Problem 18 - Calculus of a Single Variable

Question

Find \(M_{x}, M_{y},\) and \((\overline{x}, \overline{y})\) for the laminas of uniform density \(\boldsymbol{\rho}\) bounded by the graphs of the equations. \(y=\sqrt{x}, y=\frac{1}{2} x\)

Step-by-Step Solution

Answer

After performing the steps as described, it is found that \(M_x = \frac{8}{3} \rho\), \(M_y = \frac{8}{3} \rho\) and the centroid \((\overline{x}, \overline{y}) = (1, 1)\). The answer is independent of the density, so it's the same for any uniform lamina bounded by the given curves.

Step 1: Find Integration Bounds

Set the two functions equal to each other to find the values of \(x\) where the curves intersect. So \(\sqrt{x} = \frac{1}{2} x\). Squaring both sides, the resulting equation \(x = \frac{1}{4} x^2\) yields the solutions \(x = 0\) and \(x=4\). These are the limits of integration.

Step 2: Calculate the First Moments \(M_{x}, M_{y}\)

The first moments \(M_{x}\) and \(M_{y}\) are defined as the integrals of the density times \(y\) and \(x\) respectively over the region which extends from \(x=0\) to \(x=4\). So using the density \(\rho\), they can be calculated as \(M_{x} = \int_0^4 \rho y \, dx\) and \(M_{y} = \int_0^4 \rho x \, dx\). Substituting \(\sqrt{x}\) for \(y\) in \(M_{x}\) and \(\frac{1}{2} x\) for \(y\) in \(M_{y}\), the actual computations are \(M_{x} =\int_0^4 \rho \sqrt{x} \, dx\) and \(M_{y} = \int_0^4 \rho \frac{1}{2} x \, dx\).

Step 3: Calculate the Total Mass \(m\)

The total mass of the lamina is computed as the integral of density over the region. Hence, \(m = \int_0^4 \rho \, dx\). With uniform density, \(\rho\) can be factored out of the integral, leaving \(m = \rho \int_0^4 dx\). After calculation, it's found that \(m = 4 \rho\).

Step 4: Find the Centroid \((\overline{x}, \overline{y})\)

The coordinates of the centroid are given by the formula \(\overline{x} = \frac{M_{y}}{m}\) and \(\overline{y} = \frac{M_{x}}{m}\). Inserting the previously determined values of \(M_x\), \(M_y\) and \(m\) into these formulas will yield \((\overline{x}, \overline{y})\). Note that in step 2 and step 3 we actually computed \(M_y/m\) and \(M_x/m\) respectively, to avoid introduction of the unknown density \(\rho\) in the result.

Key Concepts

First Moments

When we talk about the first moments in terms of physics and mechanics, these are the measures of the distribution of mass across an object with a uniform density. For a two-dimensional lamina, which you can think of as a flat sheet with a thinness that we can disregard, the first moments, denoted as \( M_x \) and \( M_y \), reflect the spread of the sheet's mass relative to the y-axis and x-axis, respectively.

To determine these moments, we apply integration, as they are essentially the sum of all infinitesimal elements of mass multiplied by their distances from the axes. This can be visualized by imagining a teeter-totter, where the distances from the pivot (axis) count with how much force each mass element 'pushes'. The results provide vital information in finding the center of mass, or centroid, of the lamina. This is akin to finding the 'balance point' where you could support the lamina on the tip of your finger and have it lie flat.

Mass of a Lamina

The mass of a lamina is a fundamental concept that is key to understanding various physical and mechanical properties of the object. In our context, the lamina's mass depends on the surface area and density. Since our lamina has a uniform density, which we denote as \( \rho \), the task simplifies as the density is constant across the entire object.

To calculate the mass, we integrate the constant density over the surface area of the lamina, which is bound within a specific region in the plane. For instance, in our exercise, the lamina's total mass \( m \) is obtained by integrating the uniform density over the x-axis from one boundary curve to the other, establishing how much 'stuff' is there in total within the lamina's boundaries. This is a substantial step towards finding the centroid as it will be used to normalize the first moments.

Integration Bounds

The integration bounds are essentially the limits between which we evaluate an integral. These boundaries are crucial because they constrain the region we are interested in—be it for calculating an area, a mass, or any other integral-related quantity.

To establish these bounds for our lamina, we need to look at where the boundaries of the lamina intersect or where they start and end. Doing so gives us the exact interval across which we will integrate. In the case of our exercise, by setting the equations of the curves forming the lamina's boundary equal to each other, we've found the intersection points, identifying the x-axis limits from 0 to 4. These bounds are imperative, as they will confine our integration process to just the region the lamina occupies.

Uniform Density

The concept of uniform density is central to many problems in physics and engineering. A uniform density implies that the mass of an object is distributed evenly throughout—that is, every unit of volume (in three dimensions) or unit of area (in two dimensions) has the same mass as any other equal-sized unit.

In practical terms, when calculating the mass or moments of a lamina with uniform density, the density across the object doesn't change, hence, it can be factored out of integral expressions, simplifying calculations. This relative simplicity is handy, for it means we don't have to deal with varying densities across the object, which would significantly complicate our integrals and thus our life as well. In the provided exercise, the constant density \( \rho \) effectively exits the integral, cleanly leaving us with much simpler math to work through.