Problem 21 - 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. \(y=x^{2 / 3}, y=0, x=8\)

Step-by-Step Solution

Answer

In order to find \(M_{x}, M_{y},\) and \((\overline{x}, \overline{y})\), the steps involve finding the area of the lamina, performing integral calculations to find the moments and finally finding the centroid by dividing the moments by the area.

Step 1: Find the Area A

First, calculate the area of the region which is defined by the equations given to find the value of \(A\). As the lamina is bounded by the graph \(y=x^{2 / 3}\), \(y=0\) and \(x=8\), the area is determined by:\[A = \int_0^8 x^{2/3} dx\]

Step 2: Evaluate the Integral for Area

Performing integration, the area \(A\) is:\[A = \left[ \frac{3}{5} x^{5/3} \right]_0^8 = \frac{3}{5} \times 8^{5/3} - \frac{3}{5} \times 0^{5/3} = \frac{384}{5}\]

Step 3: Find Moments \(M_x\) and \(M_y\)

Next, calculate moments \(M_x\) and \(M_y\) using the formulas given. As \(f(x, y) = \rho\), for uniform density \(\rho\), \(f(x, y) = 1\). So,\[M_x = \int_0^8 y \times x^{2/3} dx\] and \[M_y = \int_0^8 x \times x^{2/3} dx\]. \nEvaluate these integrals to obtain \(M_x\) and \(M_y\).

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

Finally, calculate the position of the centroid by \(\overline{x} = \frac{M_y}{A}\) and \(\overline{y} = \frac{M_x}{A}\) given \(M_x\) and \(M_y\) found in Step 3 and \(A\) from Step 2.

Key Concepts

Moments Mx and My

Understanding the moments Mx and My is crucial when we aim to locate the centroid of a lamina with uniform density. The moment of a lamina about an axis is the integral of the density function times the distance from the axis. For our context, a lamina is simply a flat, two-dimensional object.

For the moment about the y-axis (Mx), we integrate the product of the density and the distance from the y-axis (which is simply x) along the entire region. If the density is uniform and can be taken as 1, then the integral simplifies to the area under the curve times x. The moment Mx thus becomes:\[M_{x} = \int_{a}^{b} y \times x dx\]
Similarly, My is the moment about the x-axis, which involves integrating the product of density and the distance from the x-axis (which is y) over the same region:\[M_{y} = \int_{a}^{b} x \times y dy\]
In layman's terms, you can think of moments as a way to measure how much 'weight' there is at different distances from an axis, weighted by the shape of the region we're looking at.

Integration for Area

Area calculation via integration is a fundamental technique in calculus. Imagine slicing a shape into infinitely thin vertical strips (if integrating with respect to x) or horizontal strips (if with respect to y) and then summing up the areas of those strips.

In mathematical terms, for a curve bounded by the function y=f(x), a straight line y=0 (the x-axis), and vertical lines at x=a and x=b, the area A under the curve from x=a to x=b is:\[A = \int_{a}^{b} f(x) dx\]
This formula is the foundation on which we build more complex area computations, like finding the areas between curves or under polar graphs. By understanding and applying this principle, we can solve a wide range of geometric problems in calculus.

Uniform Density

When we speak of a lamina with uniform density, we imply that its mass is spread out equally across its entire area. This is an important assumption because it simplifies many calculations. In physics and engineering, density (usually denoted as ρ) is mass per unit area. A uniform density means that no matter where you measure on the lamina, you get the same value for density.

For most shapes of laminae, we can assume a uniform density to find the center of mass – which coincides with the geometric centroid when density is uniform. This assumption also allows us to use 1 as the value for density when integrating to find moments, as the density factor would cancel out anyway when calculating the centroid.

Centroid Calculation

The centroid of a lamina is often referred to as its 'center of gravity,' and it's the point at which you could balance the lamina on the tip of a pencil without it tilting to any side. To calculate the centroid \((\overline{x}, \overline{y})\), we divide the moment about the y-axis (Mx) by the total area (A) to get the x-coordinate, and the moment about the x-axis (My) by the area to get the y-coordinate:\[\overline{x} = \frac{M_{y}}{A}\quad and \quad \overline{y} = \frac{M_{x}}{A}\]
By applying these formulas after calculating Mx, My, and A, we'll find the centroid's precise location on the plane. This is a pivotal concept not only in mathematics but also in fields such as mechanical design, where balance and symmetry are essential.