Problem 135 - Calculus Volume 1

Question

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and y = 0 around the x-axis. $$x=\frac{1+y^{2}}{y}, y=0, \text { and } y=2$$

Step-by-Step Solution

Answer

The volume is \(\frac{28\pi}{3}\).

Step 1: Understand the Problem

We are given a curve defined by the equation \(x = \frac{1 + y^2}{y}\), and we need to find the volume of the solid formed by rotating this curve around the x-axis, between \(y = 0\) and \(y = 2\). We will use the method of cylindrical shells.

Step 2: Set Up the Shells Method Formula

The formula for volume using cylindrical shells when rotating around the x-axis is given by: \( V = 2\pi \int_a^b y \cdot f(y) \, dy \), where \(f(y)\) represents the x value of the curve. In this case, \( f(y) = \frac{1 + y^2}{y} \).

Step 3: Substitute Into the Integral

Substitute \(f(y) = \frac{1 + y^2}{y}\) into the shells formula so that our volume integral becomes: \[ V = 2\pi \int_0^2 y \left(\frac{1+y^2}{y}\right) \, dy = 2\pi \int_0^2 (1 + y^2) \, dy \].

Step 4: Simplify the Integral Expression

Simplifying, we get: \[ V = 2\pi \int_0^2 (1 + y^2) \, dy = 2\pi \left(\int_0^2 1 \, dy + \int_0^2 y^2 \, dy\right) \]. Evaluate each part separately.

Step 5: Evaluate the Integral

Evaluate each component: \( \int_0^2 1 \, dy = 2 - 0 = 2 \) and \( \int_0^2 y^2 \, dy = \left.\frac{y^3}{3}\right|_0^2 = \frac{8}{3} - 0 = \frac{8}{3} \).

Step 6: Calculate the Total Volume

Substitute these values back into the integral: \[ V = 2\pi (2 + \frac{8}{3}) = 2\pi \left(\frac{6}{3} + \frac{8}{3}\right) = 2\pi \cdot \frac{14}{3} = \frac{28\pi}{3} \].

Step 7: Conclusion

Thus, the volume of the solid formed by rotating the region between the given curve and \(y = 0\) around the x-axis, from \(y = 0\) to \(y = 2\), is \(\frac{28\pi}{3}\).

Key Concepts

Volume of Revolution

When you think of volume of revolution, imagine a 2D shape spinning around an axis and forming a 3D shape. This is how we find volumes of some complex shapes.
One popular method to calculate such volume is using the **Cylindrical Shells Method**.
Here’s how it works:

• Envision sections of the shape being wrapped around to form layered, cylindrical slices (or shells).
• Calculate the volume of these shells using integration techniques from calculus.

When doing this, it’s crucial to choose the right axis and set up a proper function to integrate. This allows you to correctly determine how each small "sliver" adds up to the full volume.

Calculus

Calculus is essential for understanding changes and is divided into differential and integral calculus. We are focusing here on calculus techniques that involve finding areas or volumes.
**Differential calculus** helps understand rates of change and slopes of curves.
**Integral calculus**, like in our problem, allows us to sum up an infinite number of tiny parts to find areas under a curve or volumes as in the case of revolved regions.
For students learning calculus, it's key to:

• Understand the fundamental theorems that define calculus.
• Practice solving a variety of problems that require setting up and solving integrals.

This not only helps in finding volumes but connects deeply with many aspects of both pure and applied mathematics.

Integral Calculus

Integral calculus is the study of integrals and their applications. It’s about summing everything up or finding the whole from its parts.
The integral, represented by the symbol \(\int\), is a tool that calculates the total size or value when a function changes.
Here's what to keep in mind about integrals:

• Definite integrals compute the accumulated quantity, such as area or volume.
• Indefinite integrals relate to finding antiderivatives or general solutions of functions.

In applying the integral calculus to the cylindrical shells method, you are calculating a definite integral. This involves setting the limits within which you are revolving the area, and finding the integral of your function across these limits. Understanding this process is crucial for working with many practical physical problems.