Question

Find the volume generated by rotating the region bounded by the given curves about the specified line. Express the answers in exact form or approximate to the number of decimal places indicated. $$ y=\sin x, y=0, x=2 \pi, x=3 \pi \text { about the } y \text { -axis (Express the answer in exact form.) } $$

Short answer

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

Step-by-step solution

Understanding the Problem

We need to find the volume of the solid of revolution generated by rotating the area bounded by the curves \(y = \sin x\), \(y = 0\), \(x = 2\pi\), and \(x = 3\pi\) about the \(y\)-axis.

Setup the Integral

To find the volume about the \(y\)-axis, we will use the method of cylindrical shells. The formula for volume using cylindrical shells when revolving around the \(y\)-axis is \[ V = \int_{a}^{b} 2\pi x f(x) \, dx \]In our case, \(f(x) = \sin x\), \(a = 2\pi\), and \(b = 3\pi\).

Substitute and Simplify

Substitute the given functions and limits into the formula:\[ V = \int_{2\pi}^{3\pi} 2\pi x \sin x \, dx \]This integral will calculate the volume of the solid.

Integration by Parts

To solve the integral \(\int 2\pi x \sin x \, dx\), use integration by parts where:Let \(u = x\), then \(du = dx\);and \(dv = \sin x \, dx\), then \(v = -\cos x\).The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\).

Applying Integration by Parts

Applying the formula, we have:\[ \int 2\pi x \sin x \, dx = 2\pi \left(-x \cos x + \int \cos x \, dx \right) \]Evaluating \(\int \cos x \, dx\) gives \(\sin x\), so:\[ \int 2\pi x \sin x \, dx = 2\pi \left(-x \cos x + \sin x\right) + C \]

Evaluate the Definite Integral

Now evaluate the integral from \(2\pi\) to \(3\pi\):\[ V = 2\pi \left[ (-x \cos x + \sin x) \right]_{2\pi}^{3\pi} \]Calculate the expression at \(x = 3\pi\) and subtract the expression at \(x = 2\pi\).

Performing the Calculations

Putting \(x = 3\pi\) and \(x = 2\pi\):\[ V = 2\pi \left( -(3\pi) \cdot (-1) + 0 - (-(2\pi) \cdot (-1) + 0) \right) \]\[ V = 2\pi (3\pi - 2\pi) = 2\pi \cdot \pi = 2\pi^2 \]

Final Result

Thus, the volume of the solid generated by revolving the region about the \(y\)-axis is \(2\pi^2\).

Key concepts

Cylindrical Shells Method

The Cylindrical Shells Method is an ingenious approach to finding the volume of a solid of revolution. We use this technique when the region is rotated around an axis, typically when it's more convenient than the disk or washer method. The beauty of the cylindrical shells method lies in how it slices the region vertically and creates cylindrical shells.
When revolving around the y-axis, each vertical slice of the region translates into a thin-walled cylinder or shell. The volume of each infinitesimally thin shell can be described by the formula:

• The height of the shell represented as the function value, which is the distance from the x-axis to the curve, denoted by \(f(x)\).
• The radius of the shell, given by the x-coordinate as \(x\).
• The thickness of the shell, represented by a small change in x, \(dx\).

Thus, the formula for the volume of a shell is: \[ V = \int_{a}^{b} 2\pi x f(x) \, dx \]
For our exercise, we rotated the sine function \(y = \sin x\) from \(x = 2\pi\) to \(x = 3\pi\) around the y-axis. Through the cylindrical shells method, we efficiently computed the volume of the resulting shape.

Integration by Parts

Integration by Parts is a powerful technique used in calculus to solve integrals, particularly when the product of two functions is involved. It is derived from the product rule of differentiation and flips the problem into a more manageable form. The formula for Integration by Parts is:

• \(\int u \, dv = uv - \int v \, du\)
• Choose \(u\) typically as the algebraic function and the rest as \(dv\).

In this problem, we needed to find \(\int 2\pi x \sin x \, dx\). By letting \(u = x\) and \(dv = \sin x \, dx\), we then find:

• \(du = dx\)
• \(v = -\cos x\)

The integration then transforms into:\[-x \cos x + \int \cos x \, dx\] After evaluating \(\int \cos x \, dx\) to be \(\sin x\), we substitute back to find the integral expression. This method effectively simplifies solving challenging integrals involving sine and polynomial terms.

Definite Integrals

Definite Integrals are essential in calculating the area under a curve and, in this context, the volume of a solid of revolution. They represent the cumulative sum of an infinite number of infinitesimally small contributions over a specific interval. It is helpful in providing exact values over ranges using upper and lower limits \([a, b]\).
For calculating the volume in our exercise, we used the definite integral:\[ V = \int_{2\pi}^{3\pi} 2\pi x \sin x \, dx \] This integral represents the exact volume of the solid obtained by rotating the curve between these limits. By evaluating, we calculate:

• At \(x = 3\pi\) and \(x = 2\pi\) substituting back the expression found via Integration by Parts.
• Subtract the results to find \(2\pi\) times the differences of these limits after substituting back the expression.

The definite integral calculation ultimately leads to the exact volume of \(2\pi^2\), demonstrating how integration delivers precise solutions to real-world geometrical problems.