Problem 38 - Calculus Single Variable

Question

Find the center of mass of the given region \(\mathcal{R},\) assuming that it has uniform unit mass density. \(\mathcal{R}\) is the region bounded above by \(y=\ln (x),\) below by the \(x\) -axis, and on the sides by \(x=1\) and \(x=e\).

Step-by-Step Solution

Answer

The center of mass is \(\left(\frac{3e^2 - 1}{4(e-1)}, \frac{1}{2(e-1)}\right).\)

Step 1: Understanding the Problem

The region \(\mathcal{R}\) is bounded by the curve \(y = \ln(x)\), the \(x\)-axis (where \(y = 0\)), and vertical lines \(x=1\) and \(x=e\). We need to find the center of mass of this area assuming a uniform density, which simplifies as the problem is primarily about geometry.

Step 2: Setting Up the Center of Mass Coordinates

The center of mass (\((\overline{x}, \overline{y})\)) of a lamina with uniform density is given by \[ \overline{x} = \frac{1}{A} \int_{a}^{b} x(\ln(x)) \, dx \] and \[ \overline{y} = \frac{1}{A} \int_{a}^{b} \frac{1}{2}(\ln(x))^2 \, dx \], where \(A\) is the area of the region and the limits \(a\) and \(b\) are \(1\) and \(e\) respectively.

Step 3: Calculating the Area of\(\mathcal{R}\)

The area \(A\) under the curve \(y = \ln(x)\) from \(x = 1\) to \(x = e\) is given by \(A = \int_{1}^{e} \ln(x) \, dx\). Using integration by parts with \( u = \ln(x) \) and \( dv = dx \), we find that \( A = [x \ln(x) - x]_1^e = e - 1.\)

Step 4: Calculating \(\overline{x}\)

Using \(A\), calculate \(\overline{x} = \frac{1}{e-1} \int_{1}^{e} x \ln(x) \, dx\). Again, apply integration by parts where \(u = \ln(x)\) and \(dv = x \, dx\). This yields \((x^2/2) \ln(x) - x^2/4 \big]_{1}^{e}\). Evaluating, we have \( \overline{x} = \frac{1}{e-1} \left( \frac{e^2}{2} - \frac{e^2}{4} - \frac{1}{2} + \frac{1}{4} \right) = \frac{3e^2 - 1}{4(e-1)}.\)

Step 5: Calculating \(\overline{y}\)

For \(\overline{y}\), compute \( \overline{y} = \frac{1}{e-1} \int_{1}^{e} \frac{1}{2} (\ln(x))^2 \, dx\). This requires more integration by parts used on \(v = (1/2)(\ln(x))^2\). Eventually, solving gives \( \overline{y} = \frac{1}{2(e-1)} \left( (e(\ln(e)^2 - 2\ln(e) + 2)) - (1(-2 + 1)) \right) = \frac{1}{2(e-1)}.\)

Step 6: Final Center of Mass Coordinates

With both \(\overline{x}\) and \(\overline{y}\) computed, the center of mass is at \(\left(\frac{3e^2 - 1}{4(e-1)}, \frac{1}{2(e-1)}\right).\)

Key Concepts

Integration by Parts

Integration by parts is an essential method in calculus for integrating the product of two functions. It's based on the product rule for differentiation and allows us to transform the integral of a product into a more manageable form. The formula is given by:

• \( \int u \, dv = uv - \int v \, du \).

Integration by parts is useful when choice of functions allows the resulting integrals to be easily solved. For calculating geometrical properties such as the center of mass, this technique plays a crucial role. In our problem, it was applied to find the area under the curve and in evaluating complex expressions like the coordinates of the center of mass (\( \overline{x}, \overline{y} \)). By choosing suitable \( u \) and \( dv \) from the given function, we simplify the integration process.
It turns complicated integrals into simpler ones, which are more straightforward to solve.

Area Under a Curve

The area under a curve is a basic calculation in calculus, representing the integral of a function over an interval. For the curve \( y = \ln(x) \), this means determining how much space it covers between \( x=1 \) and \( x=e \).
The area can be found using the definite integral \( \int_{1}^{e} \ln(x) \, dx \).
This integral results in the total area beneath the curve and above the x-axis for the specified interval.
In our exercise, determining this area was pivotal for finding the center of mass. It's because the center of mass formulas divide by the area \( \( A \) \), ensuring that we're calculating the average position.
Calculating such areas allows us to work with volumes or even find probabilities in more advanced applications.

Uniform Density

The concept of uniform density simplifies many physics and geometry problems. It implies that the density is consistent across the entire region, leading to a straightforward calculation for center of mass. In our problem, the area \( \mathcal{R} \) was assumed to have uniform unit mass density, which simplifies calculations as the mass is directly proportional to the area.
When dealing with uniform density:

• Every part of the region has the same mass per unit area.
• We only need to calculate geometrical properties.

Thus, the complexity associated with varying densities is avoided. Uniform density ensures that the calculations focus on determining average positions instead of dealing with intricate mass distributions.

Natural Logarithm

The natural logarithm, represented by \( \ln(x) \), is a fundamental mathematical function. Its base is the constant \( e \), and it's inverse to the exponential function \( e^x \). In calculus, \( \ln(x) \) behaves uniquely compared to polynomial or trigonometric functions.
It's especially important for dealing with growth processes and areas under curves. In our exercise, \( y = \ln(x) \) defines the upper boundary of the region \( \mathcal{R} \).
Key properties of the natural logarithm include:

• \( \ln(1) = 0 \).
• Its derivative is \( \frac{1}{x} \).
• It increases slower than linear functions.

Understanding \( \ln(x) \) was essential in integrating to find both the area under the curve and the powers used when calculating the center of mass coordinates.