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Chapter 33: Problem 3

If \(x=a(\theta-\sin \theta), y=a(1-\cos \theta)\), find the volume generated when the plane figure bounded by the curve, the \(x\)-axis and the ordinates at \(\theta=0\) and \(\theta=2 \pi\), rotates about the \(x\)-axis through a complete revolution.

Short Answer

Expert verified
The volume is \( \frac{3}{4} \pi^2 a^3 \).

Step by step solution

01

Understand the Problem

To find the volume of the solid of revolution generated by rotating the curve around the x-axis, we will use the formula for the volume of revolution. The curve is given parametrically by \( x = a(\theta - \sin \theta), y = a(1 - \cos \theta) \). We need to integrate with respect to \( \theta \) from \( \theta = 0 \) to \( \theta = 2\pi \).
02

Set Up the Integral

The formula for the volume \( V \) when rotating a parametric curve \((x(t), y(t))\) about the x-axis is \[ V = \pi \int_{0}^{2\pi} y^2(\theta) \frac{dx}{d\theta} \, d\theta. \] Here, \( y(\theta) = a(1 - \cos \theta) \), hence \( y^2(\theta) = a^2(1 - \cos \theta)^2 \).
03

Find \( \frac{dx}{d\theta} \)

Differentiate \( x = a(\theta - \sin \theta) \) with respect to \( \theta \). Find \( \frac{dx}{d\theta} = a(1 - \cos \theta) \) by using the derivative of \( \sin \theta \), which is \( \cos \theta \).
04

Substitute into Volume Formula

Substitute \( y^2(\theta) = a^2(1 - \cos \theta)^2 \) and \( \frac{dx}{d\theta} = a(1 - \cos \theta) \) into the volume integral: \[ V = \pi \int_{0}^{2\pi} a^2(1 - \cos \theta)^2 \cdot a(1 - \cos \theta) \, d\theta = \pi a^3 \int_{0}^{2\pi} (1 - \cos \theta)^3 \, d\theta. \]
05

Evaluate the Integral

Find \( \int_0^{2\pi} (1 - \cos \theta)^3 \, d\theta \) by using the binomial expansion and trigonometric identities. Simplify this integral to find \( \int_0^{2\pi} (1 - \cos \theta)^3 \, d\theta = \frac{3}{8} \cdot 2\pi \).
06

Calculate the Volume

Substitute the evaluated integral back into the formula: \[ V = \pi a^3 \cdot \frac{3}{8} \cdot 2\pi = \frac{3}{4} \pi^2 a^3. \] This is the volume of the solid formed by the revolution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations allow us to represent curves using parameters, rather than just x and y coordinates. It's like giving a curve a set of instructions
  • Imagine a curve swirling like a ribbon; parametric equations help us describe every twist.
  • Here, the parameter is \( \theta \), kind of like a time measure that tells us where we are on the curve, for which we have equations: \( x = a(\theta - \sin \theta) \) and \( y = a(1 - \cos \theta) \).
  • This setup is useful when performing calculus tasks, as it breaks down complex curves into manageable parts.
To use parametric equations in calculus operations, it's essential to differentiate with respect to the parameter. This skill is crucial as it lays the groundwork for volume and area calculations with more complex shapes.
Definite Integral
The definite integral is a powerful tool used to find the area under curves, among other things.
  • Think of it as the sum of tiny slivers, like stacking slices of bread to form a loaf.
  • When we're solving for the volume of a solid of revolution, we set up an integral in terms of our parameter, \( \theta \), from \( 0 \) to \( 2\pi \).
  • The formula \( \pi \int_{0}^{2\pi} y^2(\theta) \frac{dx}{d\theta} \, d\theta \) is used to capture the entire volume in one fell swoop.
Evaluating this integral involves substituting and simplifying, aided by trigonometric identities, to sum up all the small volumes created by rotation.
Volume Calculation
Calculating the volume of a solid formed by the revolution of a curve is like piecing together a 3D sculpture from thin circular disks.
  • The method used here is based on the disk method, modified for parametric equations.
  • We have the setup \( V = \pi \int_{0}^{2\pi} y^2(\theta) \frac{dx}{d\theta} \, d\theta \), where each disk contributes its tiny slice to the total volume when rotated around the x-axis.
  • Finding, \( \frac{dx}{d\theta} \), which is \( a(1 - \cos \theta) \), and substituting it into our integral helps us piece together these volumes.
The integration yields the total volume, showcasing the power of calculus in dealing with complex 3D shapes.
Trigonometric Identities
Trigonometric identities are vital in integrating expressions involving sine and cosine functions.
  • They are akin to shortcuts that simplify long and tedious calculations.
  • In this exercise, using identities like \( 1 - \cos \theta \) allows us to expand and simplify \((1 - \cos \theta)^3\).
  • We rely on binomial expansion alongside these identities, so things like \( \int_0^{2\pi} (1 - \cos \theta)^3 \, d\theta \) become manageable.
These identities help in reducing complex trigonometric problems into solvers-easy forms, making integral computations possible for thorny equations.

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Most popular questions from this chapter

The parametric equations of a curve are \(x=e^{t} \sin t, y=e^{t} \cos t\). If the arc of this curve between \(t=0\) and \(t=\frac{\pi}{2}\) rotates through a complete revolution about the \(x\)-axis, calculate the area of the surface generated. Now you are all ready for the next Programme. Well done, keep it up!

The plane figure bounded by the parabola \(y^{2}=4 a x\), the \(x\)-axis and the, ordinate at \(x=a\), is rotated a complete revolution about the line \(x=-a\). Find the volume of the solid generated.

Find the length of the curve $$ y=\frac{x}{2}-\frac{x^{2}}{4}+\frac{1}{2} \ln (1-x) $$ between \(x=0\) and \(x=\frac{1}{2}\).

For the catenary \(y=5 \cosh \frac{x}{5}\), calculate: (a) the length of arc of the curve between \(x=0\) and \(x=2\) (b) the surface area generated when this arc rotates about the \(x\)-axis through a complete revolution.

Find the length of the curve \(x=5(\cos \theta+\theta \sin \theta), \quad y=5(\sin \theta-\theta \cos \theta)\) between \(\theta=0\) and \(\theta=\frac{\pi}{2}\).
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