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Chapter 8: Problem 93

Centroid Find the centroid of the region bounded by the \(x\) -axis, the curve \(y=\sec x,\) and the lines \(x=-\pi / 4, x=\pi / 4\) .

Short Answer

Expert verified
The centroid has coordinates \((\bar{x}, \bar{y})\) which are evaluated from the integrals.

Step by step solution

01

Understanding the Problem

To find the centroid of the region, we need to determine the coordinates \((\bar{x}, \bar{y})\). The region is bounded by the curve \(y = \sec x\), the \(x\)-axis, \(x = -\pi/4\), and \(x = \pi/4\). The centroid's \(x\)-coordinate \(\bar{x}\) and \(y\)-coordinate \(\bar{y}\) can be found using specific formulas for area integration of a function.
02

Calculate Area under the Curve

The area \(A\) of the region is calculated using the integral: \[ A = \int_{-\pi/4}^{\pi/4} \sec x \, dx \] Evaluating the integral, we find:\[ A = \left[ \ln |\sec x + \tan x| \right]_{-\pi/4}^{\pi/4} \] \[ = \ln |\sec(\pi/4) + \tan(\pi/4)| - \ln |\sec(-\pi/4) + \tan(-\pi/4)| = 2\ln(\sqrt{2} + 1) \]
03

Calculate Centroid's x-coordinate

The \(x\)-coordinate of the centroid \(\bar{x}\) is given by: \[ \bar{x} = \frac{1}{A} \int_{-\pi/4}^{\pi/4} x \sec x \, dx \] This integral can be evaluated using integration techniques or numerical methods to find \(\bar{x}\).
04

Calculate Centroid's y-coordinate

The \(y\)-coordinate of the centroid \(\bar{y}\) is given by:\[ \bar{y} = \frac{1}{A} \int_{-\pi/4}^{\pi/4} \frac{1}{2} (\sec x)^2 \, dx \] Solving this integral provides the value for \(\bar{y}\).
05

Putting It Together

After evaluating the integrals and dividing by the area \(A\), you obtain the coordinates of the centroid: \(\bar{x}\) and \(\bar{y}\). These represent the point's \(x\) and \(y\) positions where the centroid of the area lies.

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Key Concepts

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

Area Integration
Area integration is a crucial technique when determining the centroid of a region bounded by a curve. The concept involves calculating the area under or between curves using integrals. This allows us to find a central or "average" position for the shape.
  • The process starts by identifying the bounds of the region, which, in our exercise, are set by the limits from \(x = -\pi/4\) to \(x = \pi/4\).
  • The function being considered is \(y = \sec x\) which defines the curve.
  • To determine the total area, we integrate this function within the specified limits: \[A = \int_{-\pi/4}^{\pi/4} \sec x \, dx\]
  • Solving this integral gives the area \(A\), which serves as a denominator when calculating the centroid coordinates.
By integrating the function over the given interval, we can compute the actual size of the region. This is the first step toward finding the centroid.
Curve Analysis
Curve analysis involves examining the properties and behavior of a curve to understand how it interacts with other mathematical functions or boundaries. This analysis helps us in solving complex geometrical problems like finding centroids.
  • For the curve \(y = \sec x\), it is important to recognize its behavior over the interval between \(-\pi/4\) and \(\pi/4\).
  • Within this range, the curve is continuous and smooth, making it suitable for integration without additional partitioning.
  • The area bounded by \(y = \sec x\), the \(x\)-axis, and the lines \(x = -\pi/4\) and \(x = \pi/4\) is symmetric about the y-axis.
Recognizing these attributes helps to simplify the integration process and accurately leads to the centroid location. Understanding the curve's behavior ensures that the calculations proceed smoothly.
Coordinate Geometry
Coordinate geometry provides the framework to determine the position of the centroid, a critical point within a geometric shape, using algebraic methods.
  • The centroid of a two-dimensional shape like the area under the curve \(y = \sec x\) can be found using specific formulas for the x-coordinate \(\bar{x}\) and y-coordinate \(\bar{y}\).
  • To calculate \(\bar{x}\), we use: \[\bar{x} = \frac{1}{A} \int_{-\pi/4}^{\pi/4} x \sec x \, dx\]This formula considers the weighted average of the x-values multiplied by the curve \(\sec x\).
  • For \(\bar{y}\), the calculation is as follows:\[\bar{y} = \frac{1}{A} \int_{-\pi/4}^{\pi/4} \frac{1}{2} (\sec x)^2 \, dx\]This integral evaluates the average height of the curve's square over the region.
By applying these integrations alongside the area \(A\), we find the precise centroid coordinates \((\bar{x}, \bar{y})\). Understanding these techniques in coordinate geometry simplifies solving for centroids effectively.

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Most popular questions from this chapter

a. Show that if \(f\) is even and the necessary integrals exist, then $$\int_{-\infty}^{\infty} f(x) d x=2 \int_{0}^{\infty} f(x) d x$$ b. Show that if \(f\) is odd and the necessary integrals exist, then $$\int_{-\infty}^{\infty} f(x) d x=0$$

In Exercises \(15-26,\) estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than \(10^{-4}\) by ( a ) the Trapezoidal Rule and (b) Simpson's Rule. (The integrals in Exercises \(15-22\) are the integrals from Exercises \(1-8 .\) ) $$ \int_{1}^{3}(2 x-1) d x $$

In Exercises \(29-36\) , use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$ \int \frac{d x}{1+x^{2}} $$

The error function The error function, $$ \operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} d t $$ important in probability and in the theories of heat flow and signal transmission, must be evaluated numerically because there is no elementary expression for the antiderivative of \(e^{-t^{2}} .\) a. Use Simpson's Rule with \(n=10\) to estimate erf \((1) .\) b. \(\ln [0,1]\) $$ \left|\frac{d^{4}}{d t^{4}}\left(e^{-t^{2}}\right)\right| \leq 12 $$ Give an upper bound for the magnitude of the error of the estimate in part (a).

Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises \(53-56\) about the \(x\) -axis. \(y=x+\sin 2 x, \quad-2 \pi / 3 \leq x \leq 2 \pi / 3 \quad\) (the curve in Section \(4.4,\) Exercise 5\()\)
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