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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Use Exercise 22 to find the centroid of a quarter-circular region of radius a.

To determine

To find: The centroid of the quarter-circular region of radius a.

Explanation

Given data:

Radius is a.

Formula used:

Write the expression for centroid (x¯,y¯) .

x¯=12ACx2dy (1)

y¯=12ACy2dx (2)

The region of quarter-circle region is shown in Figure 1.

Consider a quarter-circle as C. From Figure 1, the curve C is,

C=C1+C2+C3

Here,

C1,C2andC3 are curves.

Consider the curves as,

C1:x=t,y=0,0ta;C2:x=acost,y=asint,0tπ2;C3:x=0,y=at,0ta;

Find the value of dx for curve C1 .

ddt(x)=ddt(t)dxdt=1 {ddx(x)=1}dx=dt

Find the value of dy for curve C1 .

ddt(y)=ddt(0)dydt=0 {ddx(0)=0}dy=0dt

Find the value of dx for curve C2 .

ddt(x)=ddt(acost)dxdt=a(sint) {ddx(cosx)=sinx}dx=asintdt

Find the value of dy for curve C2 .

ddt(y)=ddt(asint)dydt=acost {ddx(sint)=cost}dy=acostdt

Find the value of dx for curve C3 .

ddt(x)=ddt(0)dxdt=0 {ddx(0)=0}dx=0dt

Find the value of dy for curve C3 .

ddt(y)=ddt(at)dydt=01 {ddx(k)=0,ddx(x)=1}dy=dt

Write the expression for area of quarter-circle (A) .

A=14πr2

Here,

r is radius.

Substitute a for r,

A=14πa2

Substitute 14πa2 for A and C1+C2+C3 for C in equation (1),

x¯=12(14πa2)C1+C2+C3x2dy=1(12πa2)C1+C2+C3x2dy

x¯=2πa2(C1x2dy+C2x2dy+C3x2dy) (3)

Find the value of C1x2dy+C2x2dy+C3x2dy .

C1x2dy+C2x2dy+C3x2dy=0a(t)2(0dt)+0π2(acost)2(acostdt)+0a(0)2(dt)=0+0π2(a3cos3t)dt+0=a30π2(cos2t)costdt=a30π2(1sin2t)costdt{cos2t=1sin2t}

C1x2dy+C2x2dy+C3x2dy=a30π2(costsin2tcost)dt=a3[sint13sin3t]0π2=a3[(sin(π2)13sin3(π2))(sin(0)13sin3(0))]=a3[(113(1)3)(013(0)3)]

Simplify the equation

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Chapter 16 Solutions

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