Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-10, sketch the region \(R\) bounded by the graphs of the given equations, and show a typical vertical slice. Then find the volume of the solid generated by revolving \(R\) about the \(x\)-axis. $$ y=x^{2 / 3}, y=0, \text { between } x=1 \text { and } x=27 $$

Step-by-Step Solution

Answer

The volume is \( \frac{6558\pi}{7} \)."

Step 1: Understand the Problem

We are given the functions \( y = x^{2/3} \) and \( y = 0 \) with the interval \( x = 1 \) to \( x = 27 \). The region \( R \) is the area under the curve \( y = x^{2/3} \) from \( x = 1 \) to \( x = 27 \). Our goal is to find the volume of the solid formed by revolving this region around the \( x \)-axis.

Step 2: Sketch the Region

Draw the curve \( y = x^{2/3} \), which resembles the cubic root function but squared, starting from \( x = 1 \) to \( x = 27 \). The line \( y = 0 \) corresponds to the x-axis. The region \( R \) is the area between this curve and the x-axis from \( x = 1 \) to \( x = 27 \).

Step 3: Set up the Integral for the Volume

To find the volume of the solid when the region \( R \) is revolved around the x-axis, we use the disk method. The volume \( V \) is given by the integral of \( \pi [f(x)]^2 \) over the interval. Therefore, \[ V = \pi \int_{1}^{27} [x^{2/3}]^2 \, dx. \]

Step 4: Simplify the Function

Within the integrand \( [x^{2/3}]^2 \), simplify it to get \( x^{4/3} \). The integral now becomes \[ V = \pi \int_{1}^{27} x^{4/3} \, dx. \]

Step 5: Integrate the Function

Apply the power rule for integration \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). For the integral of \( x^{4/3} \), it becomes \[ \frac{x^{4/3 + 1}}{4/3 + 1} = \frac{x^{7/3}}{7/3} = \frac{3}{7}x^{7/3}. \]

Step 6: Evaluate the Definite Integral

Compute \[ \pi \cdot \left[ \frac{3}{7}x^{7/3} \right]_{1}^{27} = \pi \cdot \left( \frac{3}{7} \cdot 27^{7/3} - \frac{3}{7} \cdot 1^{7/3} \right). \]

Step 7: Calculate the Final Volume

First, compute \( 27^{7/3} = (27^{1/3})^7 = 3^7 = 2187 \). Thus the evaluation becomes \[ \pi \cdot \left( \frac{3}{7} \cdot 2187 - \frac{3}{7} \cdot 1 \right) = \pi \cdot \left( \frac{6561}{7} - \frac{3}{7} \right) = \pi \cdot \frac{6558}{7}. \] The final volume is \( \frac{6558\pi}{7} \).

Key Concepts

Disk Method

The Disk Method is a technique used in calculus to find the volume of a solid of revolution. When you revolve a region around an axis, it creates a three-dimensional shape. With the disk method, this shape consists of many tiny disks stacked together. Each of these disks has a small height and a circular face.

To use this method, follow these steps:

• Identify the function or functions that bound the region.
• Determine which axis you are rotating around, which for this problem is the x-axis.
• Set up the integral using the formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where [f(x)] represents the distance from the curve to the axis of rotation.

This approach works by approximating the region into thin slices, each contributing a volume of a disk with radius determined by the function value and thickness given by a tiny change in x.

Definite Integral

In calculus, the definite integral is a fundamental concept used for computing accumulated quantities, such as areas under curves or volumes of solids of revolution. In this exercise, it plays a pivotal role in determining the volume of the solid obtained from the rotation.

The definite integral from a to b of a function f(x) is expressed as:

• \[ \int_{a}^{b} f(x) \, dx \]
• This represents the accumulation of values that f(x) takes between x = a and x = b.
• The result of a definite integral in our context is a specific numerical value representing total volume.

In our formula for the volume, \(\pi \int_{1}^{27} [x^{4/3}] \, dx\), each tiny part of this integral represents a disk's contribution to the entire volume.

Solid of Revolution

A solid of revolution is a three-dimensional object formed by rotating a two-dimensional area around an axis. Think of it as creating a shape by spinning a flat region, like drawing a circle by twirling a line.

Here's how you can visualize the formation:

• Identify the region to revolve, which in this problem is the area under the curve from x = 1 to x = 27.
• Imagine this flat region rotated around the x-axis.
• The resulting solid will have circular cross-sections perpendicular to the x-axis.

Understanding solids of revolution helps it to link geometric visualization with calculus, enabling the use of techniques like the disk method to find volumes.

Calculus Problem Solving

Calculus problem solving is about finding systematic methods to tackle complex mathematical problems, often involving dynamic changes. Each step involves precision and logical structuring of the problem.

The process for this exercise involves:

• Recognizing the problem: Identify the boundaries and the shape of the region involved.
• Using visualization: Sketching the region helps to conceptualize the problem.
• Breaking down the problem: Start from sketching, move to setting up integrals, and apply methods like the disk method.
• Performing calculations: Simplify and solve integrals to find specific numeric results.

Through each of these steps, the abstraction of calculus turns into tangible solutions, allowing transitions from two-dimensional shapes to real-world volumes efficiently.