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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

If a wire with linear density ρ(x, y) lies along a plane curve C, its moments of inertia about the x- and y-axes are defined as

Ix =C y2 ρ(x, y) ds Iy =C x2 ρ(x, y) ds

Find the moments of inertia for the wire in Example 3.

To determine

To find: The moment of inertia for a wire about the x-axis, and y-axis.

Explanation

Given data:

The density ρ(x,y) of the wire lies along a plane curve C .

The moment of inertia of the wire about x-axis is defined as follows.

Ix=Cy2ρ(x,y)ds (1)

The moment of inertia of the wire about y-axis is defined as follows.

Iy=Cx2ρ(x,y)ds (2)

Formula used:

Write the expression for density function as follows.

ρ(x,y)=k(1y) (3)

Here,

k is the density constant.

Write the expression for ds .

ds=(dxdt)2+(dydt)2dt (4)

Consider the parametric equations of the plane curve C as follows.

x=cost,y=sint,0tπ

Calculation of density function:

Substitute sint for y in equation (3),

ρ(x,y)=k(1sint)

Calculation of ds :

Substitute cost for x and sint for in equation (4),

ds=(ddtcost)2+(ddtsint)2dt=(sint)2+(cost)2dt=sin2t+cos2tdt=1dt {cos2θ+sin2θ=1}

ds=dt

Calculation of moment of inertia about x-axis Ix :

Substitute sint for y , k(1sint) for ρ(x,y) , dt for ds , 0 for lower limit, and π for upper limit of integral in equation (1),

Ix=0π(sint)2k(1sint)dt=k0π(sin2tsin3t)dt=k0π(sin2t)dtk0π(sin3t)dt=k0π(1cos2t2)dtk0π(sin2t)(sint)dt

Simplify the expression as follows.

Ix=k20π(1cos2t)dtk0π(1cos2t)(sint)dt=k2[tsin2t2]0πk0π(1cos2t)(sint)dt=k2[(πsin2π2)(0sin2(0)2)]k0π(1cos2t)(sint)dt=k2[(π02)(0)]k0π(1cos2t)(sint)dt

Simplify the expression as follows.

Ix=kπ2+k0π(1cos2t)(sint)dt (5)

Apply substitution method.

u=costdu=sintdt

Find new limits

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Chapter 16 Solutions

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