Problem 8 - Calculus: Early Transcendentals

Question

Let \(S\) be the region of the xy-plane bounded above by the curve \(x^{3} y=64,\) below by the line \(y=1,\) on the left by the line \(x=2,\) and on the right by the line \(x=4 .\) Find the volume of the solid obtained by rotating \(S\) around: (a) the \(x\) -axis; (b) the line \(y=1\); (c) the y-axis; and (d) the line \(x=2\).

Step-by-Step Solution

Answer

(a) 242π, (b) 6.4π, (c) 36π, (d) 16π.

Step 1: Identify the region

The region is bounded by the curve \(x^3 y = 64\), the line \(y=1\), and the vertical lines \(x=2\) and \(x=4\). We can express the curve as \(y = \frac{64}{x^3}\). The region is below \(y = \frac{64}{x^3}\) and above \(y=1\) between \(x=2\) and \(x=4\).

Step 2: Set up the integral for rotation around the x-axis

For rotation around the x-axis, we use the disk method. The outer radius is \(R(x) = \frac{64}{x^3}\) and the inner radius is \(r(x) = 1\). The volume is given by:\[V = \pi \int_{2}^{4} \left(R^2(x) - r^2(x)\right) \, dx = \pi \int_{2}^{4} \left(\left(\frac{64}{x^3}\right)^2 - 1^2\right) \, dx\]

Step 3: Calculate the integral for part (a)

Evaluating the integral for the volume, we have:\[V = \pi \int_{2}^{4} \left(\left(\frac{64}{x^3}\right)^2 - 1\right) \, dx \]Simplify to:\[V = \pi \int_{2}^{4} \left(\frac{4096}{x^6} - 1\right) \, dx\]This becomes:\[= \pi \left[ \frac{-8192}{x^5} - x \right]_{2}^{4}\]Calculating the definite integral, substitute the limits:\[= \pi \left( \left(\frac{-8192}{4^5} - 4\right) - \left(\frac{-8192}{2^5} - 2\right) \right)\]\[= \pi \left( -\frac{8192}{1024} - 4 + \frac{8192}{32} - 2 \right)\]\[= \pi \left( -8 - 4 + 256 - 2 \right)\]\[= \pi \cdot 242\]\[= 242\pi\]

Step 4: Set up the integral for rotation around the line y=1

For rotation around the line \(y=1\), the disk method changes slightly. Here, the outer and inner radii are switched, the solid is just a cylinder of height 3 (\([\frac{64}{x^3} - 1] - 0\)) but from the same bounds of \(x=2\) to \(x=4\). Calculate as:\[V = \pi \int_{2}^{4} \left( \left( \frac{64}{x^3} - 1 \right)^2 - 0^2 \right) \, dx\]

Step 5: Calculate the integral for part (b)

Simplify and calculate:\[V = \pi \int_{2}^{4} \left( \left( \frac{64}{x^3} - 1 \right)^2 \right) \, dx \ = \pi \cdot \int_{2}^{4} (4096x^{-6} - \frac{128}{x^3} + 1) \, dx\]\[= \pi \cdot \left[ \frac{-8192}{5x^5} + \frac{64}{x^2} + x \right]_{2}^{4}\]Compute with limits:\[ = \pi \left[ (\frac{-8192}{5 \times 1024} + \frac{64}{16} + 4) - (\frac{-8192}{5 \times 32} + \frac{64}{4} + 2) \right]\]\[= \pi \left( -\frac{8192}{5120} + 4 + 4 - (-\frac{8192}{160} + 16 + 2) \right)\]\[= \pi \left( -1.6 + 4 + 4 - (-51.2 + 16 + 2) \right)\]\[= 6.4\pi\] (Switch step titles would be needed, this could be a challenge.)

Step 6: Set up the integral for rotation around the y-axis

For rotating around the y-axis, use the shell method. The radius of a shell is \(x\), and height is \(\frac{64}{x^3} - 1\). The volume of rotation is:\[V = 2 \pi \int_{2}^{4} x \left(\frac{64}{x^3} - 1\right) \, dx\]

Step 7: Calculate the integral for part (c)

Expand and simplify:\[V = 2\pi \int_{2}^{4} (64x^{-2} - x) \, dx \]Integrate:\[V = 2\pi \left[ -64x^{-1} - \frac{x^2}{2} \right]_{2}^{4}\]Substitute the bounds:\[= 2\pi \left( (-16 - 8) - (-32 - 2) \right)\]\[= 2\pi (18) = 36\pi\]

Step 8: Set up the integral for rotation around the line x=2

For rotation around \(x=2\), the shell radius becomes \(x - 2\), and height \(\frac{64}{x^3} - 1\). The volume integral is:\[V = 2 \pi \int_{2}^{4} (x-2) \left(\frac{64}{x^3} - 1\right) \, dx\]

Step 9: Calculate the integral for part (d)

This results in:\[V = 2\pi \int_{2}^{4} \left( (x-2)(\frac{64}{x^3} - 1) \right) \, dx\]Simplify:\[V = 2\pi \int_{2}^{4} \left( \frac{64x}{x^3} - 64x^{-3} - x + 2\right) \, dx \]Integrate:\[V = 2\pi \left[ x^{-1} - \frac{64}{2}x^{-2} - \frac{x^2}{2} + 2x \right]_{2}^{4}\]\[= 2\pi \left( \left(\frac{128}{3} + 8 - [16] \right) \right)\]\[= 2\pi (8) = 16\pi\]

Step 10: Conclusion

The calculated volumes for each part of the problem are: (a) \(242\pi\), (b) \(6.4\pi\), (c) \(36\pi\), and (d) \(16\pi\).

Key Concepts

Disk Method

The Disk Method is a straightforward way to determine the volume of a solid of revolution around an axis. Essentially, imagine slicing the solid into numerous thin, disk-shaped slices, and then summing the volume of each disk. Each disk has a thickness, \(dx\), which represents a small change along the x-axis when rotating around the x-axis.
- **Formula**: The volume of one disk is \( ext{Volume} = \pi \cdot R^2(x) \, dx\), where \(R(x)\) is the outer radius of the disk. If there's an inner radius \(r(x)\), the volume of the washer (disk with a hole) is given by: \[V = \pi \int_{a}^{b} \left(R^2(x) - r^2(x)\right) \, dx\] - **Application**: This method is ideal for functions like \(y = f(x)\) when the region is revolved around the x-axis or a horizontal line that doesn’t intersect the region. The region in this problem, bounded by \(x^3 y = 64\) and other lines, can be treated using this approach for rotation around the x-axis.

Shell Method

When the Disk Method becomes cumbersome, particularly when revolving around the y-axis or a vertical line, consider using the Shell Method. Instead of disks, this method involves imagining cylindrical shells that make up the solid. It’s convenient for functions \(y = f(x)\) rotated around the y-axis.
- **Structure**: Picture the solid as comprising concentric cylindrical shells with a radius, \(x\). The height corresponds to the function value minus any lower bound like \(y = 1\). The thickness is \(dx\).- **Volume Formula**: Calculate the volume of each shell by: \[V = 2 \pi \int_{a}^{b} x \cdot (f(x) - g(x)) \, dx\] - **Contextual Use**: This method was applied to find the volume when rotating the bounded region around the y-axis. Notably, the Shell Method handled rotations around vertical lines effectively within our specified interval between \(x = 2\) and \(x = 4\).

Definite Integral

The Definite Integral plays a massive role in solving real-world volume problems by summing infinitely many tiny quantities—like the slices in a solid. It expresses the integral of a function over a closed interval. This process allows us to calculate the total accumulated quantity, such as volume, between two bounds.
- **Understanding the Concept**: - The notation \(\int_{a}^{b} f(x) \, dx\) represents the area between the function and the x-axis from \(x = a\) to \(x = b\). - Each application in this exercise required setting up integrals with specific limits depending on the region's bounds.- **Practical Application**: In our solutions, definite integrals were used creatively to calculate the volumes of three-dimensional objects formed by revolving two-dimensional regions. They were pivotal in integrating the different methods (Disk and Shell)—each requiring different but related setups.

Bounded Regions

Defining bounded regions is your starting point in solving volume-related problems, as they outline the area being revolved. A clear understanding of edges and intersections ensures accurate application of methods like the Disk and Shell Methods.
- **Definition & Identification**: - Bounded regions are defined by one or more curves, lines, or axes enclosing a space. - Identify the upper/lower bounds and where relevant lines intersect a curve.- **Importance in Calculation**: - For this exercise, the region is enclosed between \(x^3 y = 64\), \(y = 1\), and lines \(x = 2\) and \(x = 4\). These boundaries played a critical role in determining the limits of integration for our volume formulas. - Understanding the bounded region ensures you apply the right method and aligns your integration bounds with the physical constraints of the problem.