9. Line Integrals

c. Applications of Scalar Line Integrals

2. Average Value

The centroid of an object can be thought of as the average values of \(x\), \(y\) and \(z\) within the object, while the center of mass gives the "weighted" averages of \(x\), \(y\) and \(z\). These ideas can be generalized to give the average and weighted average of any function over the object.

Average Value

In Calculus 2, we learned how to compute the the average value of a function on an interval. This can be generalized to the average value of a function on any curve.

The average value of a function \(f(x,y,z)\) along the curve \(\vec r(t)=(x(t),y(t),z(t))\) between \(A=\vec r(a)\) and \(B=\vec r(b)\) is \[ f_\text{ave}=\dfrac{\displaystyle \int_A^B f\,ds}{\displaystyle \int_A^B 1\,ds} =\dfrac{\displaystyle \int_a^b f(\vec r(t))\,|\vec v|\,dt}{\displaystyle \int_a^b |\vec v|\,dt} \] Of course the integral in the denominator is just the length of the curve.

A temperature probe is shaped like the helix \(\vec r(\theta)=(4\cos\theta,4\sin\theta,3\theta)\) for \(0 \le \theta \le 2\pi\). It is immersed in a hot fluid whose temperature is \(T(x,y,z)=(275+x^2+y^2+z^2)^\circ\text{K}\). Find the average temperature measured by the probe.

The average value of a tempersture \(T(x,y,z)\) along the curve \(\vec r(t)=(x(t),y(t),z(t))\) between \(A=\vec r(a)\) and \(B=\vec r(b)\) is \[ T_\text{ave} =\dfrac{\displaystyle \int_A^B T\,ds} {\displaystyle \int_A^B 1\,ds} =\dfrac{\displaystyle \int_a^b T(\vec r(\theta))\,|\vec v|\,d\theta} {\displaystyle \int_a^b |\vec v|\,d\theta} \]

\(T_\text{ave}=291+12\pi^2\approx409.4^\circ\text{K}\)

To find the average temperature we divide the integral of the temperature function over the curve, by the total length of the curve itself. Thus the average temperature is \[ T_\text{ave} =\dfrac{\displaystyle \int_A^B T\,ds} {\displaystyle \int_A^B 1\,ds} =\dfrac{\displaystyle \int_0^{2\pi} T(\vec r(\theta))\,|\vec v|\,d\theta} {\displaystyle \int_0^{2\pi} |\vec v|\,d\theta} \] where \(\vec r(\theta)=(4\cos\theta,4\sin\theta,3\theta)\) and \(A=\vec r(0)=(4,0,0)\), and \(B=\vec r(2\pi)=(4,0,6\pi)\). So \(\vec v(\theta)=(-4\sin\theta,4\cos\theta,3)\) and its magnitude is \(|\vec v|=5\). The denominator is the length of the curve: \[ \int_A^B |\vec v|\,d\theta=\int_0^{2\pi} 5\,dt=10\pi \] The restriction of the temperature to the curve is: \[\begin{aligned} T(\vec r(t))&=275+x(t)^2+y(t)^2+z(t)^2 \\ &=275+16\cos^2\theta+16\sin^2\theta+9\theta^2 =291+9\theta^2 \end{aligned}\] So the numerator is: \[\begin{aligned} \int_A^B &T(\vec r(t))\,|\vec v|\,d\theta =\int_0^{2\pi} (291+9\theta^2)\,(5)\,d\theta \\ &=5\left[291\theta+3\theta^3\rule{0pt}{10pt}\right]_0^{2\pi} =5(291(2\pi)+3(2\pi)^3) \\ &=2910\pi+120\pi^3 \end{aligned}\] So, the average temperature measured by the probe is \[ T_\text{ave}=\dfrac{2910\pi+120\pi^3}{10\pi} =291+12\pi^2\approx409.4^\circ\text{K} \]

Weighted Average

Sometimes we want to count a function more heavily on some parts of the curve than others. (Say on the parts of the curve which have more mass, i.e. have more weight, i.e. is heavier.)

The weighted average of a function \(f(x,y,z)\) along the curve \(\vec r(t)=(x(t),y(t),z(t))\) between \(A=\vec r(a)\) and \(B=\vec r(b)\) with weight factor \(\delta(x,y,z)\) is \[ f_{\delta\text{-ave}} =\dfrac{\displaystyle \int_A^B f\delta\,ds} {\displaystyle \int_A^B \delta\,ds} =\dfrac{\displaystyle \int_a^b f(\vec r(t))\,\delta(\vec r(t))\,|\vec v|\,dt} {\displaystyle \int_a^b \delta(\vec r(t))\,|\vec v|\,dt} \] The integral in the denominator is the total weight (or mass) of the curve.

A straight wire extends from \(P=(3,4,2)\) to \(Q=(1,3,4)\). It's mass increases linearly along its length with density \(\delta=z\). Its temperature varies along its length according to \(T=(276+x-y)^\circ\text{K}\). What is the average temperature on the wire, weighted more heavily where the wire is more dense?

Recall the parametric equation of a line is \(X=P+t\vec v\). So the wire may be parametrized as \[ \vec r(t)=P+t\overrightarrow{PQ} \] What are \(\delta\) and \(T\) on the wire?

\( T_{\delta\text{-ave}}=\dfrac{2470}{9} \approx274.4^\circ\text{K} \)

The wire extends from \(P\) to \(Q\). So its velocity is \(\vec v=\overrightarrow{PQ}=\langle -2,-1,2\rangle\). Then. it may be parametrized as: \[\begin{aligned} \vec r(t)&=P+t\overrightarrow{PQ}=(3,4,2)+t(-2,-1,2) \\ &=(3-2t,4-t,2+2t) \end{aligned}\] for \(0 \le t \le 1\). The speed is \(|\vec v|=\sqrt{4+1+4}=3\). On the wire: \[ \delta=z=2+2t \] and \[ T=276+x-y=276+(3-2t)-(4-t)=275-t \] To find the weighted average temperature we must compute: \[ T_{\delta\text{-ave}} =\dfrac{\displaystyle \int_P^Q T\delta\,ds} {\displaystyle \int_P^Q \delta\,ds} \] The denominator is the total mass: \[ \int_P^Q \delta\,ds =\int_0^1 (2+2t)3\,dt =3\left[2t+t^2\rule{0pt}{10pt}\right]_0^1 =9 \] The numerator is: \[\begin{aligned} \int_P^Q &T\,\delta\,ds =\int_0^1 (275-t)(2+2t)3\,dt \\ &=3\int_0^1 (550+548t-2t^2)\,dt \\ &=3\left[550t+274t^2-\dfrac{2}{3}t^3\right]_0^1 =2470 \end{aligned}\] So, the average weighted temperature along the wire is \[ T_{\delta\text{-ave}}=\dfrac{2470}{9} \approx274.4^\circ\text{K} \]

The temperatures and densities at the endpoints are: \[\begin{aligned} T(P)&=275 &\delta(P)&=2 \\ T(Q)&=274 &\delta(Q)&=4 \end{aligned}\] The average temperatue \(T_{\delta\text{-ave}}\approx274. 44^\circ\text{K}\) is reasonable because it is between the temperatures at the endpoints but closer to that at the endpoint which is denser.

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