Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the solid generated by revolving about the \(x\) -axis the region bounded by the line \(y=6 x\) and the parabola \(y=6 x^{2}\).

Step-by-Step Solution

Answer

The volume is \(1.2\pi\) cubic units.

Step 1: Identify the Intersection Points

Find the intersection of the line \(y = 6x\) and the parabola \(y = 6x^2\). Set \(6x = 6x^2\) and solve for \(x\).\[6x = 6x^2\]\[x(6 - 6x) = 0\] gives two solutions, \(x = 0\) and \(x = 1\). The curves intersect at the points (0,0) and (1,6).

Step 2: Set Up the Volume Integral

Use the disk method to find the volume of the solid. The radius of a disk is \(R = 6x - 6x^2\) (the line minus the parabola). The volume of each disk is \(\pi R^2\), thus the volume of the solid is given by the integral: \[V = \int_{0}^{1} \pi (6x - 6x^2)^2\, dx\].

Step 3: Simplify the Expression Under the Integral

Expand and simplify the expression \((6x - 6x^2)^2\) to integrate it easily.\[(6x - 6x^2)^2 = (36x^2 - 72x^3 + 36x^4)\].

Step 4: Integrate Each Term

Calculate the integral\(\int_{0}^{1} \pi (36x^2 - 72x^3 + 36x^4)\, dx\). Integrate each term separately.:\[\pi \int_{0}^{1} (36x^2)\, dx = \pi [12x^3]_{0}^{1} = 12\pi\],\[\pi \int_{0}^{1} (-72x^3)\, dx = \pi [-18x^4]_{0}^{1} = -18\pi\],\[\pi \int_{0}^{1} (36x^4)\, dx = \pi [7.2x^5]_{0}^{1} = 7.2\pi\].

Step 5: Compute the Volume

Add the results to find the total volume.\( V = 12\pi - 18\pi + 7.2\pi = 1.2\pi \).

Step 6: Conclusion

The volume of the solid generated by revolving the region enclosed by the line and the parabola around the \(x\)-axis is \(1.2\pi\) cubic units.

Key Concepts

Disk Method

The disk method is a powerful technique used in calculus to find the volume of a solid of revolution. It is particularly handy when a region is revolved around an axis. This method visualizes the solid as a stack of thin, flat disks or washers, piled up from one end of the solid to the other.

When using the disk method, each disk has a certain radius, which is dependent on the function defining one side of the region being revolved. In this exercise, the radius is the difference between the line and the parabola:

• The outer radius: determined by the line function, here it is: \(6x\).
• The inner radius: determined by the parabola function, here it is: \(6x^2\).

The area of each disk is then given by \(\pi R^2\). Therefore, the volume of one disk is \(\pi (6x - 6x^2)^2\), and integrating this expression over the interval from \(x = 0\) to \(x = 1\) gives the total volume of the solid.

Volume Integration

Volume integration involves setting up and evaluating an integral to find the volume of a solid. With the disk method, we integrate the area of a disk across a certain interval, accounting for all disks from start to end in the solid.

The integral set up for the given problem is: \[ V = \int_{0}^{1} \pi (6x - 6x^2)^2\, dx \]To perform the integration, we first expand the squared expression

• Expand \((6x - 6x^2)^2\) to get \(36x^2 - 72x^3 + 36x^4\)
• Integrate each term individually within the limits from \(0\) to \(1\)
• The result for each integral accounts for the contribution to the volume from each slice of the solid

Once these integrals are computed, their sum gives the total volume of the solid of revolution.

Intersection of Curves

Finding where two curves intersect is crucial in defining the bounds of the region for volume calculations. In this problem, the intersection of the line \(y = 6x\) and the parabola \(y = 6x^2\) determines the limits for integration.

To find these points:

• Set the equations equal: \(6x = 6x^2\)
• Solve for \(x\) by factoring: \(x(6 - 6x) = 0\)
• This results in \(x = 0\) and \(x = 1\), indicating intersections at \((0,0)\) and \((1,6)\)

These intersection points effectively mark the boundary of the area to be revolved. They are crucial for accurately setting up the integration limits, ensuring that the entire region is included in the volume calculation.

Calculus Problem Solving

Tackling calculus problems, particularly those involving volumes of solids of revolution, follows a systematic approach. It begins with identifying the relevant functions and finding their intersecting points. From there, setting up the integral and employing the correct method, such as the disk method, becomes straightforward.

When solving these problems:

• Comprehend the problem and visualize the region to be revolved.
• Determine which axis the rotation occurs around to select the correct volume method.
• Use the intersection points to define the limits of integration.
• Simplify and expand expressions as necessary to make integration manageable.
• Integrate carefully, and calculate the sum of all parts to get the final volume.

Though involved, this structured approach simplifies complex calculus problems, making them more approachable with practice.