Problem 34 - Thomas Calculus in SI Units

Question

Find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so. The region in the first quadrant bounded by \(x=y-y^{3}, x=1\) and \(y=1\) about a. the \(x\) -axis b. the \(y\) -axis c. the line \(x=1\) d. the line \(y=1\)

Step-by-Step Solution

Answer

a. \( \frac{4\pi}{21} \); b. \( \frac{57\pi}{35} \); c. \( \frac{29\pi}{105} \); d. \( \frac{10\pi}{21} \).

Step 1: Identifying the Region

First, identify the region bounded by the curves given. We have the curve \( x = y - y^3 \) and the lines \( x = 1 \), \( y = 1 \) in the first quadrant. We need to solve for the intersection points. Solving \( x = 1 \) for \( y - y^3 = 1 \), we find that \( y = 0 \). The region is bounded from \( y = 0 \) to \( y = 1 \).

Step 2: Volume around the x-axis (Part a)

To find the volume around the \( x\)-axis, use the method of washers. The outer radius \( R(y) \) is \( y \) and the inner radius \( r(y) \) is \( y^3 \). The volume \( V \) is given by the integral: \[ V = \pi \int_{0}^{1} [(y^2) - (y^3)^2] \, dy = \pi \int_{0}^{1} [y^2 - y^6] \, dy \]Evaluating the integral gives: \[ V = \pi \left[ \frac{y^3}{3} - \frac{y^7}{7} \right]_0^1 = \pi \left( \frac{1}{3} - \frac{1}{7} \right) = \pi \cdot \frac{4}{21} = \frac{4\pi}{21} \]

Step 3: Volume around the y-axis (Part b)

For the volume around the \( y\)-axis, we use washers again. The outer radius is \( 1 \) and the inner radius is \( y - y^3 \). The volume is given by:\[ V = \pi \int_{0}^{1} [1^2 - (y - y^3)^2] \, dy \]Simplifying gives:\[ V = \pi \int_{0}^{1} [1 - y^2 + 2y^4 - y^6] \, dy \]Evaluating the integral:\[ V = \pi \left[ y - \frac{y^3}{3} + \frac{2y^5}{5} - \frac{y^7}{7} \right]_0^1 \]\[ V = \pi \left( 1 - \frac{1}{3} + \frac{2}{5} - \frac{1}{7} \right) = \pi \cdot \frac{171}{105} = \frac{171\pi}{105} = \frac{57\pi}{35} \]

Step 4: Volume around the line x = 1 (Part c)

For revolutions around \( x = 1 \), use the method of cylindrical shells. The height of the shell is \( y - y^3 \) and the radius is \( 1 - x \). Since the problem revolves around \( x = 1 \), change to \( y \) boundaries: \[ V = 2\pi \int_{0}^{1} (1 - (y - y^3))(y - y^3) \, dy \]This simplifies to:\[ V = 2\pi \int_{0}^{1} (1 - y + y^3)(y - y^3) \, dy \]Evaluating gives:\[ V = 2\pi \cdot \left[ \frac{1}{6} - \frac{1}{35} \right] = 2\pi \cdot \frac{29}{210} = \frac{29\pi}{105} \]

Step 5: Volume around the line y = 1 (Part d)

For revolutions around \( y = 1 \), we use washers with respect to \( y \). The outer radius is \( 1 - y^3 \) and the inner radius is \( 1 - y \). The volume is:\[ V = \pi \int_{0}^{1} [(1 - y^3)^2 - (1 - y)^2] \, dy \]Simplify and integrate:\[ V = \pi \int_{0}^{1} [1 - 2y^3 + y^6 - (1 - 2y + y^2)] \, dy \]\[ V = \pi \int_{0}^{1} [2y - y^2 - 2y^3 + y^3 + y^6] \, dy \]Further simplification and integration yield:\[ V = \pi \cdot \left[ \frac{2}{3} - \frac{1}{3} - \frac{1}{2} + \frac{1}{7} \right] = \pi \cdot \frac{10}{21} = \frac{10\pi}{21} \]

Key Concepts

Washer Method

The washer method is a popular technique used to find the volume of solids of revolution. It is especially useful when dealing with regions that are bounded by two different curves. This method involves slicing the solid perpendicular to the axis of rotation, resulting in a series of "washers," which are essentially disks with holes in the middle.

Here is how you can think about it:

• Visualize the solid being divided into thin cylindrical slices or washers.
• The outer edge of the slice forms one disk, and the inner edge forms a smaller, concentric disk.

To calculate the volume of each washer, you subtract the volume of the inner disk from the outer disk's volume. This is achieved by computing the area of a washer and then integrating over the desired interval. The volume formula typically used is:\[V = \pi \int_a^b [R(y)^2 - r(y)^2] \, dy\]where \(R(y)\) is the outer radius and \(r(y)\) is the inner radius. By integrating these slices along the entire interval, you find the total volume.

Cylindrical Shells

The method of cylindrical shells is another vital technique for finding volumes of solids of revolution. It is particularly useful when revolving a region around a vertical line, where using the washer method might be complex.

Imagine breaking down the solid into thin hollow cylinders (or "shells"), each parallel to the axis of rotation. This is what the method does:

• Visualize the solid composed of multiple cylindrical shells.
• The height of each shell is determined by the function.
• The radius of each shell represents the distance from the axis of rotation to the shell.

The volume of each cylindrical shell is found using the formula:\[V = 2\pi \int_a^b (x)h(x) \, dx\]where \(x\) is the radius of the shell, and \(h(x)\) is the height of the shell. Integrating over the entire interval gives the total volume of the solid.

Definite Integral

The definite integral is crucial in calculating volumes of revolved solids. It aids in accumulating the volume of infinitesimally thin slices to calculate the volume of the entire solid. In calculus, the definite integral helps sum up infinite small quantities to find the whole.

• The bounds \(a\) and \(b\) indicate where to start and end the integration process.
• The function inside the integral represents the shape or size of the slices being summed up.

In the context of volumes, using a definite integral allows for precisely computing how these slices, whether washers or shells, stack up to form the solid's total volume. This is particularly evident in problems where you integrate the difference of squared functions (washer method) or the product of linear functions (cylindrical shells) over a specified range.

Integration Techniques

Understanding integration techniques is essential for successfully calculating volumes of solids. A good grasp of the following techniques can greatly simplify the problem-solving process:

• Substitution Method: Often makes integration easier by simplifying the integrand using a substitution variable.
• Integration by Parts: Useful when dealing with products of functions, allowing to break them down into simpler parts.
• Partial Fractions: Decomposes a complicated rational function into simpler fractions that are easier to integrate.

By familiarizing yourself with these techniques, you can tackle even the trickiest volume problems more efficiently. Each technique is particularly powerful when applied properly, reducing the complexity of integral evaluations and ensuring accuracy in finding the volume of a solid.