Problem 20 - Calculus Single Variable

Question

In each of Exercises 19-24, use the method of washers to calculate the volume \(V\) obtained by rotating the given planar region \(\mathcal{R}\) about the \(y\) -axis. \(\mathcal{R}\) is the region between the curves \(x=y^{2}\) and \(x=y^{3}\) \(0 \leq y \leq 1\)

Step-by-Step Solution

Answer

The volume is \(\frac{2\pi}{35}\).

Step 1: Identify the curves and region

We have two curves: \(x = y^2\) and \(x = y^3\), both defined over the interval \(0 \leq y \leq 1\). The region \(\mathcal{R}\) is bounded by these curves for the given domain.

Step 2: Set up the integral for the volume using washers

To find the volume of the solid of revolution, we use the washer method. For rotation about the \(y\)-axis, the outer radius is the larger \(x\)-value \(y^2\) and the inner radius is the smaller \(x\)-value \(y^3\). The washer formula in terms of \(y\) gives: \[ V = \pi \int_{0}^{1} \left((y^2)^2 - (y^3)^2\right) \, dy \]

Step 3: Simplify the integrand

The expression \((y^2)^2 - (y^3)^2\) simplifies to \(y^4 - y^6\). Thus, the integral becomes: \[ V = \pi \int_{0}^{1} (y^4 - y^6) \, dy \]

Step 4: Calculate the integral

Compute the integral: 1. Integrate \(y^4\) to get \(\frac{y^5}{5}\).2. Integrate \(y^6\) to get \(\frac{y^7}{7}\).Applying these gives: \[ V = \pi \left[ \frac{y^5}{5} - \frac{y^7}{7} \right]_{0}^{1} \]

Step 5: Evaluate the definite integral

Substitute the limits \(y = 1\) and \(y = 0\) into the integral result: \[ \frac{1^5}{5} - \frac{1^7}{7} = \frac{1}{5} - \frac{1}{7} = \frac{7}{35} - \frac{5}{35} = \frac{2}{35} \] The evaluated volume is: \[ V = \pi \times \frac{2}{35} = \frac{2\pi}{35} \]

Step 6: Conclude the solution

The volume \(V\) of the solid obtained by rotating the region \(\mathcal{R}\) around the \(y\)-axis is \(\frac{2\pi}{35}\).

Key Concepts

Volume of Revolution

When we talk about the "Volume of Revolution," we are referring to the volume of a 3D solid formed by rotating a 2D region around a given axis. In many calculus problems, this involves rotating a function's graph around either the x or y axis. Let's break this down with a simple process:

• First, identify the curves that outline the 2D region.
• Then, decide which axis you are rotating around. This could be the x-axis, y-axis, or another line.
• Choose the appropriate method to find the volume, such as the washer method, if needed.

By understanding what each component in your region is doing, you can accurately compute how it behaves when spun around an axis to form a solid. It's a bit like imagining how a pottery wheel turns a flat piece of clay into a pot.

Integral Calculus

Integral calculus helps us find quantities like total area or volume, especially when we deal with curves instead of straight lines. This is crucial when working with calculus problems involving volumes like the one discussed here. When you calculate a volume of revolution using the washer method, you set up an integral that represents the area of each thin washer (or disc) and adds these infinitesimally small volumes together over the given range.

• Identify your integrand, which represents the shape of the curves. For washers, it's typically the difference between the outer and inner radius squared.
• Set bounds for your integral. These are typically the range over which your volume is calculated, which stems from the given curves.
• Simplify and compute the integral, which effectively sums up the area of these washers to calculate the total volume.

Integral calculus allows you to neatly sum up areas and volumes and find answers to complex questions in a succinct manner.

Definite Integrals

Definite Integrals play a key role when we want to find exact values such as volumes or areas in mathematics. Unlike indefinite integrals, which represent a family of functions, definite integrals give us a fixed value for a specific interval. For calculating the volume of revolution using the washer method, we utilize the definite integral. Here’s a simple breakdown:

• The definite integral is set with specific limits, which are the range of your variable of interest. For our problem, it's from 0 to 1 as defined by the curves.
• These limits tell you precisely where to start and stop the integration, making the integral "definite," as opposed to the general solution of an indefinite integral.
• Compute these limits in the integrand to finalize the calculation, often involving algebraic simplification, to get your final answer.

Using definite integrals, we can determine precise values for real-world problems, providing exact quantities like the volume of a solid formed through rotation.