Problem 20 - Calculus: Early Transcendentals

Question

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. \( x = 2y^2 \) , \( x = y^2 + 1 \) ; about \( y = -2 \)

Step-by-Step Solution

Answer

The volume is \( \frac{16\pi}{3} \).

Step 1: Identify the Curves and the Axis of Rotation

Firstly, we have the curves given by the equations \( x = 2y^2 \) and \( x = y^2 + 1 \). These curves form the boundaries of the region we are interested in. We need to rotate this bounded region around the horizontal line \( y = -2 \).

Step 2: Find the Points of Intersection

To find the limits of integration, determine where the curves intersect by setting \( 2y^2 = y^2 + 1 \). Solving gives \( y^2 = 1 \), hence \( y = \pm 1 \). The points of intersection in terms of \( x \) are \( (2, 1) \) and \( (2, -1) \).

Step 3: Set Up the Volume Integral Using the Shell Method

The shell method involves integrating in terms of \( y \). The radius of each shell is the distance from \( y = -2 \), which is \( y + 2 \). The height of each shell is the difference between the \( x \)-values of the two curves, which is \( (2y^2) - (y^2 + 1) = y^2 - 1 \). The volume \( V \) is given by the integral:\[V = 2\pi \int_{-1}^{1} (y + 2)(y^2 - 1) \, dy.\]

Step 4: Expand the Integrand

Expand the integrand of the integral:\[(y + 2)(y^2 - 1) = y(y^2 - 1) + 2(y^2 - 1) = y^3 - y + 2y^2 - 2.\]

Step 5: Evaluate the Integral

Now calculate the integral:\[V = 2\pi \int_{-1}^{1} (y^3 - y + 2y^2 - 2) \, dy.\]Calculate each term separately and apply the limits:\[V = 2\pi \left[\int_{-1}^{1} y^3 \, dy - \int_{-1}^{1} y \, dy + 2\int_{-1}^{1} y^2 \, dy - 2\int_{-1}^{1} 1 \, dy\right].\]Evaluating these integrals gives:\[\int_{-1}^{1} y^3 \, dy = 0, \quad \int_{-1}^{1} y \, dy = 0, \quad 2\int_{-1}^{1} y^2 \, dy = 2\left[\frac{y^3}{3}\right]_{-1}^{1} = \frac{4}{3}, \quad 2\int_{-1}^{1} 1 \, dy = 4.\]Thus,\[V = 2\pi (0 + 0 + \frac{4}{3} - 4) = 2\pi (\frac{4}{3} - 4) = 2\pi \left(\frac{4}{3} - \frac{12}{3}\right) = 2\pi \left(-\frac{8}{3}\right) = -\frac{16\pi}{3}.\]However, since volumes cannot be negative, consider the magnitude which is \( \frac{16\pi}{3} \).

Step 6: Conclusion

The volume of the solid generated by rotating the region bounded by the curves around \( y = -2 \) is \( \frac{16\pi}{3} \).

Key Concepts

Volume of Solids of Revolution

The volume of solids of revolution refers to the space occupied by a 3D object formed when a 2D area is revolved around a line or axis. This method is particularly useful in calculus to find volumes of objects with circular or cylindrical symmetry. In our problem, we're examining a region bounded by the curves described by the equations \( x = 2y^2 \) and \( x = y^2 + 1 \) and rotating this region around the line \( y = -2 \).
The cylindrical shells method is a common technique to calculate such volumes. It involves imagining the solid is composed of a series of nested cylindrical shells. As the region rotates around the axis, these shells expand away from the axis, forming the solid's volume. By integrating the contribution of each shell over the range of interest, we can compute the full volume of the solid.

Curve Intersection

Understanding where the given curves intersect is crucial for finding the limits of integration. In our problem, we have two curves: \( x = 2y^2 \) and \( x = y^2 + 1 \). By setting these equations equal, we determine the points where the two curves meet. Solving \( 2y^2 = y^2 + 1 \), we find that they intersect at the points where \( y^2 = 1 \), hence at \( y = \pm 1 \).
These intersection points correspond to the critical limits needed for our definite integral, as they delimit the region that will be revolved. Specifically, for this exercise, the intersection points lead to creating bounds from \( y = -1 \) to \( y = 1 \). Identifying these points ensures that our computation for the area to be rotated is both accurate and meaningful.

Definite Integrals

Definite integrals play a pivotal role in calculating the volume generated by rotating a region around an axis. They provide a way to accumulate infinitely small quantities, which is essential when dealing with volumes in calculus. In our exercise, the definite integral is used to sum up the volumes of cylindrical shells formed from rotating the region between the curves from \( y = -1 \) to \( y = 1 \).
The setup of the integral involves calculating the radius and height for each shell, then integrating over the limits defined by the points of intersection. The radius in this case is \( y + 2 \), the distance from the axis of rotation \( y = -2 \). The height is given by the difference in \( x \)-values of the curves, or \( y^2 - 1 \). We then compute the integral \[2\pi \int_{-1}^{1} (y + 2)(y^2 - 1) \, dy\] to find the volume.

Calculus Problem Solving

Calculus problem solving often involves translating a complex scenario into a series of simpler mathematical steps. This involves setting up equations, identifying boundaries, and solving integrals effectively—skills crucial for tackling problems like finding volumes of solids of revolution. For our particular problem, each step led to a clearer understanding of the task.

• Recognizing the problem type: This problem involves calculating volume using the cylindrical shells method.
• Finding curve intersections: Necessary to identify the limits of integration.
• Setting up and evaluating the integral: Essential to solving the volume accurately using the shells' dimensions.
• Checking results: Confirming calculations, especially ensuring the volume is non-negative.

With practice, calculus methods like these become powerful tools for solving a diverse range of mathematically rich problems.