Problem 56 - Calculus

Question

Use shells to find the volumes in \(55-63 .\) The rotated regions lei between the curve and \(x\) axis. $$ y=1 / x, 1 \leqslant x \leqslant 100 \text { (around the } y \text { axis) } $$

Step-by-Step Solution

Answer

The volume is \( 198\pi \).

Step 1: Identify the region and the axis of rotation.

The given curve is \( y = \frac{1}{x} \), and the region is bounded between \( 1 \leq x \leq 100 \), rotating around the \( y \)-axis. We will use cylindrical shells to find the volume.

Step 2: Write the formula for volume using cylindrical shells.

The formula for the volume of a solid of revolution using the method of cylindrical shells is: \[ V = 2\pi \int_{a}^{b} x \, f(x) \, dx \] where \( f(x) \) is the function representing the curve.

Step 3: Set up the integral for the given problem.

Substitute \( f(x) = \frac{1}{x} \), and bounds \( a = 1 \), \( b = 100 \) into the volume formula:\[ V = 2\pi \int_{1}^{100} x \cdot \frac{1}{x} \, dx \] Simplifying, we get:\[ V = 2\pi \int_{1}^{100} 1 \, dx \]

Step 4: Integrate the function.

The integral \( \int 1 \, dx \) is straightforward. It evaluates to \( x \): \[ V = 2\pi [x]_{1}^{100} \] This means we take \( x \) evaluated at 100 minus \( x \) evaluated at 1.

Step 5: Evaluate the definite integral.

Substitute the bounds into the integrated function: \[ V = 2\pi (100 - 1) \] This simplifies to:\[ V = 2\pi \times 99 \] Calculate to find the volume.

Step 6: Calculate the final volume.

Perform the final calculation: \[ V = 198\pi \] Thus, the volume of the solid is \( 198\pi \).

Key Concepts

Volume of Solid of Revolution

The concept of finding the volume of a solid of revolution is fundamentally about determining the space a shape occupies when it revolves around an axis. In the context of using cylindrical shells, we aim to revolve a region bounded by a curve around an axis—typically the x or y-axis—and calculate its resulting volume. For our problem, we are rotating the region under the curve \( y = \frac{1}{x} \), from \( x = 1 \) to \( x = 100 \), around the y-axis."
When applying the cylindrical shells method, envision the solid being formed as countless infinitesimally thin cylindrical shells, each contributing a small ring-shaped volume to the total. By summing up these rings from an integral perspective, we can capture the entire volume of the solid. This technique helps us appreciate the geometric transformation from a flat 2D region to a voluminous 3D object.

Definite Integral

The definite integral plays a crucial role in calculating volumes of solids of revolution. It helps us accumulate tiny volumes in a specified interval. In our example, the integral spans from \( x = 1 \) to \( x = 100 \). By using a definite integral, we efficiently determine the volume of the rotated region."
To apply it in this method, we incorporate the function representing the curve into the formula for cylindrical shells:

• The formula for volume: \[ V = 2\pi \int_{a}^{b} x \, f(x) \, dx \]
• Here, \( f(x) = \frac{1}{x} \) is our curve's equation.
• The bounds \( a = 1 \) and \( b = 100 \) define where our integral will accumulate volume from the beginning to the end of the specified interval.

This definite integral provides a systematic way to add up the shell volumes from the smallest x-value to the largest, resulting in an accurate measure of the total volume.

Curve Rotation

Curve rotation, also known as revolving a region, is about turning a 2D shape around a line to form a 3D object. For our task with the curve \( y = \frac{1}{x} \), this line—called the axis of rotation—is the y-axis. This entire region under the curve flips around the axis, creating a solid figure."
Visualize this process like spinning a potter's wheel: the shape stretches out from the axis, forming the solid's circular profile. Utilizing this visualization aids in understanding how the 2D shape expands and how every point on the curve generates a part of this three-dimensional body."
When doing such problems, always:

• Identify the curve and region's borders you need to rotate.
• Determine the axis of rotation to know how the curve moves.
• Visualize or sketch the resulting 3D form for better comprehension.

With this mindset, solving curve rotation problems becomes clearer and more methodical.

Integral Bounds Evaluation

A critical step in problems involving definite integrals is evaluating the integral bounds—this determines where the accumulation of volume starts and ends. For our specific example, these bounds are \( x = 1 \) and \( x = 100 \). After setting up our integral, we substitute these bounds to find the definite contribution to volume from each segment."
Once we integrate our function, we need to substitute and compute the differences:

• The integral of \( 1 \) with respect to \( x \) results in \( x \).
• Evaluating at the bounds, we have: \[ x \bigg|_1^{100} = 100 - 1 \] leads to a result of \( 99 \).
• Multiply this by \( 2\pi \) to calculate the final volume, resulting in \( 198\pi \).

This systematic evaluation ensures that we correctly account for the entire span from the initial point to the terminal point along the x-axis, providing a precise calculation of the volume of the solid formed.