Problem 29 - Thomas Calculus in SI Units

Question

For some regions, both the washer and shell methods work well for the solid generated by revolving the region about the coordinate axes, but this is not always the case. When a region is revolved about the \(y\) -axis, for example, and washers are used, we must integrate with respect to \(y .\) It may not be possible, however, to express the integrand in terms of \(y .\) In such a case, the shell method allows us to integrate with respect to \(x\) instead. Compute the volume of the solid generated by revolving the region bounded by \(y=x\) and \(y=x^{2}\) about each coordinate axis using a. the shell method. b. the washer method.

Step-by-Step Solution

Answer

The volume by both methods is \(\frac{\pi}{6}\) for the y-axis and \(\frac{2\pi}{15}\) for the x-axis.

Step 1: Set up Shell Method Integral for Revolving About y-axis

The shell method involves integrating along the axis perpendicular to the axis of revolution. When revolving about the y-axis, the typical shell has radius \(x\) and height \( (x - x^2) \). The volume \(V\) is given by \[ V = 2\pi \int_{0}^{1} x (x - x^2) \, dx \].

Step 2: Simplify and Integrate Shell Method

First, simplify the integrand: \(x(x - x^2) = x^2 - x^3 \). The integral becomes \( V = 2\pi \int_{0}^{1} (x^2 - x^3) \, dx \). Evaluate the integral: \( \int_{0}^{1} (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \). Then, multiply by \(2\pi\) to find \( V = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6} \).

Step 3: Set up Washer Method Integral for Revolving About y-axis

For the washer method about the y-axis, express \(x\) as a function of \(y\): \(x = y^{1/2}\) and \(x = y\). The volume \(V\) is given by the integral \[ V = \pi \int_{0}^{1} ((y^{1/2})^2 - y^2) \, dy \].

Step 4: Simplify and Integrate Washer Method

Simplify the integrand: \( (y^{1/2})^2 = y \), thus the integral becomes \( V = \pi \int_{0}^{1} (y - y^2) \, dy \). Evaluate the integral: \( \int_{0}^{1} (y - y^2) \, dy = \left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \). Then, multiply by \(\pi\) to get \( V = \pi \cdot \frac{1}{6} = \frac{\pi}{6} \).

Step 5: Set up Shell Method Integral for Revolving About x-axis

Using the shell method about the x-axis, the shell has radius \(y\) and height \(x - x^2 = y - y^2\), since \(y = x\). So, the volume is given by \[ V = 2\pi \int_{0}^{1} y (y - y^2) \, dy \].

Step 6: Simplify and Integrate Shell Method about x-axis

Simplify the integrand: \(y(y - y^2) = y^2 - y^3\). The integral becomes \( V = 2\pi \int_{0}^{1} (y^2 - y^3) \, dy \). Evaluate: \( \int_{0}^{1} (y^2 - y^3) \, dy = \left[ \frac{y^3}{3} - \frac{y^4}{4} \right]_0^1 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \). Then, \( V = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6} \).

Step 7: Set up Washer Method Integral for Revolving About x-axis

For the washer method about the x-axis, express the functions in terms of \(x\) as follows \(y = x\) and \(y = x^{2}\). Therefore, the volume \(V\) is determined by \[ V = \pi \int_{0}^{1} (x^2 - (x^2)^2) \, dx \].

Step 8: Simplify and Integrate Washer Method about x-axis

Simplify the integrand: \( x^2 - x^4 \). The integral becomes \( V = \pi \int_{0}^{1} (x^2 - x^4) \, dx \). Evaluate: \( \int_{0}^{1} (x^2 - x^4) \, dx = \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 = \frac{1}{3} - \frac{1}{5} = \frac{2}{15} \). Therefore, the volume \( V = \pi \cdot \frac{2}{15} = \frac{2\pi}{15} \).

Key Concepts

Shell Method

The shell method is a powerful technique in calculus used to find the volume of a solid of revolution. It is particularly advantageous when revolving around the y-axis if the integrand becomes complicated when expressed with respect to y. With the shell method, we integrate along the axis perpendicular to the rotation. Let's explore how this works:

Imagine you're wrapping cylindrical shells around the y-axis. The key idea is that each shell contributes a small slice of the solid's volume. These shells have:

• A radius equal to the distance from the shell to the axis of rotation, denoted as \( x \) for the y-axis.
• A height determined by the vertical difference between the upper and lower functions, such as \( (x - x^2) \) in this case.
• The thickness of the shell, which is an infinitesimal change in x, denoted as \( dx \).

The formula for the volume using the shell method is:
\[V = 2\pi \int_{a}^{b} x (\text{height}) \, dx\]
In our example, revolving around the y-axis meant evaluating the integral:
\[V = 2\pi \int_{0}^{1} x (x - x^2) \, dx\]
The result is \( \frac{\pi}{6} \) after simplifying and integrating the expression, neatly summing up the stack of all cylindrical shells!

Washer Method

The washer method is another integral calculus technique used to find volumes of solids of revolution. It's ideal for shapes with holes—or washers—formed by revolving a region about an axis. To illustrate how this works, let's consider washers around the y-axis.

In this context, the washer method stacks washers vertically to compose the entire solid. Here's the breakdown:

• Washers have a small thickness \( dy \) when integrated with respect to y.
• Each washer has an outer radius and an inner radius. For example, if you revolve from \( y = x \) to \( y = x^2 \), you find the radii as functions of y.
• The volume of each washer is computed with the formula \( \pi (R^2 - r^2) \) where \( R \) and \( r \) are the outer and inner radii respectively.

The integral that gives the volume of the solid resembles:
\[V = \pi \int_{a}^{b} (R^2 - r^2) \, dy\]
In the example provided, substituting appropriate values and integrating gives us another beautiful form of \( \frac{\pi}{6} \) for the revolute solid generated around the y-axis.

Integral Calculus

Integral calculus is an essential branch of mathematics dealing with areas and accumulation, such as calculating volumes like those above. It focuses on solving integrals, which fundamentally asks how quantities relate cumulatively across regions.

In the realms of solids of revolution, integral calculus becomes vital as it's used to compute the volumes of 3D objects. By integration, the continuous sum of infinitesimal elements creates a complete picture of volume. Here's a closer look:

• Definite integrals are used to calculate the precise area or volume between specified bounds \( a \) and \( b \).
• The fundamental theorem of calculus bridges derivatives and integrals, showing that integration is essentially the summing up of infinite instantaneous rates of change.
• In forms like the shell and washer method, integral calculus determines how slices—or infinitesimally thin elements—contribute to the size and shape of a solid.

The concepts of integration are frequently practiced with bounds that represent spatial limits. Whether you're rotating around an axis or stacking blocks, integral calculus paves the way for insight into complex shapes and forms. Unlocking the potential of these integrals gives tangible results, like volumes \( \frac{\pi}{6} \) and \( \frac{2\pi}{15} \). It's the calculus that turns theory into practical reality!