91Ó°ÊÓ

The Disk Method is a fantastic technique to find the volume of a solid of revolution. Imagine spinning a flat shape around an axis. The shape will create a three-dimensional object, which is the solid of revolution. In our example, the shape is bounded by the curve given by the equation \(y = \sin(x^2 - 1)\) and the \(x\)-axis.

We are specifically rotating this shape around the \(y\)-axis. To calculate the volume, we use the Disk Method formula:

• \( V = \int_{a}^{b} 2\pi x y \, dx \)

This formula works by slicing the object into thin disks perpendicular to the axis of rotation. Each disk's volume is approximately the area of the circle \(\pi (radius^2)\) times its width \(dx\).

The integral sums up the volumes of all these disks from \(x = a\) to \(x = b\), which in our problem is from \(1\) to \(\sqrt{1+\pi}\). This particular rotation involves integrating around the \(y\)-axis, which is why we multiply by \(2\pi x\) in the integral.

A solid of revolution is created when you revolve a planar region around a line, such as an axis. Picture a flat 2D region spinning around a central axis, like twisting a paper fan around its center; the resulting shape is what we call a solid of revolution.

In this exercise, revolving the curve \(y = \sin(x^2 - 1)\) around the \(y\)-axis over a specified interval creates a solid that resembles a cake ring, also known as a Bundt cake. This solid's volume is the goal of our calculation.

The process involves effectively capturing the volume of this 3D shape by slicing it into infinitely thin disks. These slices or disks account for the changing function values along the \(x\)-axis and their distance from the center or \(y\)-axis. By summing up all these infinitesimally small volumes via integration, we fully capture and calculate the volume of the entire shape.

Integration techniques are essential for solving complex calculus problems, especially when dealing with solids of revolution. In this context, integration allows us to sum up continuously changing slices to get the volume of a solid.

To solve our problem, we needed a substitution method to simplify the integral \(\int_{1}^{\sqrt{1+\pi}} 2\pi x \sin(x^2 - 1) \, dx\).

By using substitution, we can transform a difficult integral into something more manageable. Here’s how we did it:

• Let \( u = x^2 - 1 \)
• Then, \( du = 2x \, dx \)

This change of variables converted our integral into \(\int_{0}^{\pi} \pi \sin u \, du\).

Substitution is only one of many powerful methods available in the toolbox of integration techniques, but it is particularly helpful here because it simplifies our integral considerably, allowing easy evaluation of the antiderivative and thus leading us to the final solution for the volume: \(2\pi\).