9. Line Integrals

c. Applications of Scalar Line Integrals

1. Mass, Center of Mass and Centroid

In Calculus 2, we learned how to compute

  1. the mass and center of mass of a discrete collection of masses
  2. the mass and center of mass of a straight bar with linear density \(\delta(x)\) (with units of \(\dfrac{\text{mass}}{\text{length}}\))

This can be generalized to a curved bar or wire by using line integrals of scalars.

Total Mass and Center of Mass

If a wire has the shape of the curve \(\vec r(t)\) with linear density \(\delta(x,y,z)\) between \(A=\vec r(a)\) and \(B=\vec r(b)\) then

  1. The total mass of the wire is \[ M=\int_A^B \delta\,ds =\int_a^b \delta(r(t))|\vec v|\,dt \]
  2. The first moments of the mass of the wire in the \(x\), \(y\) and \(z\) directions are \[\begin{aligned} M_x&=\int_A^B x\delta\,ds =\int_a^b x(t)\,\delta(r(t))\,|\vec v|\,dt \\ M_y&=\int_A^B y\delta\,ds =\int_a^b y(t)\,\delta(r(t))\,|\vec v|\,dt \\ M_z&=\int_A^B z\delta\,ds =\int_a^b z(t)\,\delta(r(t))\,|\vec v|\,dt \end{aligned}\]
  3. And the center of mass is \[ (\bar{x},\bar{y},\bar{z})=\left(\dfrac{M_x}{M},\dfrac{M_y}{M},\dfrac{M_z}{M}\right) \]

Find the mass and center of mass of a wire in the shape of the semicircle \(y=\sqrt{4-x^2}\) if its linear density is given by \(\delta(x,y)=x^2y\).

eg_Mass_semicirc

The semicircle may be parametrized by \(\vec r(\theta)=(2\cos\theta,2\sin\theta)\) for \(0 \le \theta \le \pi\). Its velocity is \(\vec v=(-2\sin\theta,2\cos\theta)\) whose length is \(|\vec v|=\sqrt{4\sin^2\theta+4\cos^2\theta}=2\). So the mass is \[\begin{aligned} M&=\int_A^B \delta\,ds =\int_0^\pi x^2y|\vec v|\,d\theta \\ &=\int_0^\pi 4\cos^2\theta\,2\sin\theta\,2\,d\theta =16\left[\dfrac{-\cos^3\theta}{3}\right]_0^\pi =\dfrac{32}{3} \end{aligned}\] The first moments are \[\begin{aligned} M_{1x}&=\int_A^B x\delta\,ds =\int_0^\pi x^3y|\vec v|\,d\theta \\ &=\int_0^\pi 8\cos^3\theta\,2\sin\theta\,2\,d\theta =32\left[\dfrac{-\cos^4\theta}{4}\right]_0^\pi=0 \\ M_{1y}&=\int_A^B y\delta\,ds =\int_0^\pi x^2y^2|\vec v|\,d\theta \\ &=\int_0^\pi 4\cos^2\theta\,4\sin^2\theta\,2\,d\theta =8\int_0^\pi \sin^2(2\theta)\,d\theta \\ &=8\int_0^\pi \dfrac{1-\cos(4\theta)}{2}\,d\theta=4 \left[\theta-\,\dfrac{\sin(4\theta)}{4}\right]_0^\pi=4\pi \end{aligned}\] where we used the identities \(\sin(2\theta)=2\sin\theta\cos\theta\) and \(\sin^2 A=\dfrac{1-\cos(2A)}{2}\). So the center of mass is \[ (\bar{x},\bar{y}) =\left(\dfrac{0}{32/3},\dfrac{4\pi}{32/3}\right) =\left(0,\dfrac{3\pi}{8}\right) \]

This answer makes sense since the wire and the density are symmetric about the \(y\)-axis which implies \(\bar{x}=0\). Further, \(\bar{y}\approx 1.2\) which is within the range of the wire which is \(0 \le y \le 2\). Here is a plot with an \(\times\) at the center of mass.

eg_CM_semicirc

Find the mass and the \(x\)-component of the center of mass of the quartic curve: \[ \vec r(t)=\left(\dfrac{1}{2}t^2,\dfrac{2}{3}t^3,\dfrac{1}{2}t^4\right) \quad \text{for} \quad 0 \le t \le 1 \] if the linear density is \(\delta=x\).

The quantity in the square root is a perfect square.

\(M=\dfrac{7}{24}\)   \(M_x=\dfrac{5}{48}\)   \(\bar{x}=\dfrac{5}{14}\approx0.36\)

The velocity is \(\vec v=\langle t,2t^2,2t^3\rangle\) and its magnitude is: \[ |\vec v|=\sqrt{t^2+4t^4+4t^6}=\sqrt{(t+2t^3)^2}=t+2t^3 \] The density is \(\delta=x=\dfrac{1}{2}t^2\). So the mass is \[\begin{aligned} M&=\int_A^B \delta\,ds =\int_0^1 \dfrac{1}{2}t^2(t+2t^3)\,dt \\ &=\left[\dfrac{1}{8}t^4+\dfrac{1}{6}t^6\right]_0^1 =\dfrac{1}{8}+\dfrac{1}{6}=\dfrac{7}{24} \end{aligned}\] For the \(x\) moment, we multiply by \(x=\dfrac{1}{2}t^2\) in the integrand: \[\begin{aligned} M_x &=\int_A^B x\delta\,ds =\int_0^1 \dfrac{1}{2}t^2\dfrac{1}{2}t^2(t+2t^3)\,dt \\ &=\left[\dfrac{1}{24}t^6+\dfrac{1}{16}t^8\right]_0^1 =\dfrac{1}{24}+\dfrac{1}{16}=\dfrac{5}{48} \end{aligned}\] Finally, we divide the moment by the mass to get the \(x\)-component of the center of mass: \[ \bar x=\dfrac{M_x}{M}=\dfrac{5}{48}\dfrac{24}{7}=\dfrac{5}{14}\approx0.36 \]

Similarly, we can find the \(y\)- and \(z\)-moments and the \(y\)- and \(z\)-components of the center of mass: \[\begin{aligned} M_y&=\dfrac{23}{189} \quad &\bar y&=\dfrac{184}{441}\approx0.42 \\ M_z&=\dfrac{13}{160} \quad &\bar z&=\dfrac{39}{140}\approx0.28 \end{aligned}\] This appears to be near the center of the wire as shown in the plot.

Centroid

If the linear density along a curve is a constant, which we will take as \(\delta=1\), then

  1. The total mass of the curve is just its arclength: \[ M=\int_A^B \delta\,ds=\int_A^B 1\,ds=L \]
  2. The first moments of the mass are called the first moments of arc length: \[\begin{aligned} M_x&=\int_A^B x\delta\,ds=\int_A^B x\,ds=L_x \\ M_y&=\int_A^B y\delta\,ds=\int_A^B y\,ds=L_y \\ M_z&=\int_A^B z\delta\,ds=\int_A^B z\,ds=L_z \end{aligned}\]
  3. And the center of mass is called the centroid: \[ (\bar{x},\bar{y},\bar{z}) =\left(\dfrac{L_x}{L},\dfrac{L_y}{L},\dfrac{L_z}{L}\right) \]

Find the centroid of one and a half loops of the helix \(\vec r(\theta)=(4\cos\theta,4\sin\theta,3\theta)\) for \(0 \le \theta \le 3\pi\). Rotate the plot with your mouse.

The velocity is \(\vec v=(-4\sin\theta,4\cos\theta,3)\) and its length is \(|\vec v|=\sqrt{16\sin^2\theta+16\cos^2\theta+9}=5\). So the length is \[ L=\int_A^B\,ds=\int_0^{3\pi} |\vec v|\,d\theta =\int_0^{3\pi} 5\,d\theta=\left[5\theta\right]_0^{3\pi}=15\pi \] The first moments are \[\begin{aligned} L_x&=\int_A^B x\,ds =\int_0^{3\pi} 4\cos\theta\,5\,dt =\left[20\sin\theta\rule{0pt}{10pt}\right]_0^{3\pi}=0 \\ L_y&=\int_A^B y\,ds =\int_0^{3\pi} 4\sin\theta\,5\,dt =\left[-20\cos\theta\rule{0pt}{10pt}\right]_0^{3\pi} \\ &=-20(-1)--20(1)=40 \\ L_z&=\int_A^B z\,ds= \int_0^{3\pi} 3\theta\,5\,dt =\left[15\dfrac{\theta^2}{2}\right]_0^{3\pi}=\dfrac{135}{2}\pi^2 \end{aligned}\] And so the center of mass is \[\begin{aligned} (\bar{x},\bar{y},\bar{z}) &=\left(\dfrac{L_x}{L},\dfrac{L_y}{L},\dfrac{L_z}{L}\right) \\ &=\left(0,\dfrac{40}{15\pi},\dfrac{135\pi^2}{2\cdot15\pi}\right) =\left(0,\dfrac{8}{3\pi},\dfrac{9\pi}{2}\right) \end{aligned}\]

It is no surprise that \(\bar{x}=0\) since half of the helix has positive \(x\)'s and half has negative \(x\)'s.
It is no surprise that \(\bar{z}=\dfrac{9\pi}{2}\) since this is half of the height of the helix which is \(9\pi\).
Further, \(\bar{y}=\dfrac{8}{3\pi}\approx 0.85\) makes sense since there is more helix for positive than negative \(y\)'s and this is within the bounds \(-4 \le y \le 4\).

Find the centroid of the twisted cubic \(\vec r(t)=\left(t,t^2,\dfrac{2}{3}t^3\right)\) from \(t=0\) to \(t=1\).

Use the first moments of arc length: \[ L_x=\int_A^B x\,ds \qquad L_y=\int_A^B y\,ds \qquad L_z=\int_A^B z\,ds \] to find the centroid: \[ (\bar{x},\bar{y},\bar{z}) =\left(\dfrac{L_x}{L},\dfrac{L_y}{L},\dfrac{L_z}{L}\right) \]

\((\bar{x},\bar{y},\bar{z}) =\left(\dfrac{3}{5},\dfrac{11}{25},\dfrac{7}{30}\right)\)

First, we have already found the magnitude of the velocity vector is \(|\vec v|=1+2t^2\). The arc length is \[ L=\int_A^B\,ds=\int_0^1 (1+2t^2)\,dt =\left[t+2\dfrac{t^3}{3}\right]_0^1=\dfrac{5}{3} \] For the \(x\) moment of arc length, we we multiply by \(x=t\) in the integrand: \[ L_x=\int_A^B x\,ds=\int_0^1 t(1+2t^2)\,dt =\left[\dfrac{t^2}{2}+2\dfrac{t^4}{4}\right]_0^1=1 \] Next, for the \(y\) moment of arc length, we multiply by \(y=t^2\) in the integrand: \[ L_y=\int_A^B y\,ds=\int_0^1 t^2(1+2t^2)\,dt =\left[\dfrac{t^3}{3}+2\dfrac{t^5}{5}\right]_0^1=\dfrac{11}{15} \] Finally, for the \(z\) moment of arc length, we we multiply by \(z=\dfrac{2}{3}t^3\) in the integrand: \[ L_z=\int_A^B z\,ds=\int_0^1 \dfrac{2}{3}t^3(1+2t^2)\,dt =\dfrac{2}{3}\left[\dfrac{t^4}{4}+2\dfrac{t^6}{6} \right]_0^1=\dfrac{7}{18} \] Now we divide each moment by the arc length: \[ \bar{x}=1\cdot\dfrac{3}{5}=\dfrac{3}{5} \qquad \bar{y}=\dfrac{11}{15}\dfrac{3}{5}=\dfrac{11}{25} \qquad \bar{z}=\dfrac{7}{18}\dfrac{3}{5}=\dfrac{7}{30} \] (Notice how we divide by the arc length by multiplying by its reciprocal.) So the centroid is \[ (\bar{x},\bar{y},\bar{z}) =\left(\dfrac{3}{5},\dfrac{11}{25},\dfrac{7}{30}\right) \]

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