Problem 26 - Thomas Calculus

Question

Find the volume of the solid generated by revolving the region about the given line. \begin{equation} \begin{array}{l}{\text { The region in the first quadrant bounded above by the line } y=2} \\ {\text { below by the curve } y=2 \sin x, 0 \leq x \leq \pi / 2, \text { and on the left by }} \\ {\text { the } y \text { -axis, about the line } y=2}\end{array} \end{equation}

Step-by-Step Solution

Answer

The volume is \( 8\pi - 4\pi^2 \).

Step 1: Visualize the Problem

The region we are interested in is bounded above by the line \( y = 2 \), below by the curve \( y = 2 \sin x \), from \( x = 0 \) to \( x = \frac{\pi}{2} \), and on the left by the \( y \)-axis. We need to determine the volume of the solid formed by revolving this region around the line \( y = 2 \).

Step 2: Use the Shell Method

Since we are revolving around a horizontal line \( y = 2 \), we use the shell method. The shell method involves calculating the volume of cylindrical shells. For a shell at position \( x \), the radius from the axis of revolution is \( 2 - y \) (since \( y = 2 - y \text{, where}\ y = 2 \sin x \)).

Step 3: Determine Shell Radius and Height

The height of each shell is from \( y = 2 \sin x \) to \( y = 2 \), so the height is \( 2 - 2 \sin x \). The radius is \( y - 0 = y \).

Step 4: Write the Shell Volume Formula

The volume \( V \) of each cylindrical shell can be determined with the formula: \[ V = \int_0^{\frac{\pi}{2}} 2\pi (\text{radius})(\text{height}) \, dx \] Substituting for the radius and height, we have: \[ V = \int_0^{\frac{\pi}{2}} 2\pi (2 - 2\sin x)(2 \sin x) \, dx \]

Step 5: Simplify and Evaluate the Integral

Expand the expression inside the integral: \( 2\pi \int_0^{\frac{\pi}{2}} (4\sin x - 4\sin^2 x) \, dx \). Then split into separate integrals: \[ 2\pi \left( \int_0^{\frac{\pi}{2}} 4\sin x \, dx - \int_0^{\frac{\pi}{2}} 4\sin^2 x \, dx \right) \] Evaluate each integral separately using techniques such as integration by substitution or trigonometric identities.

Step 6: Solve the Integrals

First, for the integral \( \int_0^{\frac{\pi}{2}} 4\sin x \, dx \), substitute \( u = \cos x \) resulting in \( -4\cos x \big|_0^{\frac{\pi}{2}} = 4\). For \( \int_0^{\frac{\pi}{2}} 4\sin^2 x \, dx \), use the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \) to simplify and evaluate, resulting in \( 2\pi \).

Step 7: Compute the Final Volume

Substitute these results back into the overall formula: \[ V = 2\pi (4 - 2\pi) \]. Thus, the volume of the solid is \( 8\pi - 4\pi^2 \).

Key Concepts

Shell Method

The shell method is an intuitive way to calculate the volume of a solid of revolution. It involves using cylindrical shells to break down the volume calculation into manageable parts. When you revolve a region around an axis, typically a horizontal or vertical line, cylindrical shells are formed which, when summed up, give the total volume of the solid. Here, the axis is the line \( y=2 \) and we are revolving the region around it.

• Visualize the revolution as a series of concentric hollow cylinders or shells.
• Focus on calculating the volume of a single thin shell at position \( x \).
• To find the total volume, integrate these shells over the relevant range of \( x \).

The shell method is particularly useful when the axis of rotation is parallel to the area of the shape being revolved.

Cylindrical Shells

Cylindrical shells are fundamental to the shell method. They are essentially thin, hollow cylinders whose collective volume can represent the entire solid of revolution. When using the shell method, each shell has its own radius and height, which are crucial to determining the shell's volume.

• Each shell's radius is calculated as the distance from the axis of revolution to the shell itself.
• The height of each shell corresponds to the function value over the interval, like \( 2 - 2\sin x \) here.
• The formula \( V = \int 2\pi (\text{radius})(\text{height}) \, dx \) helps compute their cumulative volume.

Thus, for the defined area, the cylindrical shell formulation simplifies the process of integrating volumes.

Trigonometric Integration

Trigonometric integration is a powerful technique when dealing with functions involving trigonometric expressions. In this exercise, the need arises for integrating \( \sin x \) and \( \sin^2 x \), both of which are trigonometric functions.

• To integrate \( \sin x \), a common technique is substitution: set \( u = \cos x \), leading to an easier integration process.
• For \( \sin^2 x \), use the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \). This identity simplifies the integral by transforming it into a sum of simpler integrable terms.

By applying these techniques, one calculates definite integrals with trigonometric functions effectively.

Definite Integral Evaluation

Definite integral evaluation is the process of calculating the accumulated value of a function over a specific interval. In this exercise, you perform definite integrals to find the overall volume.

• First, split the integral into more manageable parts, like \( \int 4\sin x \, dx \) and \( \int 4\sin^2 x \, dx \).
• As you solve these, apply integration techniques that fit each case.
• Compute each part's integral over the specified bounds \( 0 \) to \( \frac{\pi}{2} \) after substituting or simplifying using trigonometric identities.

Finally, combine these evaluated pieces to get the total volume, \( 8\pi - 4\pi^2 \), ensuring a precise and correct solution.