Problem 42 - Thomas Calculus

Question

A Bundt cake, well known for having a ringed shape, is formed by revolving around the \(y\) -axis the region bounded by the graph of \(y=\sin \left(x^{2}-1\right)\) and the \(x\) -axis over the interval \(1 \leq x \leq\) \(\sqrt{1+\pi} .\) Find the volume of the cake.

Step-by-Step Solution

Answer

The volume of the cake is \( 2\pi \) cubic units.

Step 1: Understand the problem setup

The exercise asks for the volume of a solid of revolution. The solid is generated by rotating the area bounded by the curve \( y = \sin(x^2 - 1) \), the x-axis, and between \( x = 1 \) and \( x = \sqrt{1 + \pi} \) around the y-axis.

Step 2: Set up the integral for volume

To find the volume by revolving around the y-axis, we use the method of cylindrical shells. The formula for the volume \( V \) of a solid of revolution by the shell method is \( V = \int_a^b 2\pi \cdot x \cdot f(x) \ dx \). Here, \( f(x) = \sin(x^2 - 1) \), and \( a = 1 \), \( b = \sqrt{1 + \pi} \).

Step 3: Integrate the shell formula

Calculate the integral: \[ V = \int_1^{\sqrt{1 + \pi}} 2\pi x \sin(x^2 - 1) \ dx. \] We use substitution: let \( u = x^2 - 1 \), so \( du = 2x \ dx \). Adjust integral limits accordingly: when \( x = 1 \), \( u = 0 \); when \( x = \sqrt{1 + \pi} \), \( u = \pi \). Thus, \[ V = \pi \int_0^{\pi} \sin(u) \ du. \]

Step 4: Solve the integral

The integral \( \int \sin(u) \ du \) evaluates to \(-\cos(u)\). Using the limits from 0 to \( \pi \): \[ \pi \left[ -\cos(u) \right]_0^{\pi} = \pi \left( -\cos(\pi) + \cos(0) \right) = \pi (1 + 1) = 2\pi. \]

Step 5: Conclude with the volume

The volume of the cake, the solid of revolution, is calculated as \( 2\pi \) cubic units.

Key Concepts

Understanding the Cylindrical Shell Method

The cylindrical shell method is a technique used to find the volume of a solid of revolution. It involves revolving a region around an axis, often used when revolving a region around the y-axis. This method forms "cylindrical shells" instead of disks or washers.
In this context, imagine thin vertical strips of the region being rotated around the y-axis. These form hollow cylindrical shapes that stack together to fill the solid.
The volume of each shell is calculated using the formula:

• Volume of shell = Height of shell \( \times \) Circumference of shell \( \times \) Thickness of shell

For integration purposes, the formula is expressed as:

• \[V = \int_a^b 2\pi \cdot x \cdot f(x) \, dx\]

where:

• \(2\pi x\) is the circumference of the shell,
• \(f(x)\) is the height,
• \(dx\) is the thickness of the shell.

This method is particularly useful when the function is described in terms of x and rotation occurs around the y-axis.

The Role of Integration Techniques in Volume Calculation

To find the volume of a solid using the cylindrical shell method, you need to compute a specific integral. This requires the use of various integration techniques.
In our problem, the integrand involves a trigonometric function, \(\sin(x^2 - 1)\), and to simplify this, we use the substitution method.**Substitution Method in Integration**Substitution transforms a complicated integral into a simpler one:

• Set a new variable: let \(u = x^2 - 1\)
• Calculate the differential: \(du = 2x \, dx\)

This substitution helps because it aligns with an existing term in the integral, \(2x \, dx\).
After substitution, the integral becomes:

• \\[V = \pi \int_0^{\pi} \sin(u) \, du\]

The bounds of integration also change according to substitutions of the original x-values, making computation easier.

Trigonometric Substitution Simplified

Trigonometric substitution is a method used in integration to simplify expressions involving square roots. Although not directly used in this problem, understanding it can enhance comprehension.
A common scenario would involve substituting trigonometric identities to simplify an integral related to circles or ellipses. It leverages identities such as \( \sin^2(x) + \cos^2(x) = 1 \) to rewrite integrals in simpler terms.Though not used in every integral involving trigonometric functions, it's a handy tool in your calculus toolbox when dealing with integrands that resemble trigonometric forms.
This technique is a powerful ally when facing challenging integrals. It allows for manipulation that can break down even the most complex of integrands into manageable pieces. It works by reducing complex relationships into simpler trigonometric relationships, benefitting especially those revolving around geometry.