Problem 24 - Thomas Calculus in SI Units

Question

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the given curves about the given lines. \(y=x^{3}, \quad y=8, \quad x=0\) a. The \(y\) -axis b. The line \(x=3\) c. The line \(x=-2\) d. The \(x\) -axis e. The line \(y=8\) f. The line \(y=-1\)

Step-by-Step Solution

Answer

a) \( \frac{128\pi}{5} \), b) \( \frac{696\pi}{5} \), c) \( \frac{504\pi}{5} \), d) \( \frac{256\pi}{5} \), e) \( \frac{128\pi}{7} \), f) \( \frac{608\pi}{7} \).

Step 1: Understand the Problem

We need to find the volume of the solid generated by rotating the region bounded by the curves around different axes using the shell method. The shell method uses cylindrical shells to compute the volume of the solid generated.

Step 2: Describe the Region and Setup Integration

The region is bounded by the curve \(y = x^3\), the line \(y = 8\), and the line \(x = 0\). Thus, we consider the strip from \(x = 0\) to \(x = 2\), where \(y = x^3\) intersects \(y = 8\). Since \(y = x^3\), we find \(x = 2\), for \(y = 8\).

Step 3: Apply the Shell Method

The volume of a shell is calculated as \( V = \int_{a}^{b} 2\pi \cdot \text{(radius)} \cdot \text{(height)} \cdot dx \). The radius will depend on the line of revolution, and the height will be determined by the function and bounding lines.

Step 4a: Rotation Around the y-axis

For rotation around the \(y\)-axis, the radius is \(x\) and the height is \(8 - x^3\). The volume becomes \(V = \int_{0}^{2} 2\pi x (8 - x^3) \, dx\). Simplify and integrate: \(V = \int_{0}^{2} 2\pi (8x - x^4) \, dx\). This evaluates to \(V = \left[16\pi x^2 - \frac{2\pi}{5} x^5 \right]_{0}^{2} \), resulting in \(V = \frac{128\pi}{5}\).

Step 4b: Rotation Around the Line x=3

Now, the radius is \((3 - x)\) and the height remains \(8 - x^3\). The integral is \(V = \int_{0}^{2} 2\pi (3 - x)(8 - x^3) \, dx\). Expanding and integrating yields \(V = \frac{696\pi}{5}\).

Step 4c: Rotation Around the Line x=-2

The radius is \((x + 2)\), and we use the same height, \(8 - x^3\). Integrate \(V = \int_{0}^{2} 2\pi (x + 2)(8 - x^3) \, dx\). Solve to find \(V = \frac{504\pi}{5}\).

Step 4d: Rotation Around the x-axis

For rotation around the \(x\)-axis, we must express \(x\) in terms of \(y\) and adjust the integral. The radius is \(y\) and the height is \((8^{1/3} - y^{1/3})\). Integrate \(V = \int_{0}^{8} 2\pi y(8^{1/3} - y^{1/3}) \, dy\). Simplify and evaluate to \(V = \frac{256\pi}{5}\).

Step 4e: Rotation Around the Line y=8

Rotation around \(y = 8\) involves radius \(8 - y\) and height as the horizontal span \(y^{1/3} - 0\). Integrate \(V = \int_{0}^{8} 2\pi (8 - y) y^{1/3} \, dy\), resulting in the volume \(V = \frac{128\pi}{7}\).

Step 4f: Rotation Around the Line y=-1

Consider \(y + 1\) as the new radius with height \(y^{1/3}\). Set up the integral \(V = \int_{0}^{8} 2\pi (y + 1) y^{1/3} \, dy\), and evaluate it, resulting in \(V = \frac{608\pi}{7}\).

Key Concepts

Volume of Revolution

The volume of revolution is a fascinating concept in calculus that involves rotating a two-dimensional area around an axis to create a three-dimensional solid. Imagine taking a flat region bounded by curves and, by spinning it around a line, forming a physical object like a vase or a bell. In this exercise, our bounded region is outlined by the curve \( y = x^3 \), the line \( y = 8 \), and \( x = 0 \). We explore several axes of revolution, dissecting how each rotation affects our solid's shape and size.

This area, upon being revolved, transforms into a solid whose volume we can calculate using integration techniques. By the end of this exploration, you'll see how changes in rotation axes impact the resulting solid's characteristics, bringing abstract calculus concepts into tangible understanding.

Integration Techniques

Integration is a staple of calculus, providing the tools needed to determine areas, calculate growth, and, importantly, compute volumes like those of our three-dimensional solids. For volumes of revolution, integration lets us accumulate a series of measurements—infinitely many tiny steps—to reach the full size of the solid.

When applying integration to problems of rotation, we often have to express functions in terms of other variables. For example, if we revolve around the \( x \)-axis, it's useful to reframe \( x \) in terms of \( y \). Likewise, integrating requires understanding related techniques like expanding products for ease of computation.

• Understand the function's role as a height marker in our integral—specifically how the bounding lines influence this.
• Consider the rotational axis to define the radius in our integral, which might transform depending on position.

Mastering these techniques makes handling intricate volumes clear and systematic.

Cylindrical Shells

The shell method is a valuable technique for finding volumes of revolution, particularly useful when the solid has a radial symmetry around a line. It breaks down the volume into a collection of cylindrical shells, which are thin-walled tubes that together form the solid.

Imagine each shell as part of a nested family, where their rotations around the axis reflect the original region being rotated. The essence of the shell method revolves around the formula:

\[ V = \int_{a}^{b} 2\pi \cdot (\text{radius}) \cdot (\text{height}) \cdot dx \]

This formula identifies each shell's contribution with:

• Radius: The distance from the axis of rotation to the shell.
• Height: Determined by the function defining the region.
• Thickness: Defined by small increments \(dx\), allowing for precise integration.

The shell method's power lies in its straightforward application, ideal for scenarios where adjusting the axis of revolution allows for defining a natural thickness of the shells.

Calculus Problem Solving

Resolving calculus problems like this one demands both a strong grasp of fundamental concepts and the ability to apply them creatively. Calculus problem-solving involves breaking the problem into manageable parts, examining the geometry, and selecting the appropriate technique.

This task reflects on boundary identification, function comparison, and strategic choice between methods—shell versus disk, for instance. Each rotation axis presents a unique challenge, demanding close consideration of its effects on volume components. Setup requires ingenuity in capturing the problem's core geometry.

Solving such problems hones your skills and enriches your understanding of calculus applications in real-world scenarios.

• Clarity in identifying the defining elements of the problem's region, including endpoints and intersections.
• Precision in integrating to exact any volume change due to rotation.

Ultimately, mastering these techniques and approaches transforms complex calculus scenarios into navigable challenges, boosting proficiency in mathematical reasoning.