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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

A thin wire has the shape of the first-quadrant part of the circle with center the origin and radius a. If the density function is ρ(x, y) = kxy, find the mass and center of mass of the wire.

To determine

To find: The mass of a thin wire which has the shape of the first-quadrant part of the circle with center the origin and radius a and center of mass of a thin wire which has the shape of the first-quadrant part of the circle with center the origin and radius a .

Explanation

Given data:

The mass of a thin wire which has the shape of the first-quadrant part of the circle with center the origin and radius a . The linear density function is ρ(x,y)=kxy .

Formula used:

Write the expression to find the mass of a thin wire which is in the shape of a curve C .

m=Cρ(x,y)ds (1)

Here,

ρ(x,y) is the linear density and

ds is the change in surface.

Write the expression to find ds .

ds=(dxdt)2+(dydt)2dt (2)

Write the expression to find the x-coordinate of a center of mass of the thin wire.

x¯=1mCxρ(x,y)ds (3)

Here,

m is the mass of the thin wire.

Write the expression to find the y-coordinate of a center of mass of the thin wire.

y¯=1mCyρ(x,y)ds (4)

As the wire is in the shape of the first-quadrant part of the circle with center the origin and radius a , the parametric equations and limits of the scalar parameter t are considered as follows.

x=acost,y=asint,0tπ2

Calculation of ds :

Substitute acost for x and asint for y in equation (2),

ds=(ddtacost)2+(ddtasint)2dt=(asint)2+(acost)2dt=a2sin2t+a2cos2tdt=a2(sin2t+cos2t)dt

Simplify the expression as follows.

ds=a2(1)dt {cos2θ+sin2θ=1}=adt

Calculation of mass m :

Substitute kxy for ρ(x,y) , adt for ds , 0 for lower limit, and π2 for upper limit in equation (1),

m=0π2kxy(adt)=ak0π2xydt

Substitute acost for x and asint for y ,

m=ak0π2(acost)(asint)dt=a3k0π2(cost)(sint)dt=a3k0π2(ddtsint)(sint)dt

Use the following formula to compute the expression.

[ddtf(t)][f(t)]ndt=[f(t)]n+1n+1

Compute the expression as follows.

m=a3k[sin2t2]0π2=a3k2[sin2t]0π2=a3k2[sin2(π2)sin2(0)]=a3k2[(1)20]

m=12ka3

Thus, the mass of a thin wire which has the shape of the first-quadrant part of the circle with center the origin and radius a is 12ka3_

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Chapter 16 Solutions

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