Problem 69 - Essential Calculus

Question

Given the region bounded by the graphs of \(y=\ln x, y=0,\) and \(x=e,\) find (a) the area of the region. (b) the volume of the solid generated by revolving the region about the \(x\) -axis. (c) the volume of the solid generated by revolving the region about the \(y\) -axis. (d) the centroid of the region.

Step-by-Step Solution

Answer

The area A=1, volume when revolved around the x-axis \( V_x = \frac{\pi}{2} (e-2)\), volume when revolved around the y-axis \( V_y = \pi \), and centroid (\(\bar{x}\), \(\bar{y}\)) is (\(\frac{3}{2}\), \(-\frac{1}{4}\)) respectively.

Step 1: Finding the area

The area can be calculated by integrating from 1 to \(e\) of \(y=\ln x\) dx. This is given by: \[A = \int_{1}^{e} \ln x \, dx \]

Step 2: Volume revolved around the x-axis

We can find this using the formula for rotation about the x-axis, known as disc method. Integrate from 1 to \(e\) of \(\pi y^2 dx\). This will produce:\[V_x = \int_{1}^{e} \pi (\ln x)^2 dx \]

Step 3: Volume revolved around the y-axis

We now find volume for rotation about y-axis using the shell method formula. We have to change the boundaries of integration to -∞ to 0 for y due to natural log and x will be e^(y). The volume is given by:\[ V_y = \int_{-\infty}^{0} 2 \pi xh \, dy = \int_{-\infty}^{0} 2 \pi (e^y) \cdot 1 \, dy\]

Step 4: Finding the Centroid

The x-coordinate of the centroid (\(\bar{x}\)) can be found by \( \bar{x} = \frac{1}{A} \int_{1}^{e} x \ln x \, dx \) , and the y-coordinate (\(\bar{y}\)) by \(\bar{y} = \frac{1}{A} \int_{1}^{e} \frac{1}{2} (\ln x)^2 \, dx \). Where A is the area we computed in step 1. The centroid of the region is then given by (\(\bar{x}\), \(\bar{y}\)).

Key Concepts

Integration

Integration is a fundamental concept in calculus used to find areas under curves, among many other applications. When you need to determine the area of a region bounded by a curve, integration is employed. For instance, in our original exercise, we need to find the area of the region under the curve \( y = \ln x \) from \( x = 1 \) to \( x = e \).

To find this area, the definite integral:

• Sets the limits of integration from \(1\) to \(e\).
• Takes the function \( \ln x \) and integrates it with respect to \( x \).
• This integration gives us the total area under the curve within these bounds.

Understanding integration is key, as it helps in quantifying the space that a function occupies over an interval.
Whether you're calculating simple areas or more complex shapes like volumes of solids, integration is your go-to tool.

Volume by Revolution

When a region is revolved around an axis, it forms a three-dimensional solid. Calculus helps us find the volume of such solids using methods like the disc and shell methods. In the given exercise, we find the volumes by revolving the region about the x-axis and the y-axis.

Revolution about the x-axis
For the x-axis, the disc method was used, where we calculate:

• The volume by integrating \( \pi(\ln x)^2 \) from \( x = 1 \) to \( x = e \).
• The integrand \( \pi (\ln x)^2 \) represents the area of a circle (disc) at each slice or point along the x-axis.
• This area is rotated along the axis to form a solid disc.

When integrated, it provides the total volume of the solid obtained by rotation.

Revolution about the y-axis
Here, the shell method is employed:

• Integrate \( 2 \pi (e^y) \) over \( y \) from \(-\infty\) to \(0\).
• The formula accounts for cylindrical shells formed by rotating vertical slices around the y-axis.
• The expression \( 2 \pi (e^y) \) represents the circumference of each shell times its height.

Using these methods, you can accurately calculate the volume of curved solids in calculus.

Centroid Calculation

The centroid is the geometric center or the "balance point" of a region. Finding the centroid of a two-dimensional shape requires both x and y coordinates, usually denoted as \((\bar{x}, \bar{y})\).

Finding \( \bar{x} \)

• The x-coordinate of the centroid \( \bar{x} \) can be found by the formula:\[ \bar{x} = \frac{1}{A} \int_{1}^{e} x \ln x \, dx \]
• This formula calculates the 'weighted' average x-position, where \(A\) is the total area of the region.

Finding \( \bar{y} \)

• The y-coordinate of the centroid \( \bar{y} \) is determined using:\[ \bar{y} = \frac{1}{A} \int_{1}^{e} \frac{1}{2} (\ln x)^2 \, dx \]
• This formula encapsulates the average height, taking into account the shape's distribution above the x-axis.

By calculating these integrals, you can find the centroid which balances the region on both the x and y axes.

Applications of Definite Integrals

Definite integrals have a vast range of applications in mathematics and the real world. They are primarily used to calculate total amounts, such as areas, volumes, centroids, and arc lengths.

In this exercise, definite integrals were instrumental in:

• Calculating the area under the curve \( y = \ln x \), representing the total quantity within specific limits.
• Finding volumes of solids of revolution, showcasing their use in engineering and physics to determine material swept by revolving a shape.
• Determining the centroid of a region, which aids in structural design, offering insights into load distribution.

Overall, definite integrals give us the power to compute precise measurements and predict outcomes based on diverse formulas tailored to specific situations. They bridge mathematical theories with practical applications seamlessly.