Problem 24 - APEX Calculus

Question

Find the center of mass of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\) Note: these are the same lamina as in Exercises \(11-18\). \(R\) is the circle sector bounded by \(x^{2}+y^{2}=25\) in the first quadrant; \(\delta(x, y)=\left(\sqrt{x^{2}+y^{2}}+1\right) \mathrm{kg} / \mathrm{m}^{2}\)

Step-by-Step Solution

Answer

The center of mass is approximately at (2.32, 2.32) m.

Step 1: Define the Region and Density Function

The region \(R\) is a sector of a circle with radius 5, in the first quadrant, which is bounded by the equation \(x^2 + y^2 = 25\). The density function is given as \(\delta(x, y) = \sqrt{x^2 + y^2} + 1\).

Step 2: Set Up the Integral for Mass

The mass \(M\) of the region \(R\) is found by integrating the density function over \(R\). In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\), and \(x^2 + y^2 = r^2\). Convert the density to polar form: \(\delta(r, \theta) = r + 1\). The limits for \(r\) are from 0 to 5, and for \(\theta\) from 0 to \(\frac{\pi}{2}\):\[ M = \int_0^{\frac{\pi}{2}} \int_0^5 (r + 1) r \, dr \, d\theta. \]

Step 3: Integrate to Find the Mass

First integrate with respect to \(r\):\[ \int_0^5 (r^2 + r) \, dr = \left[ \frac{r^3}{3} + \frac{r^2}{2} \right]_0^5 = \frac{125}{3} + \frac{25}{2}. \]Calculate the result:\[ \frac{125}{3} = 41.67 \qquad \frac{25}{2} = 12.5 \qquad \text{thus} \quad \frac{125}{3} + \frac{25}{2} = \frac{250}{6} = 41.67 + 12.5 = 54.17. \]Now integrate with respect to \(\theta\):\[ M = \int_0^{\frac{\pi}{2}} 54.17 \, d\theta = 54.17 \cdot \frac{\pi}{2} = 27.085\pi. \]

Step 4: Find Moment About Y-Axis

For the moment about the y-axis, \(M_y\), we use:\[ M_y = \int_0^{\frac{\pi}{2}} \int_0^5 r \cos\theta (r + 1) r \, dr \, d\theta. \]Simplify to:\[ \int_0^5 (r^3 \cos\theta + r^2 \cos\theta) \, dr.\]Integrate with respect to \(r\):\[ \int_0^5 \left( \frac{r^4}{4} + \frac{r^3}{3} \right) \cos\theta \, dr. \]

Step 5: Calculate Moment About Y-Axis

Plug in the values:\[ \left[ \frac{r^4}{4} + \frac{r^3}{3} \right]_0^5 = \frac{625}{4} \cos\theta + \frac{125}{3} \cos\theta. \]Resulting moment:\[ M_y = \int_0^{\frac{\pi}{2}} \left( \frac{625}{4} + \frac{125}{3} \right) \cos\theta \, d\theta. \]

Step 6: Simplify Moment About Y-Axis

Calculate numeric values:\[ \frac{625}{4} = 156.25 \quad \text{and} \quad \frac{125}{3} = 41.67. \]Combine and integrate:\[ M_y = \int_0^{\frac{\pi}{2}} (197.92) \cos\theta \, d\theta. \]Evaluate:\[ M_y = 197.92 \cdot 1 = 197.92. \]

Step 7: Find Moment About X-Axis

For moment about the x-axis, \(M_x\), use:\[ M_x = \int_0^{\frac{\pi}{2}} \int_0^5 r \sin\theta (r + 1) r \, dr \, d\theta. \]Simplify:\[ \int_0^5 (r^3 \sin\theta + r^2 \sin\theta) \, dr. \]

Step 8: Calculate Moment About X-Axis

Integrate over \(r\):\[ M_x = \int_0^{\frac{\pi}{2}} \sin\theta \left( \frac{625}{4} + \frac{125}{3} \right) \, d\theta. \]Integrate over \(\theta\):\[ M_x = 197.92 \cdot 1 = 197.92. \]

Step 9: Calculate Center of Mass Coordinates

The coordinates of the center of mass \(( \bar{x}, \bar{y} )\) are given by:\[ \bar{x} = \frac{M_y}{M} = \frac{197.92}{27.085\pi} \]and\[ \bar{y} = \frac{M_x}{M} = \frac{197.92}{27.085\pi}. \]

Step 10: Simplify Final Coordinates

Calculate for both, \(\bar{x}\) and \(\bar{y}\):\[ \bar{x} = \frac{197.92}{27.085\pi} \approx 2.32 \, \mathrm{m},\]\[ \bar{y} = \frac{197.92}{27.085\pi} \approx 2.32 \, \mathrm{m}.\]Thus, the center of mass is \((2.32, 2.32)\).

Key Concepts

Lamina

A lamina is a flat, two-dimensional object with a defined shape and mass distribution. In mathematics and physics, it is often used to study properties like center of mass. In this exercise, the lamina is described by a sector of a circle with a radius of 5. The term "lamina" is crucial because it simplifies three-dimensional problems into two dimensions. This allows certain computations, like integrating, to become more approachable. Understanding the shape and boundary of the lamina, such as being in the first quadrant, provides context, ensuring focus is on the correct region for calculations.

Density Function

The density function \(\delta(x, y)\) relates to how mass is distributed across the lamina. It calculates as \(\delta(x, y) = \sqrt{x^2 + y^2} + 1\), implying the mass depends on the distance from the origin, plus a constant. Density functions play an essential role in determining quantifies like mass and center of mass. When examining physical systems, different parts can have varying density levels. This function impacts the integral setup, leading to more computations. Here, conversion to polar coordinates facilitates integrating this function, since it reflects radial symmetry. Recognizing how density affects distributions provides insights into more complex systems as well.

Polar Coordinates

Polar coordinates simplify problems involving circular regions by using radius and angle instead of cartesian coordinates (x and y). For this lamina, converting the coordinates involved changing \({x, y}\) to \({r, \theta}\) with \(x = r\cos\theta\) and \(y = r\sin\theta\).This conversion simplifies integrating functions over circular paths, making radial limits more straightforward. Through polar coordinates, here \(r\) ranges from 0 to 5, which matches the circle's radius, and \(\theta\) ranges from 0 to \(\frac{\pi}{2},\) aligning with the circular sector in the first quadrant. Optimizing a problem like this illustrates polar coordinates' powerful role in simplifying integrals over curved paths.

Integrals

Integrals calculate quantities over a region, like mass or moments. For the lamina, integrals determine both mass \( M \) and moments \( M_x \) and \( M_y \). The double integral setup incorporates the density function in polar form, \( \delta(r, \theta) = r + 1 \), reflecting the lamina's mass accumulation across its area.Breaking it into steps helps manage complexities, starting with calculating the mass using:\[ M = \int_0^{\frac{\pi}{2}} \int_0^5 (r + 1) r \, dr \, d\theta. \]Moments are handled similarly but adjust based on axis considerations, such as using \( \cos\theta \) or \( \sin\theta \). The integrals establish how variables influence the overall system, showing integrals' efficacy in tying together complex parts of physics and mathematics into simpler expressions.