/*! 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 6 Finding a centroid Find the cent... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 6

Finding a centroid Find the centroid of the region between the \(x\) -axis and the arch \(y=\sin x, 0 \leq x \leq \pi\)

Short Answer

Expert verified
The centroid is at \(\left( \frac{\pi}{2}, \frac{\pi}{4} \right)\)."

Step by step solution

01

Understand the Centroid

To find the centroid of a region, we need to calculate the average location of the area. For a 2D region bounded by a curve from \(x = a\) to \(x = b\), the centroid \((\bar{x}, \bar{y})\) is given by the formulas:\[\bar{x} = \frac{1}{A} \int_a^b x f(x) \, dx\]\[\bar{y} = \frac{1}{A} \int_a^b \frac{f(x)^2}{2} \, dx\]where \(A\) is the area of the region.
02

Calculate the Area (A)

We first need to calculate the area \(A\) under the curve \(y = \sin x\) from \(x = 0\) to \(x = \pi\). This is done by integrating \(\sin x\) over the interval:\[A = \int_0^\pi \sin x \, dx\]The integral of \(\sin x\) is \(-\cos x\), so:\[A = [-\cos x]_0^\pi = -\cos(\pi) + \cos(0) = 2\]
03

Calculate \(\bar{x}\)

Use the formula for \(\bar{x}\):\[\bar{x} = \frac{1}{A} \int_0^\pi x \sin x \, dx\]Using integration by parts: let \(u = x\) and \(dv = \sin x \, dx\), then \(du = dx\) and \(v = -\cos x\). The integral becomes:\[\int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x\]Evaluating this from 0 to \(\pi\):\[\int_0^\pi x \sin x \, dx = [-x \cos x + \sin x]_0^\pi = [0 + 0] - [0 + 0] = \pi\]So: \[\bar{x} = \frac{1}{2} \cdot \pi = \frac{\pi}{2}\]
04

Calculate \(\bar{y}\)

Use the formula for \(\bar{y}\):\[\bar{y} = \frac{1}{A} \int_0^\pi \frac{(\sin x)^2}{2} \, dx\]This is:\[= \frac{1}{2A} \int_0^\pi (1 - \cos(2x))/2 \, dx\] (using the identity \(\sin^2 x = \frac{1 - \cos(2x)}{2}\))\[= \frac{1}{4} \cdot \left[ x - \frac{1}{2} \sin(2x) \right]_0^\pi\]\[= \frac{1}{4} [\pi - 0] = \frac{\pi}{4}\]
05

Final Centroid Coordinates

The coordinates of the centroid are the values of \(\bar{x}\) and \(\bar{y}\) calculated previously:\(\bar{x} = \frac{\pi}{2}\) and \(\bar{y} = \frac{\pi}{4}\).Therefore, the centroid of the region is located at \(\left( \frac{\pi}{2}, \frac{\pi}{4} \right)\).

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Key Concepts

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

Integration by Parts
Integration by Parts is a technique based on the product rule for differentiation. It's useful when integrating the product of two functions. The formula for Integration by Parts is:\[\int u \, dv = uv - \int v \, du\]Here:
  • \(u\) is a function that we can differentiate easily.
  • \(dv\) is a part of the integral we can integrate quickly.
For example, in finding \(\bar{x}\) for the centroid, we set \(u = x\) and \(dv = \sin x \, dx\). Then:
  • We differentiate \(u\) to get \(du = dx\).
  • We integrate \(dv\) to get \(v = -\cos x\).
Substituting, we get the integral as:\[-x \cos x + \int \cos x \, dx\]When evaluating the integral limits from 0 to \(\pi\), it shows how Integration by Parts simplifies the process, providing a neat solution.
Area under Curve
The area under the curve is a fundamental concept in calculus. It represents the total space beneath a curve and above the x-axis between two points. For the curve \(y = \sin x\), the area \(A\) is found by computing the definite integral from 0 to \(\pi\):\[A = \int_0^\pi \sin x \, dx\]The integral \(\int \sin x \, dx\) leads to \(-\cos x\). At the limits, this results in:\[-\cos(\pi) + \cos(0) = 2\]Remember, calculating the area accurately is crucial for centroid problems because accurate values of \(A\) ensure correct calculations for \(\bar{x}\) and \(\bar{y}\). Without this prerequisite, centroid computations would be incorrect.
Trigonometric Integrals
Trigonometric Integrals can appear intimidating, but with the right identities and techniques, they become manageable. In calculating \(\bar{y}\), the integral involves \(\sin^2 x\), which is transformed using the identity:\(\sin^2 x = \frac{1 - \cos(2x)}{2}\)Implementing this change simplifies the integral:\[\frac{1}{2A} \int_0^\pi \left(\frac{1 - \cos(2x)}{2}\right) \, dx\]Further breaking down the integral into parts makes its evaluation easier. By evaluating over the interval \([0, \pi]\), the integral becomes manageable, yielding:\[\frac{\pi}{4}\]Such identities are invaluable for making the evaluation of trigonometric integrals straightforward. Knowing and applying these can be a fantastic simplifier when dealing with sinusoids and their powers.

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Most popular questions from this chapter

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \begin{equation} \int_{0}^{2 \sqrt{\ln 3}} \int_{y / 2}^{\sqrt{\ln 3}} e^{x^{2}} d x d y \end{equation}

Use a CAS double-integral evaluator to estimate the values of the integrals. \begin{equation}\int_{0}^{1} \int_{0}^{1} \tan ^{-1} x y d y d x\end{equation}

Noncircular cylinder \(A\) solid right (noncircular) cylinder has its base \(R\) in the \(x y-\) plane and is bounded above by the paraboloid \(z=x^{2}+y^{2} .\) The cylinder's volume is \begin{equation}V=\int_{0}^{1} \int_{0}^{y}\left(x^{2}+y^{2}\right) d x d y+\int_{1}^{2} \int_{0}^{2-y}\left(x^{2}+y^{2}\right) d x d y\end{equation} Sketch the base region \(R\) and express the cylinder's volume as a single iterated integral with the order of integration reversed. Then evaluate the integral to find the volume.

Find the volume of the solid cut from the square column \(|x|+|y| \leq 1\) by the planes \(z=0\) and \(3 x+z=3 .\)

In polar coordinates, the average value of a function over a region \(R\) (Section 15.3\()\) is given by $$\frac{1}{\text { Area(R) }} \iint_{R} f(r, \theta) r d r d \theta$$ Volume of noncircular right cylinder The region enclosed by the lemniscate \(r^{2}=2 \cos 2 \theta\) is the base of a solid right cylinder whose top is bounded by the sphere \(z=\sqrt{2-r^{2}}\) . Find the cyl- inder's volume.
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