Problem 285 - Calculus Volume 1

Question

For the following exercises, use a calculator to draw the region enclosed by the curve. Find the area \(M\) and the centroid \((\overline{x}, \overline{y})\) for the given shapes. Use symmetry to help locate the center of mass whenever possible. Quarter-circle: \(y=\sqrt{1-x^{2}}, \quad y=0, \quad\) and \(x=0\)

Step-by-Step Solution

Answer

The area is \( M = \frac{\pi}{4} \), and the centroid is \( (\frac{4}{3\pi}, \frac{4}{3\pi}) \).

Step 1: Understand the Problem

We are tasked with finding the area and centroid of a quarter-circle. The curve described by \( y = \sqrt{1 - x^2} \) represents the top right quarter of a circle centered at the origin with a radius of 1. We will find the area \( M \) and the centroid \((\overline{x}, \overline{y})\) of this region, which is bounded by \( y = \sqrt{1-x^2} \), \( y = 0 \), and \( x = 0 \).

Step 2: Set Up the Area Integral

The area \( M \) of the quarter-circle can be found by integrating the function \( y = \sqrt{1-x^2} \) from \( x = 0 \) to \( x = 1 \). This integral represents the area under the curve from \( x = 0 \) to \( x = 1 \). So, we set up the integral for the area as:\[ M = \int_0^1 \sqrt{1 - x^2} \, dx \]

Step 3: Calculate the Area

To solve the integral \( \int_0^1 \sqrt{1 - x^2} \, dx \), use the formula for a quarter circle area in conjunction with the unit circle. The area of a quarter-circle with radius \( r = 1 \) can be directly computed as:\[ M = \frac{\pi r^2}{4} = \frac{\pi}{4} \]

Step 4: Set Up the Centroid Formulas

The centroid \((\overline{x}, \overline{y})\) can be found using the formulas:\[ \overline{x} = \frac{1}{M} \int_0^1 x \sqrt{1 - x^2} \, dx \] and \[ \overline{y} = \frac{1}{M} \int_0^1 \frac{1}{2} (1 - x^2) \, dx \] Here, \( M = \frac{\pi}{4} \), which is the area calculated in Step 3.

Step 5: Calculate the x-coordinate of the Centroid

Compute \( \overline{x} \) using the integral formula:\[ \overline{x} = \frac{4}{\pi} \int_0^1 x \sqrt{1-x^2} \, dx \] Substitute \( x = \sin(\theta) \) and solve the integral using trigonometric substitution or recognize the symmetry that simplifies this calculation for symmetrical geometric shapes resulting in \( \overline{x} = \frac{4}{3\pi} \).

Step 6: Calculate the y-coordinate of the Centroid

Compute \( \overline{y} \) using the integral formula:\[ \overline{y} = \frac{4}{\pi} \int_0^1 \frac{1}{2} (1 - x^2) \, dx \] This simplifies using the symmetry of the quarter circle around the x-axis leading to \( \overline{y} = \frac{4}{3\pi} \).

Step 7: Verify Using Symmetry

Given the symmetry of the region (quarter-circle), it follows that both centroid coordinates are equal \( \overline{x} = \overline{y} \). This consistency confirms our calculations.

Key Concepts

Quarter Circle

A quarter circle represents one-fourth of a full circle. This particular shape is often defined in terms of its position in a coordinate plane. It can be thought of as a slice from the circle, usually including the center and stretching to the edges of two 90-degree angles.
For this exercise, we are dealing with the quarter circle in the first quadrant of the unit circle. It is defined mathematically by the equation:

• \( y = \sqrt{1 - x^2} \)
• along with the lines \( y=0 \) and \( x=0 \).

This corresponds to the top-right section of the circle, providing a clear geometric representation for calculations. The radius of 1 simplifies many mathematical operations, such as calculating the area and centroid.
To help visualize this quarter circle, imagine drawing an arc from the origin to the coordinate (1,0) while adhering to the constraints of the x and y axes.

Area of a Curve

Determining the area under a curve is a fundamental concept in calculus. It involves integrating the function that defines the curve over a specified interval.
For our quarter circle, the function is given by \( y = \sqrt{1 - x^2} \). This function outlines the top boundary of the shape. Our task is to find the area beneath this curve from \( x=0 \) to \( x=1 \). This process involves setting up the integral:

• \[ M = \int_0^1 \sqrt{1-x^2} \, dx \]

Calculating this integral involves using knowledge about the circle's symmetry and specific geometry. The result is the area of the quarter circle, which can be derived using the formula for a circle's area, then truncated to a quarter size.
This calculation yields:

• \[ M = \frac{\pi}{4} \]

Reflect on the power of calculus to translate geometric concepts into precise numerical results.

Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin of a coordinate system. It provides a crucial framework for understanding many trigonometric and calculus concepts.
Our quarter circle is a part of this unit circle, specifically occupying the first quadrant. This context allows us to easily utilize properties of the unit circle, such as:

• The radius always being 1, which simplifies area calculations.
• Trigonometric identities can be directly applied.

Since the unit circle is symmetric, calculations like the area or centroid benefit from symmetry, minimizing the need for complex operations.
Understanding its geometry can help simplify many problems in trigonometry, calculus, and analytical geometry.

Symmetry in Calculus

Symmetry plays a pivotal role in simplifying many calculus problems. In this context, the symmetry of the quarter circle is particularly helpful.
The quarter circle is symmetric about its axes; this symmetry implies that certain calculations, like finding the centroid, can be simplified. For instance, when both dimensions of the shape are mirrored, the centroid's x and y coordinates are expected to be equal.
This is confirmed when we calculate both coordinates:

• \[ \overline{x} = \frac{4}{3\pi} \]
• \[ \overline{y} = \frac{4}{3\pi} \]

The consistent result shows the reliability of using symmetry in calculations. Harnessing symmetry allows us to avoid unnecessary complexity, achieving accurate results with a more straightforward approach. Consider symmetry an invaluable tool in your calculus toolkit.