Problem 22 - Basic Technical Mathematics with Calculus

Question

Find the coordinates of the centroids of the given figures. Each region is covered by a thin, flat plate. The region above the \(x\) -axis, bounded by the ellipse with vertices (a, 0) and \((-a, 0),\) and minor axis \(2 b\). (The area of an ellipse is \(\pi a b\).

Step-by-Step Solution

Answer

The centroid of the region is \((0, \frac{4b}{3\pi})\).

Step 1: Understand the Problem

We need to find the coordinates of the centroid for the region above the x-axis of an ellipse. Given an ellipse centered at the origin with the vertices \((a, 0)\) and \((-a, 0)\), the full ellipse equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

Step 2: Set Up the Ellipse Equation and Region Limits

The area of interest is the upper-half ellipse above the x-axis, so we consider only the top half, where \( y \geq 0\). The ellipse equation \(y\) can be written as \(y = b\sqrt{1-\frac{x^2}{a^2}}\). We need to integrate from \(-a\) to \(a\).

Step 3: Calculate the Area of the Region

Since the area of the full ellipse is given as \(\pi ab\), the area of the top half is \(\frac{1}{2}\pi ab\). This represents the total area of the region above the x-axis.

Step 4: Determine the Centroid Formula

The coordinates of the centroid \( (\bar{x}, \bar{y}) \) for such a symmetric region can be computed using symmetry properties and integrals. Here, due to symmetry of the ellipse about the y-axis, \( \bar{x} = 0 \). We need to focus now on finding \( \bar{y} \).

Step 5: Calculate \(\bar{y}\) Coordinate

The formula for \( \bar{y} \) in this scenario is \( \bar{y} = \frac{1}{Area} \int_{-a}^{a} \int_{0}^{b\sqrt{1-x^2/a^2}} y \, dy \, dx \). Compute this integral. After the integration process, we find \( \bar{y} = \frac{4b}{3\pi} \).

Step 6: Conclude the Coordinates of the Centroid

By the calculations \( \bar{x} = 0 \) due to symmetry and \( \bar{y} = \frac{4b}{3\pi} \), the centroid of the region is \( (0, \frac{4b}{3\pi}) \).

Key Concepts

Ellipse Geometry

An ellipse is an elongated circle with a unique shape characterized by its width and height. Think of it like a circle that has been stretched out.
The primary elements defining an ellipse are the **major** and **minor axes**:

• The **major axis** is the longest diameter of the ellipse, stretching from one end to the other.
• The **minor axis** is the shortest diameter, intersecting the major axis at the center of the ellipse.

Additionally, the **vertices** of an ellipse are the endpoints located on the major axis. In our given problem, the ellipse is centered at the origin with vertices at a and -a along the x-axis, making the major axis 2a in length, and the minor axis 2b in length, along the y-direction.

Integral Calculus

Integral calculus is essential for calculating areas and centroids of complex shapes like ellipses. It helps us find the total sum or 'accumulated value' over a specific interval.
For calculating the **centroid**, we utilize integrals in a double whammy – finding the area of the shape and locating its balance point. The **centroid** is like the balancing point of the shape if it were made of a thin, evenly distributed material.
We use the integral to determine \(ar{y}\), which is done by integrating over one variable first and then the other. The integral used is: \[ ar{y} = \frac{1}{\text{Area}} \int_{-a}^{a} \int_{0}^{b\sqrt{1-x^2/a^2}} y \, dy \, dx \]The first step is to tackle the inner integral, which accounts for all heights (y-values) from the x-axis to the ellipse top at a particular x. Then the outer integral adds all these up across from -a to a, the complete span of the ellipse at the x-axis.

Coordinate System

A coordinate system helps us to precisely locate points on a plane through two numbers, i.e., coordinates. These pairings (x, y) specify locations in a **Cartesian Plane**.
This system forms the backbone for our ellipse equation, which in its standard form is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\)

• The \(x\)-coordinate specifies how far a point is from the y-axis horizontally.
• The \(y\)-coordinate measures how far a point is vertically from the x-axis.

In our centroid calculation, we're focused on the upper half of the ellipse with \(y \geq 0\), our aim is to find the center of mass or average position of all points in the positive \(y\)-section. The integral calculus method integrates these coordinates to find the exact position of the centroid within the ellipse.

Symmetry in Mathematics

Symmetry simplifies many mathematical problems. It refers to the balanced and proportionate similarity of parts of a figure and helps in diverse calculations like centroids.
In our problem, the ellipse is symmetrical about the y-axis, meaning that the shape is evenly divided horizontally.

• This symmetry tells us that the \(\bar{x}\)-coordinate, which measures the centroid horizontally from the center, is 0.
• This halves our task – allowing us to focus only on finding \(\bar{y}\), the vertical center.

Symmetry reduces complexity and computation as it exploits predictable properties of shapes – letting us deduce rather than calculate every part. In many centroid determinations, symmetry ensures we only need to calculate one coordinate direction!