Problem 48 - Thomas Calculus

Question

Volume Find the volume swept out by revolving the region bounded by the \(x\) -axis and one arch of the cycloid $$ x=t-\sin t, \quad y=1-\cos t $$ about the \(x\) -axis.

Step-by-Step Solution

Answer

The volume is \(5\pi^2\).

Step 1: Understand the Problem

We are asked to find the volume of a solid generated by revolving a region around the x-axis. The region is bounded by the x-axis and one arch of the cycloid defined by the parametric equations: \(x = t - \sin t\) and \(y = 1 - \cos t \).

Step 2: Identify the Bounds for the Cycloid

An arch of a cycloid completes when the parameter \(t\) goes from 0 to \(2\pi\). Thus, the bounds for \(t\) are \(0 \le t \le 2\pi\).

Step 3: Set up the Formula for the Volume by Revolution

Use the formula for the volume of a solid of revolution around the x-axis: \[ V = \int_{a}^{b} \pi y^2 \frac{dx}{dt} \, dt \]Substitute \(y=1-\cos t\) and compute \(\frac{dx}{dt}\).

Step 4: Compute \(\frac{dx}{dt}\)

Differentiate \(x = t - \sin t\) with respect to \(t\):\[\frac{dx}{dt} = 1 - \cos t\]

Step 5: Substitute into the Volume Formula

Substitute \(y = 1-\cos t\) and \(\frac{dx}{dt} = 1 - \cos t\) into the volume formula:\[ V = \int_{0}^{2\pi} \pi (1-\cos t)^2 (1 - \cos t) \, dt \]This simplifies to:\[ V = \pi \int_{0}^{2\pi} (1-\cos t)^3 \, dt \]

Step 6: Expand and Solve the Integral

Expand \((1-\cos t)^3\):\[ (1-\cos t)^3 = 1 - 3 \cos t + 3 \cos^2 t - \cos^3 t \]Integrate each term separately over \(0\) to \(2\pi\).

Step 7: Evaluate the Integral

For each integral:- \(\int_{0}^{2\pi} 1 \, dt = 2\pi\)- \(\int_{0}^{2\pi} \cos t \, dt = 0\) (due to symmetry)- \(\int_{0}^{2\pi} \cos^2 t \, dt = \pi\) (Using identity \( \cos^2 t = \frac{1}{2}(1 + \cos 2t)\))- \(\int_{0}^{2\pi} \cos^3 t \, dt = 0\) (Using symmetry and identity)Plug these into the expanded integral:\[ V = \pi[2\pi - 3(0) + 3(\pi) - 0] = 5\pi^2 \]

Step 8: Final Answer for the Volume

The volume of the solid formed by revolving the given cycloid around the x-axis is \(5\pi^2\).

Key Concepts

Cycloid

A cycloid is a fascinating curve generated by tracing a specific point on a circle as it rolls along a straight line. Imagine placing a pencil mark on the rim of a bicycle wheel. When the wheel rolls along a path, the pencil will trace out a cycloid. It's intriguing because it combines circular motion with linear motion.

The basic parametric equations for a cycloid are:

• \( x = t - \sin t \)
• \( y = 1 - \cos t \)

In our example, the cycloid is outlined by these equations, and we're interested in finding out what happens when this curve is revolved around a straight axis, which forms a three-dimensional object. This setup is particularly common in calculus problems that deal with curves and volumes, as it offers real-world applications, like the design of gear teeth and roller coasters.

Parametric Equations

Parametric equations are a powerful way to describe complex relationships between two variables using a third parameter, often represented by \( t \). In a traditional Cartesian system, we might express one variable in terms of another, like \( y = f(x) \).

However, in parametric equations, we use a parameter to define both \( x \) and \( y \):

• \( x = f(t) \)
• \( y = g(t) \)

This method allows for more flexibility and control over how curves are drawn. It accommodates changes that involve different speeds or accelerations along the path. In the case of the cycloid, the parameter \( t \) typically ranges over the interval from \(0\) to a full rotation, like \(2\pi\), which corresponds to one full cycle or arch of the cycloid.

The beauty of parametric equations in a problem like this is that they easily model the dynamic relationship between the coordinates as time or another factor changes.

Solid of Revolution

A solid of revolution is a three-dimensional shape resulting when a two-dimensional region is revolved around an axis. This is a classic topic in calculus because it illustrates how integral calculus can be used to compute volumes of objects with circular symmetry.

To find the volume of such a solid, we use the formula:

• \[ V = \int_{a}^{b} \pi y^2 \frac{dx}{dt} \, dt \]

In this formula, \( y \) and \( x \) are expressed in terms of a parameter \( t \), and \( \frac{dx}{dt} \) is the derivative of \( x \) with respect to \( t \).

For the cycloid, we substitute these expressions into the equation and compute the integral over one full cycle of the cycloid (from \(0\) to \(2\pi\)). This setup simplifies the complex shape that could be hard to measure otherwise. By breaking down the integral into simpler parts, we can calculate the total volume of the solid formed, ensuring each step of the process is transparent and manageable.