Problem 31 - 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. $$ y=\sqrt{x}, y=0, x=4 $$

Step-by-Step Solution

Answer

The first moments about the x- and y-axes of the lamina are \(M_{x}=8\rho\) and \(M_{y}=8\rho\) respectively, and the centroid of the lamina is located at \((\bar{x}, \bar{y}) = \left(\frac{3}{2}\rho, \frac{3}{2}\rho\right)\)

Step 1: Determine the region

Firstly, the region R for the lamina must be determined. This region is bounded by the graphs of the three equations: \(y=\sqrt{x}\), \(y=0\), and \(x=4\). The region R is thus the area under the curve \(y=\sqrt{x}\) from \(x=0\) to \(x=4\).

Step 2: Setup the integrals for \(M_x\) and \(M_y\)

Next, the double integrals for the first moments \(M_x\) and \(M_y\) are set up. As the density \(\rho\) is uniform, it can be considered as a constant factor outside the integral. The first moment about the x-axis \(M_x\) is given by the double integral of y over the region R: \[M_x = \rho\int_0^4\int_0^{\sqrt{x}} y \,dy\,dx\] and the first moment about the y-axis \(M_y\) is given by the double integral of x over the region R: \[M_y = \rho\int_0^4\int_0^{\sqrt{x}} x \,dy\,dx\]

Step 3: Evaluate the integrals

The double integrals for \(M_x\) and \(M_y\) are then evaluated. This gives the first moments about the x- and y-axes respectively. Using the properties of integrals, these can be simplified as follows: \[\begin{align*} M_x & = \rho\int_0^4 \frac{1}{2}x \,dx = \frac{\rho}{2} \int_0^4x \,dx = \frac{\rho}{2} \cdot \frac{x^2}{2}\Bigg|_0^4 = \frac{16\rho}{2} = 8\rho \ M_y & = \rho\int_0^4 x \,dx = \rho \cdot \frac{x^2}{2}\Bigg|_0^4 = \frac{16\rho}{2} = 8\rho \end{align*}\]

Step 4: Calculate the Area

The total area A of the region R is calculated from the integral of 1 over the same region: \[A = \int_0^4\int_0^{\sqrt{x}} \,dy\,dx = \int_0^4 \sqrt{x} \,dx = \frac{2}{3}x^{3/2}\Bigg|_0^4 = \frac{16}{3}\]

Step 5: Calculate Centroid Coordinates

Finally, the coordinates of the centroid \((\bar{x}, \bar{y})\) are calculated by dividing the first moments by the total area: \[\bar{x} = \frac{M_y}{A} = \frac{8\rho}{16/3} = \frac{3}{2}\rho, \quad \bar{y} = \frac{M_x}{A} = \frac{8\rho}{16/3} = \frac{3}{2}\rho\] Note that the density \(\rho\) cancels out because it is uniform.

Key Concepts

Calculus Integrals

Integrals are essential in finding moments of a lamina, as they allow us to calculate quantities over a region. When solving for moments like \(M_x\) and \(M_y\), we use double integrals over a defined region. The first moment about the x-axis \(M_x\) is the integral of the lamina's height (y-value) times the density \(\rho\) over the region. For our example, the corresponding integral was \(\rho\int_0^4\int_0^{\sqrt{x}} y \,dy\,dx\). Similarly, for the moment about the y-axis \(M_y\), the x-values are used: \(\rho\int_0^4\int_0^{\sqrt{x}} x \,dy\,dx\).

• This means finding the "weighted average" position of the x and y coordinates under the given density function.
• These moments help us find the centroid - the balance point - of the lamina.

While setting up integrals, remember to pay attention to bounds, as these will be determined by the equations delimiting the region of the lamina.

Moments of Inertia

Moments of inertia, though not directly sought in this problem, are related to the moments we calculated, \(M_x\) and \(M_y\). These moments measure how the lamina's mass is distributed relative to an axis.

• \(M_x\) and \(M_y\) can be thought of as the "mass balance" with respect to the x and y axes, respectively.
• The concepts of these moments share the same foundational integral principles as moments of inertia.

In a physical sense, moments of inertia describe how difficult it is to rotate the lamina about an axis, directly linked with the integrals of mass distribution over the region. Although we don't compute it here, understanding \(M_x\) and \(M_y\) introduces these larger ideas that are crucial in physics and engineering.

Coordinate Geometry

Coordinate geometry helps us describe and visualize the boundaries of our lamina. For the exercise at hand, the region bounded by \(y=\sqrt{x}\), \(y=0\), and \(x=4\) forms the lamina's shape. Here, we deal with:

• A curved top boundary (\(y=\sqrt{x}\)), representing the square root function and symbolizing nonlinear, curved structure.
• The flat line at \(y=0\) indicates the x-axis, forming another boundary.
• The vertical line \(x=4\) further bounds the region.

By understanding these equations, students can better visualize where the lamina exists in a coordinate plane. Practicing with these types of problems enhances spatial thinking skills, crucial for calculus and physics applications.

Lamina Density

The density \(\rho\) plays a critical role in determining the lamina's physical properties. Here, the density is uniform, meaning it is constant across the lamina's entire area.

• We use density in calculating the moments \(M_x\) and \(M_y\), but it eventually cancels out when finding the centroid \((\bar{x}, \bar{y})\).
• This cancellation happens because we are dividing each moment by the total area, which includes the density term, resulting in density elimination from the final centroid formulae.

This shows the significance of lamina density in practical calculations. Understanding uniform density simplifies computations, but different density functions can offer more complex and realistic applications in varied fields.