9. Line Integrals

a. Arc Length

Recall that the arc length of a parametric curve \(\vec r(t)=(x(t),y(t),z(t))\) between \(A=\vec r(a)\) and \(B=\vec r(b)\) is given by \[ L=\int_A^B\,ds =\int_A^B \sqrt{dx^2+dy^2+dz^2} =\int_a^b |\vec v|\,dt \] where the (scalar) differential of arclength is \[\begin{aligned} ds&=\sqrt{dx^2+dy^2+dz^2} \\ &=\sqrt{\left(\dfrac{dx}{dt}\right)^2 +\left(\dfrac{dy}{dt}\right)^2+\left(\dfrac{dz}{dt}\right)^2}\,dt =|\vec v|\,dt \end{aligned}\] and \(\vec v=\left(\dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right)\) is the velocity of the curve.

Notice that we switch the limits on the integral from the abstract points \(A\) and \(B\) to the values \(a\) and \(b\) of the parameter \(t\) when we switch from the integral with the general differential \(ds\) to the integral with the parameter differential \(dt\). This is because there may be more than one way to parametrize the curve and we don't want to prejudice the choice when we write the general integral.

If the curve happens to be specified by giving \(2\) of the coordinates as functions of the third (independent) coordinate, then the independent coordinate may be taken as the parameter so that the arclength differential may be taken as \[\begin{aligned} ds&=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2+\left(\dfrac{dz}{dx}\right)^2\,}\,dx &\text{for functions of }x \\ &=\sqrt{\left(\dfrac{dx}{dy}\right)^2+1+\left(\dfrac{dz}{dy}\right)^2\,}\,dy &\text{for functions of }y \\ &=\sqrt{\left(\dfrac{dx}{dz}\right)^2+\left(\dfrac{dy}{dz}\right)^2+1\,}\,dz &\text{for functions of }z \end{aligned}\]

Several examples and exercises appear in the previous chapter. Here are a couple more:

Find the length of the curve \(\vec r(t) =\left(2\ln t,2\sqrt{2}t,t^2\right)\) from \(t=1\) to \(t=2\)

Look for a perfect square inside the square root.

\(L=2\ln2+3\)

We first compute the velocity vector: \[ \vec v(t)=\left\langle \dfrac{2}{t},2\sqrt{2},2t\right\rangle \] and its magnitude: (Notice the perfect square inside the square root.) \[\begin{aligned} |\vec v|&=\sqrt{\dfrac{4}{t^2}+8+4t^2} =\sqrt{\left(\dfrac{2}{t}+2t\right)^2} \\ &=\dfrac{2}{t}+2t \end{aligned}\] Now we can compute the length of the curve: \[\begin{aligned} L&=\int_1^2 |\vec v|\,dt =\int_1^2 \left(\dfrac{2}{t}+2t\right)\,dt \\ &=\left[2\ln t+t^2\rule{0pt}{10pt}\right]_1^2 =2\ln2+3 \end{aligned}\]

Find the length of the curve \(x=\dfrac{1}{\sqrt{2}}z^2\) and \(y=\dfrac{1}{3}z^3\) for \(z\) between \(0\) and \(3\).

Look for a perfect square inside the square root.

\(L=12\)

We first compute the derivatives: \[ \dfrac{dx}{dz}=\sqrt{2}z \qquad \text{and} \qquad \dfrac{dy}{dz}=z^2 \] So the length of the curve is: (Notice the perfect square inside the square root.) \[\begin{aligned} L&=\int_0^3 \sqrt{\left(\dfrac{dx}{dz}\right)^2+\left(\dfrac{dy}{dz}\right)^2+1\,}\,dz \\ &=\int_0^3 \sqrt{2z^2+z^4+1\,}\,dz \\ &=\int_0^3 (1+z^2)\,dz =\left[z+\dfrac{z^3}{3}\right]_0^3 =12 \end{aligned}\]

You can also practice computing the arc length of curves using the following Maplet (requires Maple on the computer where this is executed):

Arc Length in 3D Rate It

9. Line Integrals

a. Arc Length

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