Problem 18 - Basic Technical Mathematics with Calculus

Question

Find the volume generated by revolving the regions bounded by the given curves about the \(y\) -axis. Use the indicated method in each case. $$y=\sqrt{x^{2}-1}, y=0, x=3 \quad \text { (shells) }$$

Step-by-Step Solution

Answer

The volume is \( \frac{8\pi \sqrt{2}}{3} \).

Step 1: Understand the problem

We need to find the volume of the solid formed when the region bounded by the curves \( y = \sqrt{x^2 - 1} \), \( y = 0 \), and \( x = 3 \) is revolved about the \( y \)-axis. We'll use the method of cylindrical shells for this purpose.

Step 2: Express cylinder height and radius in terms of x

The height of the cylindrical shell is given by \( y = \sqrt{x^2 - 1} \) and the radius of the shell is \( x \) since we revolve around the \( y \)-axis.

Step 3: Set up the volume integral

The volume of the solid formed is given by the integral: \[ V = \int_{a}^{b} 2 \pi (\text{radius}) (\text{height}) \, dx \] Substitute the expressions for the radius and height: \[ V = \int_{1}^{3} 2 \pi x \sqrt{x^2 - 1} \, dx \]. Here, the bounds \(1\) and \(3\) are the \(x\)-limits determined from \( y = \sqrt{x^2 - 1} \).

Step 4: Perform the integration

To solve \( \int_{1}^{3} 2\pi x \sqrt{x^2 - 1} \, dx \), perform a substitution: let \( u = x^2 - 1 \) implying \( du = 2x \, dx \). Now the integral becomes: \[ V = \pi \int_{0}^{8} \sqrt{u} \, du = \pi \int_{0}^{8} u^{1/2} \, du \].

Step 5: Integrate and evaluate

The integral \( \int u^{1/2} \, du = \frac{2}{3} u^{3/2} \). Applying the limits: \[ V = \pi \left[ \frac{2}{3} u^{3/2} \right]_{0}^{8} = \pi \left( \frac{2}{3} (8^{3/2}) - \frac{2}{3} (0^{3/2}) \right) \]. Calculate \( 8^{3/2} = (8^3)^{1/2} = 4 \sqrt{2} \), thus: \[ V = \pi \cdot \frac{2}{3} \cdot 4 \sqrt{2} = \frac{8\pi \sqrt{2}}{3} \].

Key Concepts

Cylindrical Shells Method

The cylindrical shells method is a clever technique used in calculus for finding the volume of a solid of revolution. This technique is particularly handy when rotating around an axis that is parallel to the incoming variable in the given equation. Instead of cutting the solid into disks or washers, it imagines slicing the solid into thin cylindrical `shells`.

Think of each shell like a tin can—with a small height and outer circumference. The height of each shell depends on the function you are rotating, and the radius is the distance from the axis of rotation. When the shell is unrolled, it forms a rectangle whose width is the height of the function and length is its circumference.

This method is especially useful when the axis of rotation is vertical, like the y-axis, and you're integrating with respect to x. When you perform the integration, each little shell gets summed, leading to the total volume of the object.

Integral Calculus

Integral calculus is a fundamental branch of mathematics that focuses on accumulation of quantities, such as areas under a curve, total displacement from a velocity function, and, importantly, volumes of solids. In the volume of revolution problems, integral calculus allows us to precisely sum up an infinite number of infinitesimally small quantities to find a total sum (e.g., area or volume).

In these types of problems, understanding the setup of your integral is key. First, identify the bounds—these are simply the limits defining the range over which you're revolving the function. For revolving around the y-axis using cylindrical shells, your bounds will be in terms of x, spanning the region itself. The integrand, or core function of your integral, comprises any expressions for height, radius, and, typically, another differential element like dx or dy.

Knowing how to apply integration techniques, such as substitution, is pivotal for solving these problems. For instance, in our original exercise, we used substitution to simplify the integral before computing the exact volume.

Volume Calculation

When tackling a volume calculation problem, especially using integration methods like cylindrical shells, it involves a series of organized steps to ensure an accurate outcome.

• First, determine the shapes that your solid will take when revolved—whether it’s disks, washers, or cylindrical shells.
• Next, translate this problem into a mathematical integral. The formula for cylindrical shells revolves around the concept: \[ V = \int_{a}^{b} 2 \pi (\text{radius})(\text{height}) \, dx \]
• Use the correct bounds for your integral, representing the region you’re revolving. For our y-axis example, the radius was x, and the height was determined by the function given.

After setting up your integral, solve it using appropriate calculus methods. Integrating smoothly encapsulates all pieces of the solid between your limits. The final result gives you the total volume of the solid formed by revolution—from start to finish seasoned by precise application of integral calculus.