dx = \pi \left[ r^2x - \frac{x^3}{3} \right]_{r-h}^{r} \].

Evaluate the Integral

First evaluate at the upper limit: \(V_{upper} = \pi \left[ r^3 - \frac{r^3}{3} \right]\). Now, evaluate at the lower limit: \(V_{lower} = \pi \left[ r^2(r-h) - \frac{(r-h)^3}{3} \right]\). Subtract the two results, \(V = V_{upper} - V_{lower}\).

Simplify the Volume Expression

The calculation becomes \[ \pi \left[ r^3 - \frac{r^3}{3} - \left( r^2(r-h) - \frac{(r-h)^3}{3} \right) \right] = \pi \left[ \frac{2}{3}r^3 - r^2(r-h) + \frac{(r-h)^3}{3} \right].\] After expanding and simplifying, the final expression for volume can be written down.

Determine the Volume of a Spherical Segment

The expression provides the volume of the spherical segment as \[ V = \frac{\pi h^2}{3} (3r-h) \]. This is derived from further simplifying and solving the expression obtained in Step 5 after carefully performing algebraic manipulations.

Key concepts

Disk Method

The disk method is a powerful technique used in calculus to find the volume of a solid of revolution.
This method is particularly useful when a region in a plane is revolved around an axis, creating a 3D solid.
In the disk method:

• Visualize the solid as made up of numerous thin, circular disks piled upon one another.
• Each disk's volume is determined by its radius and thickness.

For the problem at hand, the region bounded in the first quadrant by a circle and a line is revolved around the x-axis.
The circle's equation is given by \(x^2 + y^2 = r^2\). When expressed in terms of \(y\), it becomes \(y = \sqrt{r^2 - x^2}\).
Each vertical slice of the region becomes a disk with:

• Radius: \(y\)
• Volume: \(\pi y^2\) times its infinitesimal thickness, \(dx\)

We integrate \(\pi y^2\) over the interval \([r-h, r]\) to find the volume of the solid.
Thus, the integral is set up as \( V = \int_{r-h}^{r} \pi (r^2 - x^2) \, dx \).
This integral calculates the sum of the volumes of all the disks, giving the total volume of the revolution solid.

Spherical Segment

A spherical segment is a portion of a sphere cut by two parallel planes.
It can also be visualized as the volume of a "cap" on a sphere.
In this exercise, we're interested in deriving the volume of a spherical segment of height \(h\) from a sphere with radius \(r\).
This happens when revolving the defined region around the \(x\)-axis.The volume formula for a spherical segment is determined by the derived expression:

• Volume \( V = \frac{\pi h^2}{3} (3r - h) \)

This formula accounts for:

• The height of the segment, \(h\)
• The sphere's radius, \(r\)

The derivation involves intricate calculus steps, culminating in a simplified elegant expression.
This formula reveals how the volume of the segment depends on its height and the sphere's size.

Calculus Problem Solving

Calculus provides the tools to solve complex geometric problems involving curves and solids.
By interpreting real-world geometric problems as mathematical equations, we can use calculus methods, like integration, to find solutions. Here's why calculus is so efficient for such tasks:

• Integration allows us to sum infinitely many infinitesimal elements, like disks or slices, to find whole volumes.
• Derivatives and related algebraic manipulations give insights into changing shapes and how they contribute to overall geometry.

In tackling the problem:

• Understand how geometry converts to calculus (forming integrals and differential elements).
• Follow precise steps, like setting up integrals with correct limits and simplifying calculations.

Remember, each step in calculus, whether setting up integral bounds or simplifying expressions, guides us closer to an accurate solution.
With practice, these methods become powerful tools for solving a wide array of mathematical problems, paving the way for greater understanding of math and its applications.