Problem 11 - Calculus

Question

Find the area \(M\) and the centroid \((\bar{x}, \bar{y})\) inside curves \(11-16\). $$ y=\sqrt{1-x^{2}}, y=0, x=0 \quad \text { (quarter-circle) } $$

Step-by-Step Solution

Answer

The area is \( \frac{\pi}{4} \) and the centroid is at \( \left( \frac{4}{3\pi}, \frac{4}{3\pi} \right) \).

Step 1: Understand the Shape and Boundaries

The problem defines a quarter of a circle in the first quadrant, bounded by the curve \( y = \sqrt{1-x^2} \), the x-axis \( y = 0 \), and the y-axis \( x = 0 \). This represents one-fourth of a circle centered at the origin, with radius 1.

Step 2: Calculate Area of the Quarter Circle

The formula for the area of a circle is \( \pi \times r^2 \). Here, the radius \( r = 1 \). Since we have one-fourth of a circle, the area \( M = \frac{1}{4} \times \pi \times (1)^2 = \frac{\pi}{4} \).

Step 3: Calculate the Centroid Coordinates

The centroid (center of mass) of a quarter circle in the first quadrant with radius \( r = 1 \) can be found using symmetry properties and known results for such geometric shapes. For a quarter circle, the centroid is located at \( \bar{x} = \bar{y} = \frac{4r}{3\pi} \). Substituting \( r = 1 \), both \( \bar{x} \) and \( \bar{y} \) are \( \frac{4}{3\pi} \).

Key Concepts

Area of Quarter-Circle

Understanding the area of a quarter-circle is essential when solving problems involving circular segments. To find the area of a full circle, we use the formula \( \pi \times r^2 \), where \( r \) is the radius. Since a quarter-circle is only one-fourth of a full circle, we take the area of the entire circle and multiply it by \( \frac{1}{4} \). This gives us the formula for the area of the quarter-circle:- \( M = \frac{1}{4} \times \pi \times r^2 \)In the given exercise, the quarter-circle has a radius of 1. Therefore, by substituting \( r = 1 \), the area \( M \) becomes \( \frac{\pi}{4} \). It’s crucial to comprehend this calculation as it demonstrates how simple proportions apply to basic geometric shapes.

Integration in Polar Coordinates

Integration in polar coordinates simplifies calculations for circular and radial shapes. It is particularly useful because polar coordinates naturally fit the shape and behavior of circular segments.To integrate in polar coordinates, we use the transformation from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\), where:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

The differential area element \( dA \) in polar coordinates is given by \( r \, dr \, d\theta \). For the quarter-circle, the range of \( r \) is from 0 to 1, and \( \theta \) is from 0 to \( \frac{\pi}{2} \). By setting up the double integral of the differential area over these bounds, we confirm our earlier area calculation through integration:\[M = \int_0^{\pi/2} \int_0^1 r \cdot r \, dr \, d\theta = \frac{\pi}{4}\]This approach provides a deeper understanding of how integration can verify and support geometric area calculations.

Symmetry in Geometry

Symmetry is a fundamental property of many geometric shapes that helps in identifying important features, such as centroids. For a quarter-circle located in the first quadrant, symmetry greatly simplifies the calculation of the centroid.The centroids of symmetrical shapes can often be determined without the need for complex computations. For instance, the centroid of a quarter-circle in the first quadrant with radius \( r = 1 \) utilizes symmetry along the axes.Through symmetry and known results for quarter-circles, the centroid coordinates \( (\bar{x}, \bar{y}) \) are determined as

• \( \bar{x} = \bar{y} = \frac{4r}{3\pi} \)

For the given problem with \( r = 1 \), this results in:- \( \bar{x} = \bar{y} = \frac{4}{3\pi} \)Recognizing symmetry in geometry not only reduces computational effort but also enriches our geometric intuition, allowing for quick identification of central points such as centroids.