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. $$ y=x^{2 / 3}, y=4 $$

Short answer

The first moments of the lamina are given by \(M_x = {9 \over 2}\rho(\sqrt[3]{4})^{5/2}\) and \(M_y = 2\rho(\sqrt[3]{4})^3\). Its centroid is located at \((\bar{x}, \bar{y}) = \left(2(\sqrt[3]{4}), {3 \over 2}\sqrt[3]{4}\right)\)

Step-by-step solution

Compute the Area of Lamina

The area A is given by the integral of \(y dx\) from \(-\sqrt[3]{4}\) to \(\sqrt[3]{4}\), which results from intersecting \(y = x^{2 / 3}\) and \(y = 4\). Hence, \(A = \rho \int_{-\sqrt[3]{4}}^{\sqrt[3]{4}} x^{2 / 3} dx\).

Compute the First Moments

The first moment about the x-axis \(M_x = \rho \int_{-\sqrt[3]{4}}^{\sqrt[3]{4}} y^2 dx\), and the first moment about the y-axis \(M_y = \rho \int_{-\sqrt[3]{4}}^{\sqrt[3]{4}} xy dx\). Substitute \(y = x^{2 / 3}\) into these equations. Hence, \(M_x = \rho \int_{-\sqrt[3]{4}}^{\sqrt[3]{4}} x^{4 / 3} dx\), and \(M_y = \rho \int_{-\sqrt[3]{4}}^{\sqrt[3]{4}} x^{5 / 3} dx\).

Evaluate the Integrals

Evaluating these integrals yields \(A = 3\rho(\sqrt[3]{4})^2\), \(M_x = {9 \over 2}\rho(\sqrt[3]{4})^{5/2}\), and \(M_y = 2\rho(\sqrt[3]{4})^3\).

Find the Centroid

The centroid's x-coordinate \(\bar{x} = {M_y \over A}\) and the y-coordinate \(\bar{y} = {M_x \over A}\). Plugging in our values for \(M_x\), \(M_y\), and \(A\) from steps 1-3, we have \(\bar{x} = {2\sqrt[3]{4}}^3 / 3\sqrt[3]{4}^2 = 2(\sqrt[3]{4})\) and \(\bar{y} = {9 \over 2}\sqrt[3]{4}^{5/2} / 3\sqrt[3]{4}^2 = {3 \over 2}\sqrt[3]{4}\).

Key concepts

Moment of a Lamina

One key concept in understanding the center of mass of a lamina, or a two-dimensional flat body, is the 'moment of a lamina'. The moment of a lamina about an axis is a measure of the distribution of the lamina's area relative to that axis.

Imagine you have a flat shape, like a piece of cardboard cut out into a specific shape. When you're trying to balance this piece on a pencil, you're searching for its centroid. Technically, what you're finding is the point where the moment balances out in all directions.

Mathematically, we calculate the moment with respect to the x-axis, denoted as \(M_x\), and the moment with respect to the y-axis, denoted as \(M_y\). These are determined by integrating the density of the lamina, \(\rho\), multiplied by the area at each point over the entire lamina. For example, to find \(M_x\), we might integrate the density times the square of the distance from the x-axis across the whole area of the lamina.

It's important to note that the moment isn't just about how much mass or area there is, but how that mass or area is distributed. That's why even lightweight materials can have significant moments if they have a large area spread out far from the axis.

Centroid Coordinates

The centroid of a lamina is analogous to the center of mass of a three-dimensional body but strictly in two dimensions. Essentially, it's the geometric center. The coordinates of the centroid, \((\bar{x}, \bar{y})\), can be found by dividing the first moments by the total area, A.

To compute the coordinates of the centroid, we use the formulae \(\bar{x} = \frac{M_y}{A}\) and \(\bar{y} = \frac{M_x}{A}\). Here, \(M_x\) is the first moment of the area about the x-axis, and \(M_y\) is the first moment of the area about the y-axis. These ratios essentially 'average out' the distribution of the area to find the central point.

In the exercise provided, after computing the first moments and the area, the centroid coordinates are calculated using the respective formulae. This process offers an important application of integral calculus, as integration is needed to compute both moments and the area. The coordinates give us the precise location on the plane where the lamina can be perfectly balanced, assuming uniform distribution of density.

Calculus in Physics

The application of calculus in physics is extensive, particularly in classical mechanics where we deal with quantities like position, velocity, acceleration, and, as in our exercise, the centroid of objects. Calculus helps us to analyze continuously changing values — which is basically what the physical world is all about.

For the computation of centroid coordinates, we use integrals, a fundamental concept in calculus. Specifically, we're using definite integrals, as we integrate over a specific interval of the lamina's dimensions. This process allows us to sum up infinite infinitesimally small products of density and distance, leading to the precise moments and area required to find the centroid.

Calculus is the mathematical tool that enables us to solve these continuous problems. When you're figuring out where the balance point, or centroid, of a strange-shaped lamina is, you're doing physics — and you're applying calculus to get there. The integration steps in finding \(M_x\), \(M_y\), and A from our exercise are perfect examples of how we use calculus to solve real-world physical problems.