/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Find the centroid of the region ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 30

Find the centroid of the region bounded by the given curves. $$ y=2-x^{2}, \quad y=x $$

Short Answer

Expert verified
The centroid is at \( \left(-\frac{1}{6}, \frac{17}{18}\right) \).

Step by step solution

01

Identify Points of Intersection

To find the points of intersection for the curves \( y = 2 - x^2 \) and \( y = x \), set the equations equal to each other: \( 2 - x^2 = x \). Rearrange to form a quadratic equation: \( x^2 + x - 2 = 0 \). Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = -2 \), we find \( x = 1 \) and \( x = -2 \). Thus, the points of intersection are \( (1, 1) \) and \( (-2, -2) \).
02

Set Up Integrals for Centroid Formula

The centroid \((\bar{x}, \bar{y})\) for a region bounded by two curves \( y = f(x) \) and \( y = g(x) \) with limits \( a \) to \( b \) is given by: \[ \bar{x} = \frac{1}{A} \int_{a}^{b} x[f(x) - g(x)] \, dx \] and \[ \bar{y} = \frac{1}{2A} \int_{a}^{b} [f(x)^2 - g(x)^2] \, dx, \] where \( A \) is the area of the region, \[ A = \int_{a}^{b} [f(x) - g(x)] \, dx. \]
03

Calculate Area (A) of the Region

Calculate the area \( A \) between the curves from \( x = -2 \) to \( x = 1 \): \[ A = \int_{-2}^{1} [(2 - x^2) - x] \, dx = \int_{-2}^{1} (2 - x^2 - x) \, dx. \] Simplify and integrate: \[ A = \int_{-2}^{1} (2 - x - x^2) \, dx = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{1}. \] Evaluate: \[ A = \left( 2(1) - \frac{(1)^2}{2} - \frac{(1)^3}{3} \right) - \left( 2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3} \right), \] which simplifies to \( A = \frac{9}{2} \).
04

Calculate \( \bar{x} \)

Calculate the x-coordinate of the centroid \( \bar{x} \): \[ \bar{x} = \frac{1}{A} \int_{-2}^{1} x[(2-x^2) - x] \, dx = \frac{1}{\frac{9}{2}} \int_{-2}^{1} (2x - x^2 - x^3) \, dx. \] Simplify and integrate: \[ \bar{x} = \frac{2}{9} \left[ x^2 - \frac{x^3}{3} - \frac{x^4}{4} \right]_{-2}^{1}. \] Evaluate: \[ \bar{x} = \frac{2}{9} \left( (1 - \frac{1}{3} - \frac{1}{4}) - (4 - \frac{-8}{3} - 4) \right), \] which results in \( \bar{x} = -\frac{1}{6} \).
05

Calculate \( \bar{y} \)

Calculate the y-coordinate of the centroid \( \bar{y} \): \[ \bar{y} = \frac{1}{2A} \int_{-2}^{1} [(2-x^2)^2 - x^2] \, dx = \frac{1}{2 \times \frac{9}{2}} \int_{-2}^{1} [(4 - 4x^2 + x^4) - x^2] \, dx. \] Simplify to \[ \bar{y} = \frac{1}{9} \int_{-2}^{1} (4 - 5x^2 + x^4) \, dx. \] Integrate: \[ \bar{y} = \frac{1}{9} \left[ 4x - \frac{5x^3}{3} + \frac{x^5}{5} \right]_{-2}^{1}. \] Evaluate: \[ \bar{y} = \frac{1}{9} \left( 4(1) - \frac{5}{3} + \frac{1}{5} - (4(-2) - \frac{5(-2)^3}{3} + \frac{(-2)^5}{5}) \right), \] which simplifies to \( \bar{y} = \frac{17}{18} \).
06

State the Centroid Coordinates

The coordinates of the centroid, where the area is equally distributed, are \( \bar{x} = -\frac{1}{6} \) and \( \bar{y} = \frac{17}{18} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Points of Intersection
Finding the points of intersection is a crucial step when determining the centroid of a region bounded by curves. In this case, we need to locate where the two curves, given by the equations \( y = 2 - x^2 \) and \( y = x \), intersect. This is done by setting the equations equal to each other:
  • \( 2 - x^2 = x \)
Rearranging the terms gives a quadratic equation:
  • \( x^2 + x - 2 = 0 \)
We can solve this using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, \) and \( c = -2 \). When calculated, the solutions are \( x = 1 \) and \( x = -2 \). These \( x \)-values help form the intersection points \((1, 1)\) and \((-2, -2)\), crucial for the next steps in centroid calculation.
Integrals for Area Calculation
To find the centroid of a region, we must first calculate the area between the curves. The area \( A \) between two curves \( y = f(x) \) and \( y = g(x) \) on an interval from \( a \) to \( b \) can be determined using an integral:
  • \[ A = \int_{a}^{b} [f(x) - g(x)] \, dx \]
This integral accounts for the vertical distance between the curves over the interval from \( x = -2 \) to \( x = 1 \), which contains both intersection points. Substituting the functions, we integrate:
  • \( A = \int_{-2}^{1} ((2 - x^2) - x) \, dx = \int_{-2}^{1} (2 - x - x^2) \, dx \)
Simplifying and evaluating this integral leads to the area \( A = \frac{9}{2} \). This area value is fundamental to finding the centroid coordinates.
Quadratic Equations
Quadratic equations are commonly encountered in various mathematical problems, including those involving points of intersection and integral bounds. In our problem, the intersection points of the curves \( y = 2 - x^2 \) and \( y = x \) necessitate rewriting the issue as a quadratic equation:
  • \( x^2 + x - 2 = 0 \)
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), which allows us to apply the quadratic formula. Here, the coefficients are \( a = 1 \), \( b = 1 \), and \( c = -2 \). Applying the formula:
  • \( x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \)
This simplification gives us roots \( x = 1 \) and \( x = -2 \). Understanding how to derive and solve these equations is essential for identifying intersection points accurately.
Centroid Formula
The centroid of a region provides the "average" position of all the points in the area. To find the coordinates \((\bar{x}, \bar{y})\) of the centroid for our bounded region, we use specific formulas. The x-coordinate and y-coordinate are given by:
  • \( \bar{x} = \frac{1}{A} \int_{a}^{b} x[f(x) - g(x)] \, dx \)
  • \( \bar{y} = \frac{1}{2A} \int_{a}^{b} [f(x)^2 - g(x)^2] \, dx \)
In our case, the area \( A = \frac{9}{2} \) previously calculated is substituted into these equations. Solve the respective integrals for \( \bar{x} \) and \( \bar{y} \):
  • \( \bar{x} = -\frac{1}{6} \)
  • \( \bar{y} = \frac{17}{18} \)
These centroid coordinates represent the balance point of the region bounded by the given curves. Calculating these integrals requires careful simplification but yields the centroid's exact location.

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