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Locate the centroid of the shaded areas:

Locate the centroid of the shaded areas:

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Locate the centroid of the shaded areas:

In this illustration, a geometric region is shown on an \(xy\)-coordinate plane. The graph depicts a shape commonly studied in integral calculus and physics for calculating areas and volumes. The region is bounded by the curves \( y^2 = x \) and \( y + x = 2 \), and it spans from \( x = 0 \) to \( x = 2 \). ### Detailed Explanation: **Coordinate Axes:** - The \(x\)-axis is horizontally oriented, while the \(y\)-axis is vertically oriented. - The area of interest lies in the first quadrant (where both \(x\) and \(y\) are positive). **Bounding Equations:** 1. **Parabolic Curve: \( y^2 = x \)** - This equation represents a parabola that opens to the right. - At \(y = 1\), \(y^2 = 1^2 = 1\), hence \(x = 1\). - The parabola is symmetric along the \(x\)-axis. 2. **Linear Line: \( y + x = 2 \)** - This linear equation can be rearranged to \( y = 2 - x \). - The line intersects the \(y\)-axis at \( (0, 2) \) and the \(x\)-axis at \( (2, 0) \). **Intersection Points:** - To find the intersection of \( y^2 = x \) and \( y + x = 2 \): 1. Solve \(y + y^2 = 2\) 2. Rearrange into the quadratic equation \( y^2 + y - 2 = 0 \). 3. The solutions are \( y = 1 \) and \( y = -2 \). Only \( y = 1 \) is relevant as the region is in the first quadrant. - Substituting \( y = 1 \) into \( y^2 = x \), we get \( x = 1 \). - Therefore, the curves intersect at \( (1, 1) \). **Dimensions:** - The figure's height extends from the \(x\)-axis (0) to 1 unit above it. - The width extends from \(x = 0\) at the leftmost point to \(x = 2\
Transcribed Image Text:In this illustration, a geometric region is shown on an \(xy\)-coordinate plane. The graph depicts a shape commonly studied in integral calculus and physics for calculating areas and volumes. The region is bounded by the curves \( y^2 = x \) and \( y + x = 2 \), and it spans from \( x = 0 \) to \( x = 2 \). ### Detailed Explanation: **Coordinate Axes:** - The \(x\)-axis is horizontally oriented, while the \(y\)-axis is vertically oriented. - The area of interest lies in the first quadrant (where both \(x\) and \(y\) are positive). **Bounding Equations:** 1. **Parabolic Curve: \( y^2 = x \)** - This equation represents a parabola that opens to the right. - At \(y = 1\), \(y^2 = 1^2 = 1\), hence \(x = 1\). - The parabola is symmetric along the \(x\)-axis. 2. **Linear Line: \( y + x = 2 \)** - This linear equation can be rearranged to \( y = 2 - x \). - The line intersects the \(y\)-axis at \( (0, 2) \) and the \(x\)-axis at \( (2, 0) \). **Intersection Points:** - To find the intersection of \( y^2 = x \) and \( y + x = 2 \): 1. Solve \(y + y^2 = 2\) 2. Rearrange into the quadratic equation \( y^2 + y - 2 = 0 \). 3. The solutions are \( y = 1 \) and \( y = -2 \). Only \( y = 1 \) is relevant as the region is in the first quadrant. - Substituting \( y = 1 \) into \( y^2 = x \), we get \( x = 1 \). - Therefore, the curves intersect at \( (1, 1) \). **Dimensions:** - The figure's height extends from the \(x\)-axis (0) to 1 unit above it. - The width extends from \(x = 0\) at the leftmost point to \(x = 2\
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