Problem 15 - Thomas Calculus

Question

In Exercises \(18,\) find the center of mass of a thin plate of constant density \(\delta\) covering the given region. \begin{equation}\begin{array}{l}{\text { a. The region cut from the first quadrant by the circle } x^{2}+y^{2}=9} \\ {\text { b. The region bounded by the } x \text { -axis and the semicircle }} \\ {y=\sqrt{9-x^{2}}} \\ {\text { Compare your answer in part (b) with the answer in part (a). }}\end{array} \end{equation}

Step-by-Step Solution

Answer

Region (a) center: \((\frac{3}{4},\frac{3}{4})\). Region (b) center: \((0,\frac{4}{\pi})\).

Step 1: Understand the Problem and Region (a)

We are tasked with finding the center of mass for a region given by the circle \( x^2 + y^2 = 9 \) in the first quadrant. This region is a quarter-circle with radius 3.

Step 2: Set Up the Integral for Region (a)

To find the center of mass \((\bar{x}, \bar{y})\) of the quarter-circle, we start by considering the formula \( \bar{x} = \frac{1}{A} \int x \; dA \) and \( \bar{y} = \frac{1}{A} \int y \; dA \), where \(A\) is the area of the quarter-circle. The area \(A\) is \(\frac{\pi(3)^2}{4} = \frac{9\pi}{4}\).

Step 3: Calculate \(\bar{x}\) for Region (a)

Using polar coordinates, the integral for \(\bar{x}\) is: \[ \bar{x} = \frac{2}{\pi} \int_{0}^{\pi/2} \int_{0}^{3} r^2 \cos(\theta) \, dr \, d\theta. \] Evaluating the integrals step-by-step, we get \( \bar{x} = \frac{3}{4} \).

Step 4: Calculate \(\bar{y}\) for Region (a)

The integral for \(\bar{y}\) in polar coordinates is similar: \[ \bar{y} = \frac{2}{\pi} \int_{0}^{\pi/2} \int_{0}^{3} r^2 \sin(\theta) \, dr \, d\theta. \] After computing this integral, we find \( \bar{y} = \frac{3}{4} \).

Step 5: Interpret Results for Region (a)

The center of mass for the quarter-circle is \( (\bar{x}, \bar{y}) = \left( \frac{3}{4}, \frac{3}{4} \right) \). This is expected due to symmetry around \(x = y\) in the first quadrant.

Step 6: Understand the Region (b)

Next, we find the center of mass of a semicircle of radius 3 bounded by the semicircle equation \( y = \sqrt{9-x^2} \) and the x-axis.

Step 7: Set Up the Formula for Region (b)

We'll use the same center of mass formulas: \( \bar{x} = \frac{1}{A} \int x \; dA \) and \( \bar{y} = \frac{1}{A} \int y \; dA \), where \(A\) is the area of the semicircle \( \frac{9\pi}{2} \).

Step 8: Calculate \(\bar{x}\) for Region (b)

Due to symmetry about the y-axis, \( \bar{x} = 0 \).

Step 9: Calculate \(\bar{y}\) for Region (b)

Using \( \bar{y} = \frac{2}{\pi} \int_{0}^{3} y \, dx \), where \( y = \sqrt{9-x^2} \). Solving the integral, we get \( \bar{y} = \frac{4R}{3\pi} = \frac{4 \times 3}{3\pi} = \frac{4}{\pi} \).

Step 10: Compare Results

For region (a), the center is \( \left( \frac{3}{4}, \frac{3}{4} \right) \). For region (b), the center of mass is \( (0, \frac{4}{\pi}) \). The differences arise due to changes in the bounds and geometry between a quadrant and a semicircle.

Key Concepts

Understanding Quarter-Circle

When we talk about a quarter-circle, we are referring to a segment of a circle that is cut into four equal parts. Imagine a circle with a radius. Splitting this circle into four equal sections along the two perpendicular axes forms quarter-circles.
The problem involves a quarter-circle that resides in what we call the 'first quadrant'. This is the upper right section where both x and y values are positive. For the circle described by the equation \(x^2 + y^2 = 9\), the radius is 3. Therefore, each quarter-circle has an area given by \( \frac{\pi r^2}{4} \), which simplifies to \(\frac{9\pi}{4}\) when \(r = 3\).
Finding the center of mass of this region involves understanding its symmetry and using the appropriate calculus principles.

Exploring the Semicircle

The term semicircle refers to half of a circle. When divided by a straight line through the center, the semicircle consists of the half that includes the circular arc and a straight diameter line. The exercise uses the top half of the circle given by the equation \(y = \sqrt{9-x^2}\).
This equation represents a semicircle in the positive y-direction from the center point at \( (0,0) \) to the points where it meets the x-axis. The radius remains 3 in this instance, and thus the area of this semicircle is \( \frac{9\pi}{2} \).
For calculating the center of mass, we leverage its symmetrical property about the y-axis. This symmetry provides insight that the center of mass along the x-axis, \(\bar{x}\), must be zero.

Introduction to Polar Coordinates

Polar coordinates offer a different way to express locations on a plane using a radius and an angle. In situations involving circles and curves, this system simplifies the calculations drastically. Instead of using horizontal and vertical lines (Cartesian coordinates), polar coordinates use a central point and an angle measured counterclockwise from the positive x-axis.
Whenever dealing with circles, like the quarter-circle or semicircle, converting equations to polar coordinates often makes the integral setup easier. This transformation changes an integral from \(x, y\) to \(r, \theta\), with equations such as \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\).
For the quarter-circle problem, the integrals \(\int x\,dA\) and \(\int y\,dA\) convert into forms that often simplify the math, saving both time and minimizing potential errors.

Role of Constant Density

Density is a measure of how much mass is packed into a particular space. In this context, when the density is constant across a thin plate, it allows us to simplify the calculation of the center of mass significantly.
This exercise assumes constant density, denoted \(\delta\), meaning the mass per unit area is uniform. As a result, the density can often be factored out of integrals in the calculation for the center of mass, since it doesn't vary over the region.
Effectively, this property allows the problem to reduce to geometry and calculus, without the additional complexity of integrating variable density across the surface. It simplifies the center of mass formulas to focus on areas and coordinates only.