Problem 5 - Thomas Calculus

Question

Finding a centroid Find the centroid of the region cut from the first quadrant by the circle \(x^{2}+y^{2}=a^{2}\) .

Step-by-Step Solution

Answer

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

Step 1: Understand the Shape

The problem involves finding the centroid of a region in the first quadrant. The region is part of the circle defined by the equation \(x^2 + y^2 = a^2\), specifically the part in the first quadrant. This means we are only working with the quarter circle that exists in the positive \(x\) and \(y\) plane axes.

Step 2: Identify the Area of the Region

Since the full circle has an area of \(\pi a^2\), the area of the quarter circle in the first quadrant is \(\frac{1}{4}\pi a^2\). This area is essential for calculating the centroid.

Step 3: Set Up the Centroid Formula

The centroid \((\bar{x}, \bar{y})\) of a region is calculated using the coordinates: \[ \bar{x} = \frac{1}{A} \int_{R} x \, dA \] \[ \bar{y} = \frac{1}{A} \int_{R} y \, dA \]where \(R\) is the region and \(A\) is the area of the region, which we already determined is \(\frac{1}{4}\pi a^2\).

Step 4: Express the Differential Area Element

In polar coordinates, the area element \(dA\) can be expressed as \(r \, dr \, d\theta\), where \(0 \leq r \leq a\) and \(0 \leq \theta \leq \frac{\pi}{2}\).

Step 5: Calculate \(\bar{x}\)

Using polar coordinates for \(x=r \cos(\theta)\):\[ \bar{x} = \frac{1}{A} \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} (r \cos \theta) (r \, dr \, d\theta) \] \[ \bar{x} = \frac{4}{\pi a^2} \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} r^2 \cos \theta \, dr \, d\theta \]Solving this, we separate the functions of \(r\) and \(\theta\) and integrate:\[ \bar{x} = \frac{4}{\pi a^2} \left( \int_{0}^{a} r^3 \, dr \right) \left( \int_{0}^{\frac{\pi}{2}} \cos \theta \, d\theta \right) \] \[ \bar{x} = \frac{4}{\pi a^2} \left( \frac{a^4}{4} \right) \left( 1 \right) \] \[ \bar{x} = \frac{4a^2}{3\pi} \]

Step 6: Calculate \(\bar{y}\)

Similarly, calculate for \(y=r \sin(\theta)\):\[ \bar{y} = \frac{1}{A} \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} (r \sin \theta) (r \, dr \, d\theta) \] \[ \bar{y} = \frac{4}{\pi a^2} \int_{0}^{a} \int_{0}^{\frac{\pi}{2}} r^2 \sin \theta \, dr \, d\theta \]Separate and solve as before:\[ \bar{y} = \frac{4}{\pi a^2} \left( \int_{0}^{a} r^3 \, dr \right) \left( \int_{0}^{\frac{\pi}{2}} \sin \theta \, d\theta \right) \] \[ \bar{y} = \frac{4}{\pi a^2} \left( \frac{a^4}{4} \right) \left( 1 \right) \] \[ \bar{y} = \frac{4a^2}{3\pi} \]

Step 7: Determine the Centroid Coordinates

The coordinates of the centroid of the quarter circle in the first quadrant are \( (\bar{x}, \bar{y}) = \left( \frac{4a}{3\pi}, \frac{4a}{3\pi} \right) \).

Key Concepts

Coordinate Geometry

Coordinate geometry is a branch of mathematics where algebraic equations are used to describe geometric shapes and their properties in a coordinate plane. In this exercise, we use coordinate geometry to find the centroid of a region. Specifically, we deal with the equation of a circle in the standard form:

• A circle with equation \(x^2 + y^2 = a^2\) has its center at the origin \((0,0)\) and radius \(a\).
• In the first quadrant, which is the section where both \(x\) and \(y\) are positive, we have a quarter circle.

Knowing about coordinate geometry enables us to understand which part of the circle relates to our problem. Using these concepts will guide us through figuring out where the entire region is placed on the plane, which helps in figuring out the exact location of the centroid.

Polar Coordinates

Polar coordinates offer a different perspective from the Cartesian coordinate system and are especially useful in dealing with circles and regions defined by angles. Rather than using \(x\) and \(y\) to define a point, polar coordinates use

• \(r\), the radial distance from the origin, and
• \(\theta\), the angle from the positive \(x\)-axis.

This form becomes useful in calculating areas and centroids of circular regions. For the quarter circle, we use:

• \(0 \leq r \leq a\), indicating radius, and
• \(0 \leq \theta \leq \frac{\pi}{2}\), covering the angle in the first quadrant.

Expressing the differential area element in terms of \(r\) and \(\theta\) is necessary when dealing with non-linear regions like circles, helping us to calculate integrals more neatly.

Integration

Integration is a crucial mathematical tool used to calculate areas, volumes, and, in this case, centroids. The centroid is essentially the "average" location of all the points in a shape.For the quarter circle, we use integration to find \(\bar{x}\) and \(\bar{y}\), the coordinates of the centroid, by integrating over the defined region's whole area:

• For \(\bar{x}\), we integrate \(r \cos(\theta)\), which represents the \(x\)-coordinate, over the whole area.
• For \(\bar{y}\), we similarly integrate \(r \sin(\theta)\), representing the \(y\)-coordinate.

The integrals extend over \(r\) from 0 to \(a\) and \(\theta\) from 0 to \(\frac{\pi}{2}\), aligning with the radial and angular limits of our quarter circle region. This way, integration pulls together information across the whole area to find where the 'center of mass' lies.

Quarter Circle

A quarter circle is a portion of a circle formed by cutting the circle into four equal parts. In this exercise, we're examining the part located in the first quadrant of the coordinate plane:

• It covers areas where \(x\) and \(y\) are both positive.
• Its boundary includes the parts of the circle defined by the equation \(x^2 + y^2 = a^2\).

The significance of focusing on the quarter circle in this exercise is so we only calculate the centroid relative to one positive quadrant. Understanding the nature of a quarter circle and its properties helps us in breaking down and managing the calculations required for finding the centroid. This understanding supports visualizing the area over which we're applying our polar coordinates and integration methods to derive the centroid accurately.