Question

Find the mass and the center of mass of a wire loop in the shape of a helix (measured in cm: x = t, y = 4 cos(t), z = 4 sin(t) for 0 ≤ t ≤ 2π), if the density (in grams/cm) of the wire at any point is equal to the square of the distance from the origin to the point

Answer 1

Mass

`√(17)(\displaystyle(8\pi^3)/(3)+32\pi)`

Center of mass

Coordinate x

`\displaystyle((\displaystyle((2\pi)^4)/(4)+32\pi))/((\displaystyle(8\pi^3)/(3)+32\pi))`

Coordinate y

`\displaystyle(16\pi)/((\displaystyle(8\pi^3)/(3)+32\pi))`

Coordinate z

`\displaystyle(-16\pi)/((\displaystyle(8\pi^3)/(3)+32\pi))`

Explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W

`m=\displaystyle\int_(W)D(x,y,z)=\displaystyle\int_(0)^(2\pi)D(x(t),y(t),z(t))||W'(t)||dt`

The density D(x,y,z) is given by

`D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16`

on the other hand

`||W'(t)||=√(1^2+(-4sin(t))^2+(4cos(t))^2)=√(1+16)=√(17)`

and we have

`m=\displaystyle\int_(W)D(x,y,z)=\displaystyle\int_(0)^(2\pi)D(x(t),y(t),z(t))||W'(t)||dt=\\\\√(17)\displaystyle\int_(0)^(2\pi)(t^2+16)dt=√(17)(\displaystyle(8\pi^3)/(3)+32\pi)`

The center of mass is the point
`(\bar x,\bar y,\bar z)`

where

`\bar x=\displaystyle(1)/(m)\displaystyle\int_(W)xD(x,y,z)\\\\\bar y=\displaystyle(1)/(m)\displaystyle\int_(W)yD(x,y,z)\\\\\bar z=\displaystyle(1)/(m)\displaystyle\int_(W)zD(x,y,z)`

We have

`\displaystyle\int_(W)xD(x,y,z)=√(17)\displaystyle\int_(0)^(2\pi)t(t^2+16)dt=\\\\=√(17)(\displaystyle((2\pi)^4)/(4)+32\pi)`

so

`\bar x=\displaystyle(√(17)(\displaystyle((2\pi)^4)/(4)+32\pi))/(√(17)(\displaystyle(8\pi^3)/(3)+32\pi))=\displaystyle((\displaystyle((2\pi)^4)/(4)+32\pi))/((\displaystyle(8\pi^3)/(3)+32\pi))`

`\displaystyle\int_(W)yD(x,y,z)=√(17)\displaystyle\int_(0)^(2\pi)4cos(t)(t^2+16)dt=\\\\=16√(17)\pi`

`\bar y=\displaystyle(16√(17)\pi)/(√(17)(\displaystyle(8\pi^3)/(3)+32\pi))=\displaystyle(16\pi)/((\displaystyle(8\pi^3)/(3)+32\pi))`

`\displaystyle\int_(W)zD(x,y,z)=4√(17)\displaystyle\int_(0)^(2\pi)sin(t)(t^2+16)dt=\\\\=-16√(17)\pi`

`\bar z=\displaystyle(-16√(17)\pi)/(√(17)(\displaystyle(8\pi^3)/(3)+32\pi))=\displaystyle(-16\pi)/((\displaystyle(8\pi^3)/(3)+32\pi))`