Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-12, find the volume of the solid generated when the region \(R\) bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region \(R\). (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. \(y=\frac{1}{x}, x=1, x=4, y=0 ;\) about the \(y\)-axis

Step-by-Step Solution

Answer

The volume of the solid is \( 6\pi \).

Step 1: Sketch the Region R

The region \( R \) is bounded by the curves \( y = \frac{1}{x} \), \( x = 1 \), \( x = 4 \), and \( y = 0 \). Since we're revolving this around the \( y \)-axis, we plot these on the xy-plane. The curve \( y = \frac{1}{x} \) is a hyperbola that decreases as \( x \) increases. The lines \( x = 1 \) and \( x = 4 \) are vertical lines which bound the region on the sides, and \( y = 0 \) (the x-axis) bounds it from below.

Step 2: Label a Typical Shell

A typical rectangular slice parallel to the y-axis in this region will extend vertically from \( y = 0 \) to \( y = \frac{1}{x} \) at a specific \( x \). When this slice is revolved around the y-axis, it forms a cylindrical shell. The height of the shell is \( y = \frac{1}{x} \) and its radius is \( x \). The thickness of the shell is \( \Delta x \).

Step 3: Write the Formula for Approximate Volume of the Shell

The volume of a thin cylindrical shell with radius \( x \), height \( \frac{1}{x} \), and thickness \( \Delta x \) is given by \( \Delta V = 2 \pi x \cdot \frac{1}{x} \cdot \Delta x = 2\pi \Delta x \).

Step 4: Set Up the Integral

To find the exact volume, we sum up these infinitesimally thin shells from \( x = 1 \) to \( x = 4 \). The integral to represent the total volume \( V \) is: \[V = \int_{1}^{4} 2 \pi \cdot 1 \cdot dx = 2 \pi \int_{1}^{4} 1 \cdot dx. \]

Step 5: Evaluate the Integral

Compute the integral: \[V = 2 \pi \left[ x \right]_{1}^{4} = 2 \pi (4 - 1) = 2 \pi \times 3 = 6 \pi.\]Thus, the volume of the solid is \( 6\pi \).

Key Concepts

Solid of Revolution

When we talk about a solid of revolution, we're referring to a 3D shape that is formed by rotating a 2D region around an axis. This axis can be either horizontal or vertical. In this particular problem, we consider the region defined by the curves \(y = \frac{1}{x}\), \(x = 1\), \(x = 4\), and \(y = 0\). Revolving this region around the \(y\)-axis creates a solid shape that sheaths the axis.

• Visualize this as a series of spun rectangles that create a tube-like form.
• Key features include understanding how each slice contributes to the overall shape.

The notion of solid of revolution is crucial because it allows us to calculate the volume of complex shapes using simpler geometric rotation concepts.

Cylindrical Shells Method

The cylindrical shells method is one technique to find the volume of a solid of revolution. It is especially useful when the solid is revolved around the y-axis. In this method, we imagine cutting the solid into thin, hollow cylinders—essentially like the wrapping of a shell around the axis.

• A typical cylindrical shell has a radius equal to the distance from a slice to the axis of rotation.
• The height of the shell is the value of the function at that slice.
• The thickness is a small width, \(\Delta x\), in this case.

Here, to calculate the shell's volume, we use the formula: \[\Delta V = 2\pi \times \text{radius} \times \text{height} \times \text{thickness} = 2\pi x \cdot \frac{1}{x} \cdot \Delta x.\]The cylindrical shells method is particularly advantageous when dealing with regions bounded vertically.

Integral Calculus

Integral calculus is the mathematical tool we use to sum up infinitesimally small quantities to find a whole, such as the volume of a solid. In our exercise, integral calculus helps determine the exact volume by integrating these small cylindrical shells over the interval of interest—from \(x = 1\) to \(x = 4\). This technique involves an integral: \[V = \int_{1}^{4} 2 \pi \cdot 1 \cdot dx.\] Integrating essentially means accumulating these small slices into one continuous sum. This integral encapsulates all tiny shells to compute the total volume of the solid. Integral calculus makes problems involving continuous volumes, areas, and lengths manageable by summing tiny parts instead of calculating them individually.

Definite Integrals

Definite integrals are a special type of integral used to find the accumulated quantity between specific bounds. They give us a precise numeric value rather than a general formula. In the task at hand, we use a definite integral to calculate the exact volume of the solid generated by the region \(R\) when rotated around the \(y\)-axis.

• The limits of integration are provided by the domain of the region, from \(x = 1\) to \(x = 4\).
• This integration adds up the contributions of all infinitesimal cylindrical shells from the start to the end of this region.

The solution to the definite integral: \[V = 2 \pi \left[ x \right]_{1}^{4} = 6 \pi,\] provides the total volume of the solid of revolution. Understanding definite integrals is crucial as they convert a continuous sum into a finite number.