Problem 28 - Thomas Calculus

Question

Find the volume of the solid generated by revolving the region about the given line. The region in the first quadrant bounded above by the line \(y=2,\) below by the curve \(y=2 \sin x, 0 \leq x \leq \pi / 2,\) and on the left by the \(y\) -axis, about the line \(y=2\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem and Setup the Integral

We need to find the volume of a solid formed by revolving the given region in the first quadrant around the line \( y = 2 \). The region is bounded above by \( y = 2 \), below by \( y = 2 \sin x \) from \( x = 0 \) to \( x = \pi/2 \), and to the left by the \( y \)-axis.

Step 2: Use the Washer Method

The washer method is applicable here since we are revolving the region around a horizontal line. For a small vertical slice at \( x \), the outer radius to the line \( y = 2 \) is \( R(x) = 2 - 0 = 2 \) (the distance from \( y = 2 \) to \( y = 0 \)), and the inner radius is \( r(x) = 2 - 2 \sin x \) (the distance from the line to the curve).

Step 3: Write the Volume Integral

The volume \( V \) is given by the integral of the washer areas from \( x = 0 \) to \( x = \pi/2 \). The area of each washer is \( \pi [R(x)^2 - r(x)^2] \). Therefore, the integral is:\[ V = \int_{0}^{\pi/2} \pi \left[ (2)^2 - (2 - 2\sin x)^2 \right] \, dx \].

Step 4: Simplify the Integral Expression

The integral becomes:\[ V = \pi \int_{0}^{\pi/2} \left[ 4 - (2 - 2\sin x)^2 \right] \, dx \].First, expand \((2 - 2\sin x)^2 \):\[(2 - 2\sin x)^2 = 4 - 8\sin x + 4\sin^2 x\].Substitute back into the integral:\[ V = \pi \int_{0}^{\pi/2} \left[ 4 - (4 - 8\sin x + 4\sin^2 x) \right] \, dx \],which simplifies to:\[ V = \pi \int_{0}^{\pi/2} (8\sin x - 4\sin^2 x) \, dx \].

Step 5: Compute the Integral

Split the integral: \[ V = \pi \left( \int_{0}^{\pi/2} 8\sin x \, dx - \int_{0}^{\pi/2} 4\sin^2 x \, dx \right) \].Calculate each:1. \( \int_{0}^{\pi/2} 8\sin x \, dx = [-8\cos x]_{0}^{\pi/2} = [0 + 8] = 8\).2. \( \int_{0}^{\pi/2} 4\sin^2 x \, dx \) can be computed using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \) and integrating \[ 4\cdot \frac{1}{2} \int_{0}^{\pi/2} (1 - \cos(2x)) \, dx = 2\left( \frac{\pi}{2} - 0 \right) = \pi \].Thus, \( V = \pi (8 - \pi) \).

Step 6: Final Answer

Upon simplifying, the volume of the solid is \( 8\pi - \pi^2 \).

Key Concepts

Volume of Revolution

Imagine you have a flat shape, like a slice of bread, and you want to spin it around an axis to form a 3D object. This process is known as finding the volume of revolution. The core idea is to take a 2D shape and revolve it around a line, creating a symmetrical 3D object. In mathematical terms, this is often done around the x-axis or y-axis, or any other line parallel to these axes.

In our example, the region is spun around the line \(y = 2\), which means every point on the curve goes in a circular path, creating a solid shape. This problem involves spinning a region defined in the first quadrant around the horizontal line \(y = 2\), resulting in a solid representing this volume of revolution.

Definite Integral

The definite integral is a powerful mathematical tool used to calculate the accumulation of quantities, such as area or volume, over a specified interval. In our problem, we use a definite integral to calculate the volume of the solid formed by the revolving region, employing limits from \(x = 0\) to \(x = \pi/2\).

The definite integral in this context translates to summing up infinitely many small pieces (washers) to make up the volume. The integral expression \[ V = \int_{0}^{\pi/2} \pi \left[ (2)^2 - (2 - 2\sin x)^2 \right] \ dx \] collects all these infinite tiny circular slices together to find the whole volume.

Trigonometric Integration

Trigonometric integration is essential when dealing with functions that involve trigonometric components, such as \( \sin x \) or \( \cos x \). In this problem, we encounter such functions while calculating the inner and outer radii of the washers.

The task involves integrating expressions containing sine and sine-squared terms, which can be complex. To help simplify, we use trigonometric identities. For example, rewriting \( \sin^2 x \) as \( \frac{1 - \cos(2x)}{2} \) simplifies the integration process significantly. This allows us to compute the integral, using known integration techniques, to determine the exact volume of the solid created.

Sinusoidal Functions

Sinusoidal functions, like sine and cosine, are fundamental in modeling periodic phenomena. In this exercise, the curve \(y = 2 \sin x\) is a sinusoidal function that helps define the boundary of the region to be revolved.

These functions have a characteristic smooth wave-like structure and repeat periodically. They play a crucial role in many areas of physics and engineering beyond pure mathematics. In our problem, the sinusoidal function determines the shape and dimensions of the region being revolved, thus influencing the volume of the solid generated. The periodic nature of the sine function here not only establishes the geometry but also presents the challenge of integrating such a function over the given interval.