Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Can you show me how to solve this problem?

**Problem Statement:**

Find the centroid of the quarter of the unit circle lying in the third quadrant.

**Solution:**

- \( x_{CM} = \) [Input Box]
- \( y_{CM} = \) [Input Box]

**Explanation:**

To find the centroid of the quarter of a unit circle in the third quadrant, you will need to integrate over the area of this specific quarter. The third quadrant refers to the section of the Cartesian plane where both x and y coordinates are negative. The centroid can be calculated using the formulas for centroids of a defined area by integrating over the region.

For the circle of radius 1 centered at the origin, the equations are:

1. Equation of the Circle: \( x^2 + y^2 = 1 \)

The centroid’s coordinates \((x_{CM}, y_{CM})\) can be found using:
- \( x_{CM} = \frac{1}{A} \int x \, dA \)
- \( y_{CM} = \frac{1}{A} \int y \, dA \)

Where \(A\) is the area of the region and \(dA\) is the differential area element.

Use this setup to calculate specific values for \(x_{CM}\) and \(y_{CM}\) for the precise location of the centroid in this segment of the unit circle.
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Transcribed Image Text:**Problem Statement:** Find the centroid of the quarter of the unit circle lying in the third quadrant. **Solution:** - \( x_{CM} = \) [Input Box] - \( y_{CM} = \) [Input Box] **Explanation:** To find the centroid of the quarter of a unit circle in the third quadrant, you will need to integrate over the area of this specific quarter. The third quadrant refers to the section of the Cartesian plane where both x and y coordinates are negative. The centroid can be calculated using the formulas for centroids of a defined area by integrating over the region. For the circle of radius 1 centered at the origin, the equations are: 1. Equation of the Circle: \( x^2 + y^2 = 1 \) The centroid’s coordinates \((x_{CM}, y_{CM})\) can be found using: - \( x_{CM} = \frac{1}{A} \int x \, dA \) - \( y_{CM} = \frac{1}{A} \int y \, dA \) Where \(A\) is the area of the region and \(dA\) is the differential area element. Use this setup to calculate specific values for \(x_{CM}\) and \(y_{CM}\) for the precise location of the centroid in this segment of the unit circle.
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