Problem 42 - Thomas Calculus in SI Units

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 \(2\pi\).

Step 1: Understand and Visualize the Region

The function given is \(y = \sin(x^2 - 1)\). We revolve the area between this function and the x-axis around the y-axis over the interval \(1 \leq x \leq \sqrt{1+\pi}\). This creates a 3D shape which is symmetric around the y-axis.

Step 2: Set Up the Volume Integral Using the Shell Method

The shell method formula for volume is \(V = 2\pi \int_a^b x \cdot h(x) \,dx\), where \(h(x)\) is the height of the strip at \(x\). Here, \(h(x) = y = \sin(x^2-1)\), so the integral becomes \(V = 2\pi \int_1^{\sqrt{1+\pi}} x \sin(x^2-1) \,dx\).

Step 3: Substitute and Simplify the Integral Expression

To solve \(\int x \sin(x^2-1) \), use the substitution \(u = x^2 - 1\), so \(du = 2x \,dx\) or \(x \,dx = \frac{1}{2} du\). Thus, the integral transforms to \(\int \frac{1}{2} \sin(u) \,du\).

Step 4: Integrate and Solve the New Expression

The integral \(\int \sin(u) \, du\) is \(-\cos(u) + C\). Therefore, \(\int \frac{1}{2} \sin(u) \, du = -\frac{1}{2} \cos(u) + C\). Substitute back \(u = x^2 - 1\) to get \(-\frac{1}{2} \cos(x^2 - 1)\).

Step 5: Evaluate the Definite Integral

Calculate the definite integral by evaluating at the bounds \(x = 1\) and \(x = \sqrt{1+\pi}\). Substitute back to find \(V = 2\pi \left[ -\frac{1}{2}\cos(x^2-1) \right]_1^{\sqrt{1+\pi}} = \pi \left[-\cos(\pi) + \cos(0)\right]\).

Step 6: Simplify with Trigonometric Values

Since \(\cos(\pi) = -1\) and \(\cos(0) = 1\), we find \(V = \pi [1 - (-1)] = \pi \times 2 = 2\pi\).

Key Concepts

Shell Method

The shell method is a powerful technique for finding the volume of 3D objects formed by revolving a region around an axis. Unlike the disk or washer methods, which use horizontal slices, the shell method utilizes cylindrical shells.
To apply the shell method, we integrate along the x-axis for the figure to be revolved around the y-axis.
In this specific problem, the important formula is:

• \( V = 2\pi \int_a^b x \cdot h(x) \,dx \)
• where \(h(x)\) is the height of the shell.

The height \(h(x)\) is equal to the distance from the x-axis to the curve, in this case, \(y = \sin(x^2-1)\). Thus the integral becomes \( V = 2\pi \int_1^{\sqrt{1+\pi}} x \sin(x^2-1) \,dx \).
The shell method simplifies complex volumes by transforming a challenging geometrical problem into a manageable calculus exercise.

Trigonometric Integration

Trigonometric integration involves solving integrals that include trigonometric functions, which can be challenging due to the oscillatory nature of these functions.
In the problem, the function to be integrated is \( \sin(x^2 - 1) \). To simplify, we use substitution. By letting \( u = x^2 - 1 \), differentiation gives \( du = 2x \, dx \).
From here, the integral \( \int x \sin(x^2-1) \, dx \) can be transformed into \( \int \frac{1}{2} \sin(u) \, du \).
The integration of \( \sin(u) \) is known to be \( -\cos(u) + C \), thus simplifying the process to evaluating \( -\frac{1}{2} \cos(u) + C \), and substituting back for \( u = x^2 - 1 \).
This substitution method is key for dealing with more complex trigonometric integrals.

Definite Integral Evaluation

The definite integral evaluation is the step where we calculate the actual volume. After integrating, we substitute back to the original variable and find the definite value by applying the bounds.
For this exercise, once you have \(-\frac{1}{2} \cos(x^2 - 1)\), you solve \( 2\pi [-\frac{1}{2}\cos(x^2 - 1)]_1^{\sqrt{1+\pi}} \).
You substitute these bounds to find \([ -\cos(\pi) + \cos(0) ]\), which evaluates to \([1 - (-1)]\).
Simplifying gives \(2\pi\), which is the volume of the shape. Understanding trigonometric values, like \(\cos(0) = 1\) and \(\cos(\pi) = -1\), ensures accuracy in this step.
Therefore, definite integrals bring us from an indefinite general solution to a specific numerical result, completing the volume assessment.