   Chapter 16.2, Problem 36E

Chapter
Section
Textbook Problem

Find the mass and center of mass of a wire in the shape of the helix x = t, y = cos t, z = sin t, 0 ⩽ t ⩽ 2π, if the density at any point is equal to the square of the distance from the origin.

To determine

To find: The mass of a thin wire which is in the shape of a helix and center of mass of a thin wire which is in the shape of a helix.

Explanation

Given data:

The parametric equations of the helix shaped thin wire are given as follows.

x=t,y=cost,z=sint,0t2π

The density function of the thin wire is the square of the distance from the origin.

Formula used:

Write the expression to find the mass of the thin wire.

m=Cρ(x,y,z)ds (1)

Here,

ρ(x,y,z) is the density function of the wire and

ds is the change in surface.

Write the expression for ds .

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

Write the expression to find density function.

ρ(x,y,z)=d2 (3)

Here,

d is the distance between any point to the wire to the origin.

Write the expression to find distance between two points (x1,y1,z1) and (x2,y2,z2) .

d=(x2x1)2+(y2y1)2+(z2z1)2 (4)

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

x¯=1mCxρ(x,y,z)ds (5)

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

y¯=1mCyρ(x,y,z)ds (6)

Write the expression to find z-coordinate of center of mass of the thin wire:

z¯=1mCzρ(x,y,z)ds (7)

Consider a point (x,y,z) on the thin wire.

As the density function ρ(x,y,z) is the square of the distance between any point on the wire to the origin, find the distance between the point (x,y,z) to the origin as follows.

Substitute x for x2 , y for y2 , z for z2 , 0 for x1 , 0 for y1 , and 0 for z1 in equation (4),

d=(x0)2+(y0)2+(z0)2=x2+y2+z2

Calculation of density function ρ(x,y,z) :

Substitute x2+y2+z2 for d in equation (3),

ρ(x,y,z)=(x2+y2+z2)2=x2+y2+z2

Substitute t for x , cost y , and sint for z ,

ρ(x,y,z)=(t)2+(cost)2+(sint)2=t2+cos2t+sin2t=t2+1 {cos2θ+sin2θ=1}

Calculation of ds :

Substitute t for x , cost for y , and sint for z in equation (2),

ds=(ddtt)2+(ddtcost)2+(ddtsint)2dt=(1)2+(sint)2+(cost)2dt=1+sin2t+cos2tdt=1+(1)dt

ds=2dt

Calculation of mass m :

Substitute (t2+1) for ρ(x,y,z) , 2dt for ds , 0 for lower limit, and 2π for upper limit in equation (1),

m=02π(t2+1)(2dt)=202π(t2+1)dt=2[t33+t]02π=2{[(2π)33+2π](033+0)}

Simplify the expression as follows.

m=2[(8π33+2π)0]=2(83π3+2π)

Thus, the mass of a helix shaped thin wire is 2(83π3+2π)_ .

Calculation of x¯ :

Substitute 2(83π3+2π) for m , t for x , (t2+1) for ρ(x,y,z) , 2dt for ds , 0 for lower limit, and 2π for upper limit in equation (5),

x¯=12(83π3+2π)02π(t)(t2+1)(2dt)=22(83π3+2π)02π(t3+t)dt=1(83π3+2π)[t44+t22]02π=34(8π3+6π)[t4+2t2]02π

Simplify the expression as follows

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