Problem 10 - Thomas Calculus in SI Units

Question

Find the center of mass of a thin plate of constant density \(\delta\) covering the given region. a. The region cut from the first quadrant by the circle \(x^{2}+y^{2}=9\) b. The region bounded by the \(x\) -axis and the semicircle \(y=\sqrt{9-x^{2}}\) Compare your answer in part (b) with the answer in part (a).

Step-by-Step Solution

Answer

Center of mass: Part (a) is off-origin; Part (b) is symmetrical about the y-axis.

Step 1: Understand the Geometry of the Region

The problem gives us two regions to deal with. For (a), the region is a quarter of a circle with radius 3 in the first quadrant. For (b), the region is a semicircle also of radius 3, situated above the x-axis.

Step 2: Setup Integral for Center of Mass (Part a)

For a quarter circle, the center of mass \(\bar{x}, \bar{y}\) is calculated as follows. We need to set up the integrals for both the x and y coordinates. The x-coordinate of the center of mass is \(\bar{x} = \frac{1}{A} \int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} x \, dy \, dx\).

Step 3: Calculate Area of Region (Part a)

The area \(A\) of the quarter circle is one fourth of \((\pi \times 3^2)\,\). This gives us: \(A = \frac{9\pi}{4}\).

Step 4: Evaluate Integral for x-coordinate (Part a)

Evaluate the integral for \(\bar{x}\).\[ \bar{x} = \frac{1}{\frac{9\pi}{4}} \int_0^3 \int_0^{\sqrt{9-x^2}} x \, dy \, dx = \frac{4}{9\pi} \int_0^3 x \cdot \sqrt{9-x^2} \, dx \] This integral can be simplified and evaluated using trigonometric substitution or a table of integrals.

Step 5: Evaluate Integral for y-coordinate (Part a)

Similarly, calculate the integral for \(\bar{y}\).\[ \bar{y} = \frac{4}{9\pi} \int_0^3 \int_0^{\sqrt{9-x^2}} y \, dy \, dx = \frac{4}{9\pi} \int_0^3 \frac{1}{2} \left(9 - x^2\right) \, dx\] After evaluating, we get \(\bar{y} = \frac{3}{2\pi}\).

Step 6: Setup Integral for Center of Mass (Part b)

For a semicircle, similarly set up the integrals. The x-coordinate of the center of mass is zero due to symmetry about the y-axis; thus, \(\bar{x} = 0\).The y-coordinate is \(\bar{y} = \frac{1}{A} \int_{-3}^{3} \int_{0}^{\sqrt{9-x^{2}}} y \, dy \, dx\).

Step 7: Calculate Area of Region (Part b)

The semicircle's area is half the area of a circle, \((\pi \times 3^2)\,\), so: \(A = \frac{9\pi}{2}\).

Step 8: Evaluate Integral for y-coordinate (Part b)

Simplify and evaluate the integral for \(\bar{y}\):\[ \bar{y} = \frac{2}{9\pi} \int_{-3}^3 \frac{1}{2} \left(9-x^2\right) \, dx\] After solving, we find \(\bar{y} = \frac{6}{9\pi} \, \cdot 2 = \frac{4}{3\pi}\).

Step 9: Conclusion: Compare Results

For part (a), the center of mass \(\bar{x}, \bar{y}\) is non-zero for both coordinates, whereas for part (b) the x-coordinate of the center of mass is zero, and the y-coordinate is different. This shows a symmetry in part (b) about the y-axis that is not present in part (a).

Key Concepts

Coordinate Geometry

Coordinate geometry, often referred to as analytic geometry, provides a bridge between algebra and geometry by using coordinates or algebraic equations to describe geometric figures. In the problem at hand, we utilize coordinate geometry to define the specific regions of interest, namely a quarter circle in the first quadrant and a semicircle above the x-axis, both derived from the circle equation \(x^2 + y^2 = 9\).

By assigning coordinates to each point on these geometric shapes, we can employ algebraic methods to calculate important physical properties like the center of mass. The use of coordinate geometry is pivotal because it allows the integration process to be applied over these regions, offering a precise mathematical method to examine these shapes.

This approach is essential in many fields of science and engineering where objects need to be analyzed in a plane using a system of coordinates.

Double Integration

Double integration is a technique used to compute a double integral, an integral of a function in two variables over a region in the xy-plane. In the context of finding the center of mass, double integration allows us to compute the average position of mass over a continuous region.

For the quarter-circle in the exercise, we integrate over both x and y coordinates to find \(\bar{x}\) and \(\bar{y}\), respective averages which indicate the center of mass along the x and y axes. Similarly, for the semicircle, we perform these integrations recognizing the boundaries defined by the circle's equation. The setup of these integrals involves identifying proper limits that match the region's geometry.

Using double integration helps in calculating physical properties that are otherwise difficult to determine analytically, especially when dealing with irregular or complex regions.

Trigonometric Substitution

Trigonometric substitution is a method for solving integrals by substituting trigonometric functions for other expressions. It's particularly useful when dealing with integrals involving roots of quadratic polynomials, such as \(\sqrt{9-x^2}\) seen in our integration for the center of mass.

In the solution, when evaluating \(\bar{x}\) and \(\bar{y}\) coordinates, trigonometric substitution simplifies the integration process. For instance, substituting \(x = 3\sin\theta\) transforms \(\sqrt{9-x^2}\) into a simpler expression \(3\cos\theta\), making the integral easier to solve.

This method can significantly ease the computation by transforming a challenging problem into a more manageable one, facilitating the integration of complex functions often encountered in calculus.

Symmetry in Calculus

Symmetry plays a crucial role in solving problems related to geometry and calculus as seen in the center of mass calculation for both part (a) and part (b) of the problem. Symmetry about an axis simplifies integral calculation by reducing the region of interest or allowing certain assumptions.

For the semicircle in part (b), symmetry about the y-axis implies that the x-coordinate of the center of mass is zero. This is because for each point on the semicircle, there is a corresponding point at an equal distance from the y-axis on the opposite side. This bilateral symmetry effectively cancels out any deviation from the center, resulting in a simplified solution.

Using symmetry, we can make more informed assumptions, reducing the need for more complex calculations, thus saving time and effort while deriving comprehensive solutions.