Problem 35 - Essential Calculus

Question

In Exercises \(31-40,\) find \(M_{x}, M_{y},\) and \((\bar{x}, \bar{y})\) for the laminas of uniform density \(\rho\) bounded by the graphs of the equations. $$ y=x^{2 / 3}, y=0, x=8 $$

Step-by-Step Solution

Answer

The mass of the lamina is \(\frac{3}{5} \rho(8^{5/3})\), the first moments with respect to x and y are \(\frac{3}{10} \rho (8^{5/3})\) and \( \frac{3}{5} \rho (8^{8/3})\) respectively, and the centroid of the lamina is \((\bar{x}, \bar{y}) = (8, \frac{1}{2})\).

Step 1: Finding Mass of the Lamina

The mass of the lamina bounded by the given curves is given by \(M = \rho \int_a^b y \, dx\). Here, our limits of integration are \(a=0\) and \(b=8\) from the given graphs. Hence, \(M = \rho \int_0^8 x^{2/3} \, dx\). Solving the integral gives: \(M = \frac{3}{5} \rho(8^{5/3})\).

Step 2: Finding First Moments \(M_x\) and \(M_y\)

The first moments with respect to x, \(M_x\), and y, \(M_y\), are given by \(M_x = \rho \int_a^b \frac{1}{2} y^2 \, dx\) and \(M_y = \rho \int_a^b x \, y \, dx\), respectively. Therefore, we have: \(M_x = \rho \int_0^8 \frac{1}{2} (x^{2/3})^2 \, dx\), and \(M_y= \rho \int_0^8 x \, x^{2/3} \, dx\). Solving these integrals gives: \(M_x = \frac{3}{10} \rho (8^{5/3})\) and \(M_y = \frac{3}{5} \rho (8^{8/3})\).

Step 3: Finding the Centroid \((\bar{x}, \bar{y})\)

The centroid of the lamina, \((\bar{x}, \bar{y})\), is found by dividing the first moment with respect to the other axis by the mass of the lamina. Hence, \(\bar{x} = \frac{M_{y}}{M}\) and \(\bar{y} = \frac{M_{x}}{M}\). On substitution, we find: \(\bar{x} = \frac{\frac{3}{5} \rho (8^{8/3})}{\frac{3}{5} \rho (8^{5/3})} = 8^1\) and \(\bar{y} = \frac{\frac{3}{10} \rho (8^{5/3})}{\frac{3}{5} \rho (8^{5/3})} = \frac{1}{2}\).

Key Concepts

Mass Calculation in Calculus

In the field of calculus, mass calculation for a lamina, which is a two-dimensional object of uniform thickness, involves integrating the density function over the area it occupies. This concept is crucial as it sets the foundation for understanding more complex physical applications involving mass distribution, such as finding the center of gravity or balance point (centroid) of a flat object.

Let's visualize the procedure with an example. If the density \(\rho\) of a lamina is constant, the total mass (M) can be expressed as \(M = \rho \int_a^b y \, dx\), where \(y\) is the height of the lamina at a certain point \(x\), and \(a\) and \(b\) are the x-limits of the boundary of the lamina.

In the exercise provided, the lamina is bound by the curve \(y = x^{2/3}\), the x-axis (\(y = 0\)), and a vertical line at \(x = 8\). To find the mass, we integrate the function \(x^{2/3}\) from 0 to 8 with respect to \(x\) and multiply by the constant density \(\rho\). Thus, the mass calculation through integration is both logical and systematic, reflecting the lamina's continuous nature across its domain.

First Moments Mx and My

The concepts of first moments, \(M_x\) and \(M_y\), in the realm of physics and calculus are essential in determining the centroid of an object. They represent a measure of the object's distribution of mass along an axis.

For a lamina with uniform density \(\rho\), the first moment about the x-axis, \(M_x\), is calculated as \(M_x = \rho \int_a^b \frac{1}{2} y^2 \, dx\), integrating the square of the height function divided by two over the region. Similarly, the first moment about the y-axis, \(M_y\), is obtained by integrating the product of \(x\) and \(y\) over \(x\), or \(M_y = \rho \int_a^b x \, y \, dx\).

These calculations give us an understanding of how far the mass of the lamina extends from the respective axis. In the step-by-step solution, we apply this concept to find \(M_x\) and \(M_y\) using integration, which tells us how the lamina's mass is spread relative to the axes. This paves the way for locating the centroid, a point where the lamina can be perfectly balanced.

Integration in Calculus

Integration in calculus is a fundamental tool that allows for the accumulation of quantities, such as area under a curve, total mass, or moments as in the previous sections. It essentially sums up infinitesimally small pieces of a quantity to calculate a whole.

In our context, integration is used threefold: to find the mass, and to calculate the first moments \(M_x\) and \(M_y\) about the x-axis and the y-axis, respectively. The process involves setting up an integral with appropriate limits, which correspond to the boundaries of the lamina.

The given exercise provides a clear demonstration, where integrals are solved with limits from 0 to 8, representing the domain over which the lamina extends. Integrating the suitable functions over this interval gives us tangible values such as mass and first moments, pivotal for understanding the physical properties of the lamina. The reliable method of integration makes these abstract concepts measurable and, by consequence, more comprehensible.