Problem 22 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\) -axis. \begin{equation} y=x-x^{2}, \quad y=0 \end{equation}

Step-by-Step Solution

Answer

The volume is \( \frac{\pi}{30} \).

Step 1: Identify the region of interest

The region bounded by the curves can be found by setting the functions equal to each other. Since we have the curve \( y = x - x^2 \) and the line \( y = 0 \), the region of interest is the area between the curve and the line from the points where \( x - x^2 = 0 \).

Step 2: Find the points of intersection

To find the intersection points, set the equation \( y = x - x^2 = 0 \). Factoring gives:\[ x(1 - x) = 0 \]This gives two critical points: \( x = 0 \) and \( x = 1 \).

Step 3: Set up the volume integral

The volume of the solid of revolution around the x-axis can be calculated using the disk method:\[ V = \, \pi \int_{a}^{b} [f(x)]^2 \, dx \]Here, \( a = 0 \), \( b = 1 \), and \( f(x) = x - x^2 \).

Step 4: Integrate to find the volume

Calculate the integral:\[ V = \, \pi \int_{0}^{1} (x - x^2)^2 \, dx \]Expand the integrand:\[ (x - x^2)^2 = x^2 - 2x^3 + x^4 \]Thus, the integral becomes:\[ \pi \int_{0}^{1} (x^2 - 2x^3 + x^4) \, dx \]Compute the antiderivative:\[ = \pi \left[ \frac{x^3}{3} - \frac{2x^4}{4} + \frac{x^5}{5} \right]_{0}^{1} \]\[ = \pi \left[ \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right] \]

Step 5: Simplify the definite integral

Evaluate the expression:\[ = \pi \left( \frac{10}{30} - \frac{15}{30} + \frac{6}{30} \right) = \pi \left( \frac{1}{30} \right) = \frac{\pi}{30} \]

Step 6: Conclusion

The volume of the solid generated by revolving the region bounded by the curves \( y = x-x^2 \) and \( y=0 \) about the x-axis is \( \frac{\pi}{30} \).

Key Concepts

Disk Method

The Disk Method is a technique used to find the volume of a solid of revolution. This solid results from rotating a particular region about an axis. In our exercise, we are revolving the region bounded by the curve \( y = x - x^2 \) and the line \( y = 0 \) about the x-axis.

This involves considering the region between these two curves. Picture it in your mind: If you slice the solid perpendicular to the x-axis, each slice will look like a disk (hence the name). The radius of each disk corresponds to the value of the function \( f(x) \) at that slice. Hence, the formula:

• Volume of each disk: \( \pi \times (\text{radius})^2 \)
• Total volume: \( \pi \int_{a}^{b} [f(x)]^2 \, dx \)

In our scenario, \( a = 0 \) and \( b = 1 \), with \( f(x) = x - x^2 \). This method transforms a potentially complex geometry problem into a matter of calculating a definite integral. Once you've nailed down \( f(x) \) and the bounds, integrating gives you the complete volume.

Definite Integral

The concept of a definite integral is fundamental in calculus, particularly when finding areas and volumes. Within the realm of volumes of solids of revolution, definite integrals allow us to calculate the exact volume of such solids over a specified interval.

In this exercise, we particularly look for the volume of the object formed by revolving the function \( y = x - x^2 \) around the x-axis, bounded between \( x = 0 \) and \( x = 1 \). The definite integral is set up as follows:

\[ V = \pi \int_{0}^{1} (x - x^2)^2 \, dx \]
This expression involves squaring the function, summing the infinitely small areas (which become disks when rotated), and scaling by \( \pi \) to respect circular areas, over the interval from 0 to 1.

The evaluation of definite integrals requires us to find an antiderivative, substitute the bounds, and then calculate the resulting expression. This ensures that we accurately calculate the volume that we seek, encapsulated perfectly by the integral's bounds and integrand.

Antiderivatives

Antiderivatives play a crucial role in solving integrals for problems like finding volumes. Essentially, an antiderivative reverses the differentiation process, helping us solve for definite integrals, as seen in the exercise.

In our given problem, the main integral to solve is:

\[ \int (x^2 - 2x^3 + x^4) \, dx \]
To compute this, we need to find the antiderivatives of each term individually:

• The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \)
• The antiderivative of \( -2x^3 \) is \( -\frac{2x^4}{4} \) or equivalently \( -\frac{x^4}{2} \)
• The antiderivative of \( x^4 \) is \( \frac{x^5}{5} \)

Once you've found the antiderivatives, you evaluate them at the bounds of the integral, in this case from 0 to 1. This helps you find the definite integral and thus the volume we are interested in. Mastering antiderivatives makes solving these types of integration problems much more approachable.