69-72. Volumes of solids Find the volume of the following solids.
70. The region bounded by y = 1/[x²(x² + 2)²], y = 0, x = 1, and x = 2 is revolved about the y-axis.

Verified step by step guidance

1. Identify the region bounded by the curves: the function \(y = \frac{1}{x^{2}(x^{2} + 2)^{2}}\), the line \(y = 0\), and the vertical lines \(x = 1\) and \(x = 2\).
2. Since the solid is formed by revolving the region about the y-axis, consider using the method of cylindrical shells. The formula for the volume using cylindrical shells is: \(V = \int_{a}^{b} 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx\).
3. In this problem, the radius of a shell is the distance from the y-axis to the shell, which is \(x\), and the height of the shell is the function value \(y = \frac{1}{x^{2}(x^{2} + 2)^{2}}\).
4. Set up the integral for the volume as: \(V = \int_{1}^{2} 2\pi x \cdot \frac{1}{x^{2}(x^{2} + 2)^{2}} \, dx\).
5. Simplify the integrand before integrating, then evaluate the integral to find the volume of the solid.

Key Concepts

Volume of Solids of Revolution

This concept involves finding the volume of a 3D solid formed by rotating a 2D region around an axis. The volume can be computed using methods like the disk/washer or shell method, depending on the axis of rotation and the shape of the region.

Shell Method

The shell method calculates volume by integrating cylindrical shells formed when a region is revolved around an axis. It is especially useful when revolving around the y-axis and involves integrating 2π(radius)(height) with respect to x or y.

Setting up Proper Integral Limits and Functions

Accurately identifying the bounds of integration and the function expressions for radius and height is crucial. For the given problem, the limits are x=1 to x=2, and the function y = 1/[x²(x² + 2)²] defines the height of the shell or the radius depending on the method.