/*! 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 21 Find the centroid of the region ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 21

Find the centroid of the region bounded by the graphs of the given equations. $$ y=x^{3}, \quad y=\sqrt[3]{x}, \quad x=0, \quad x=1 $$

Short Answer

Expert verified
The centroid of the region bounded by the given equations is \(\left(\frac{61}{56}, \frac{16}{35}\right)\).

Step by step solution

01

Identify bounding curves and set up integrals

First, let's identify the curves bounding the region: - Left boundary: \(x=0\) - Right boundary: \(x=1\) - Lower curve: \(y=x^3\) - Upper curve: \(y=\sqrt[3]{x}\) Now we will set up the integrals for the x-coordinate of the centroid, \(\bar{x}\), and the y-coordinate of the centroid, \(\bar{y}\): \[- A=\int_a^b (f(x) - g(x)) dx \] \[ \bar{x} =\frac{1}{A}\int_a^b x(f(x) - g(x)) dx \] \[ \bar{y} =\frac{1}{A}\int_a^b \frac{1}{2}(f^2(x) - g^2(x)) dx \] Where: - \(f(x)\) = upper curve, in this case, \(\sqrt[3]{x}\) - \(g(x)\) = lower curve, in this case, \(x^3\) - \(a\) = left boundary, in this case, \(0\) - \(b\) = right boundary, in this case, \(1\)
02

Find the area A

Now let's calculate the area A using the formula: \[ A = \int_0^1 (\sqrt[3]{x} - x^3) dx \] We can compute the integral: \[A = \left[\frac{3}{4}x^{4/3} - \frac{1}{4}x^4 \right]_0^1 =\left(\frac{3}{4}(1)^{4/3} - \frac{1}{4}(1)^4\right)-\left(\frac{3}{4}(0)-\frac{1}{4}(0)\right)=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\]
03

Find the x-coordinate of the centroid, \(\bar{x}\)

Next, find the x-coordinate (\(\bar{x}\)) by calculating the integral: \[ \bar{x} = \frac{1}{A}\int_0^1 x(\sqrt[3]{x} - x^3) dx = \frac{2}{1}\int_0^1 x(\sqrt[3]{x} - x^3) dx = 2\int_0^1 x(\sqrt[3]{x} - x^3) dx \] We calculate the integral: \[ 2\int_0^1 x(\sqrt[3]{x} - x^3) dx = 2\left[ \frac{3}{7}x^{7/3} - \frac{1}{8}x^8\right]_0^1 = 2\left(\frac{3}{7}(1)^{7/3} - \frac{1}{8}(1)^8 - (0)\right) =2\left(\frac{3}{7} - \frac{1}{8}\right) = \frac{61}{56} \]
04

Find the y-coordinate of the centroid, \(\bar{y}\)

Finally, find the y-coordinate (\(\bar{y}\)) of the centroid: \[ \bar{y} = \frac{1}{A}\int_0^1 \frac{1}{2}(\sqrt[3]{x^2} - x^6) dx = 2\int_0^1\frac{1}{2}(\sqrt[3]{x^2} - x^6) dx = \int_0^1 (\sqrt[3]{x^2} - x^6) dx \] We calculate the integral: \[ \int_0^1 (\sqrt[3]{x^2} - x^6) dx = \left[ \frac{3}{5}x^{5/3} - \frac{1}{7}x^7 \right]_0^1 = \left(\frac{3}{5}(1)^{5/3} - \frac{1}{7}(1)^7 \right) - \left(\frac{3}{5}(0)^{5/3} - \frac{1}{7}(0)^7 \right) = \frac{16}{35} \]
05

Find the centroid's coordinates

Now that we have the x-coordinate \(\bar{x}\) and the y-coordinate \(\bar{y}\) of the centroid, the centroid is given by the coordinates: \[(\bar{x}, \bar{y}) = \left(\frac{61}{56}, \frac{16}{35}\right) \] So the centroid of the region bounded by the graphs of the given equations is \(\left(\frac{61}{56}, \frac{16}{35}\right)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

find the given integral. \(\int \cosh (2 x+3) d x\)

Prove that the centroid of a triangular region is located at the point of intersection of the medians of the triangle. Hint: Suppose that the vertices of the triangle are located at \((0,0)\), \((a, 0)\), and \((b, h) .\)

In Example 6 we showed that the work done by an expanding gas against a piston as its volume expands from \(V_{0}\) to \(V_{1}\) is given by $$ W=\int_{V_{0}}^{v_{1}} p d V $$ where \(p\) is the pressure of the gas. If the pressure and volume of a gas are related by the equation \(p V=k\), where \(k\) is a positive constant, show that \(W=k \ln \left(V_{1} / V_{0}\right)\).

Find the area of the surface obtained by revolving the given curve about the indicated axis. $$ y=4-x^{2} \text { on }[0,2] ; \quad y \text { -axis } $$

Prove the identity. \(\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.