Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 ; \text { about the } y \text { -axis }

Step-by-Step Solution

Answer

The volume is \(6\pi\).

Step 1: Sketch the Region R

The region \( R \) is bounded by the curve \( y = \frac{1}{x} \), the lines \( x = 1 \), \( x = 4 \), and \( y = 0 \). To sketch this region, plot the hyperbola \( y = \frac{1}{x} \), which is located in the first quadrant, and draw the vertical lines from \( x = 1 \) to \( x = 4 \). The area under the curve from \( x = 1 \) to \( x = 4 \) above the x-axis represents region \( R \).

Step 2: Show a Typical Rectangular Slice

For a typical slice of width \( \Delta x \) at position \( x \), draw a vertical rectangle that extends from \( y = 0 \) up to \( y = \frac{1}{x} \). This rectangle is perpendicular to the x-axis and will form a cylindrical shell when revolved around the y-axis. Label the height of the rectangle as \( \Delta x \) and its height as \( \frac{1}{x} \).

Step 3: Write the Formula for the Shell's Volume

The formula for the volume of a cylindrical shell is \( 2 \pi \times \text{radius} \times \text{height} \times \text{thickness} \). The radius is the distance from the y-axis, which is \( x \). The height of the shell is \( \frac{1}{x} \), and the thickness is \( \Delta x \). Therefore, the volume of an approximate shell is \( 2\pi x \cdot \frac{1}{x} \cdot \Delta x = 2\pi \Delta x \).

Step 4: Set up the Integral for Total Volume

To find the total volume, integrate the formula for the volume of the shell from \( x = 1 \) to \( x = 4 \). Set up the integral as follows:\[ V = \int_{1}^{4} 2\pi \cdot 1 \cdot dx = 2\pi \int_{1}^{4} 1\, dx. \]

Step 5: Evaluate the Integral

Evaluate the integral to find the total volume. Compute: \[ V = 2\pi \left[ x \right]_{1}^{4} = 2\pi (4 - 1) = 2\pi \times 3 = 6\pi. \]

Key Concepts

Cylindrical Shells Method

The cylindrical shells method is a useful technique in calculus for finding the volume of a solid of revolution. Imagine that you're revolving a region around an axis. The resulting solid can be difficult to handle using disks or washers. This is where cylindrical shells come into play.
A cylindrical shell is like a hollow tube. Picture taking a vertical strip from your region, revolving it around the axis, and obtaining a thin cylindrical shape. The volume of each shell is given by the formula:

• Radius (r) is the distance from the axis to the strip.
• Height (h) matches the length of the strip.
• Thickness is a small change in the variable, usually written as (Δx).

The volume of each shell becomes: \(2 \pi \times \text{radius} \times \text{height} \times \text{thickness}\).We integrate these shells over the desired interval to find the total volume of the solid.

Integral Calculus

Integral calculus provides tools to find accumulations like area, volume, and other quantities. By integrating, we sum infinitesimally small parts to achieve the total quantity. For the cylindrical shells method, we rely heavily on integration.
Once you identify the volume formula for a single shell, integral calculus lets you sum all the shells together. This is done using definite integrals, where the lower and upper bounds determine the strip limits.
In our example, the integral is: \[\int_{1}^{4} 2\pi \, dx.\]

• The integral symbol shows that you are summing.
• The limits (1 to 4) signify the region along the x-axis you're considering.

Solving this gives you the total volume of the solid formed by revolving the region bounded by your function.

Revolving Regions Around an Axis

Revolving regions around an axis refers to creating 3D objects by spinning a 2D area along a line. This transformation pulls the flat shape out into a solid.
In these exercises, the region is bounded by curves, lines, or both. When the region rotates around an axis such as the y-axis, we use a technique like cylindrical shells or disks/ washers.
Consider the region bounded by the curve \(y = \frac{1}{x}\), forming a hyperbola. By revolving it around the y-axis, the region's stretch transforms into a volume.
This approach requires identifying the axis and using appropriate concepts like shell radius, height, and thickness to set up the integral that represents the desired volume.

Hyperbola Curve Integration

Hyperbola curve integration involves finding the area, volume, or other properties related to the hyperbola equation. Here, we're dealing with the curve \(y = \frac{1}{x}\).
This hyperbola opens toward the y-axis, meaning its shape spreads horizontally. When integrating this curve between the x-limits (from 1 to 4 in our exercise), you are essentially capturing the essence of how the curve behaves.
Finding the volume from this curve involves integrating it as part of a solids of revolution problem.

• This curve, when rotated, becomes vital in determining the corresponding shell height.
• Having the explicit formula \(y = \frac{1}{x}\) helps set up the integral to calculate the entire solid's volume.

Each piece of the integral relates closely to how the curve interacts with the region and axis.