# Math Insight

### Examples of scalar line integrals

#### Example 1

Let \begin{align*} \dllp(t) = (3t-2,t+1), \qquad 1 \le t \le 2 \end{align*} be a parameterization of a wire. Let the density of the wire at point $(x,y)$ be given by \begin{align*} f(x,y)= x + y. \end{align*} Compute the mass of the wire.

Solution: Mass is the integral of the density along the wire. So we must compute $\int_{\dllp} f\, ds$. Since, \begin{align*} \dllp'(t) &= (3,1),\\ \| \dllp'(t) \| &= \sqrt{3^2+1^2} = \sqrt{10},\\ f(\dllp(t)) &= (3t-2)+(t+1) =4t-1, \end{align*} the integral is \begin{align*} \int_{\dllp} f\, ds &= \int_a^b f(\dllp(t)) \| \dllp'(t)\| dt\\ &= \int_{1}^2 (4t-1) \sqrt{10} dt\\ &= (2 t^2 - t)\left.\left.\sqrt{10} \right|_{1}^2\right.\\ &= \left(8-2-(2-1)\right) \sqrt{10} = 5\sqrt{10} \end{align*}

So, if $f$ were given in grams/cm and $\dllp(t)$ were given in cm, then the mass of the wire would $5\sqrt{10}$ grams.

#### Example 2

Neither the length of the wire nor its mass can depend on parametrization. Check that you get the same answer with the parametrization \begin{align*} \adllp(t) = (9t-2,3t+1), \qquad 1/3 \le t \le 2/3. \end{align*} Not that this is the same wire as in Example 1, which is a straight line from the point $(1,2)$ to $(4,3)$.

Solution: We repeat the same calculations as in Example 1. (Recall that $f(x,y)=x+y$.) Since \begin{align*} \adllp'(t) &= (9,3),\\ \| \adllp'(t) \| &= \sqrt{9^2+3^2} = \sqrt{90} = 3 \sqrt{10},\\ f(\adllp(t)) &= (9t-2)+(3t+1) =12t-1, \end{align*} the integral is \begin{align*} \int_{\adllp} f\, ds &= \int_{1/3}^{2/3} f(\adllp(t)) \| \adllp'(t)\| dt\\ &= \int_{1/3}^{2/3} (12t-1) 3\sqrt{10} dt\\ &= (6 t^2 - t)\left.\left.3\sqrt{10} \right|_{1/3}^{2/3}\right.\\ &= \left[6\left(\frac{2}{3}\right)^2-\frac{2}{3} -\left(6\left(\frac{1}{3}\right)^2 -\frac{1}{3}\right)\right] 3\sqrt{10}\\ &= \left[\frac{24}{9} - \frac{2}{3} - \frac{6}{9} + \frac{1}{3}\right] 3\sqrt{10} = \frac{5}{3} 3\sqrt{10} = 5\sqrt{10}, \end{align*} which matches Example 1.

Note that the “speed” of the parametrization $\| \adllp'(t) \|= 3 \sqrt{10}$ was three times greater than the speed $\| \dllp'(t) \|= \sqrt{10}$ of Example 1. But the range of integration $1/3 \le t \le 2/3$ was one third the range of integration from Example 1.

#### Example 3

Here's another parametrization of the same straight-line wire from $(1,2)$ to $(4,3)$. This time, the “speed” of the parametrization is not constant, but depends on t: \begin{align*} \sadllp(t) = (3t^2-2,t^2+1), \qquad 1 \le t \le \sqrt{2}. \end{align*} Still using the density $f(x,y)=x+y$, calculate the mass of the wire.

Solution: Since \begin{align*} \sadllp'(t) &= (6t,2t),\\ \| \sadllp'(t) \| &= \sqrt{(6t)^2+(2t)^2} = \sqrt{40t^2} = 2t \sqrt{10},\\ f(\sadllp(t)) &= (3t^2-2)+(t^2+1) =4t^2-1. \end{align*} the integral is \begin{align*} \int_{\sadllp} f\, ds &= \int_{1}^{\sqrt{2}} f(\sadllp(t)) \|\sadllp'(t)\| dt\\ &= \int_{1}^{\sqrt{2}} (4t^2-1) 2t\sqrt{10} dt\\ &= \int_{1}^{\sqrt{2}} (8t^3-2t)\sqrt{10} dt\\ &= (2t^4 - t^2)\left.\left.\sqrt{10} \right|_{1}^{\sqrt{2}}\right.\\ &= \left[2\bigl(\sqrt{2}\bigr)^4 - \bigl(\sqrt{2}\bigr)^2 -2 +1\right] \sqrt{10} = 5\sqrt{10}, \end{align*} which matches Examples 1 and 2.

#### Example 4

For the slinky $\dlc$ parametrized by $\dllp(t)=(\cos t, \sin t, t)$ for $0 \le t \le 2\pi$, let its density at point $(x,y,z)$ be given by $f(x,y,z) = 1+x+z$. Find the mass of the slinky.

Solution: Since \begin{align*} f(\dllp(t)) &= 1+\cos t + t\\ \dllp'(t) &= (-\sin t, \cos t, 1)\\ \|\dllp'(t)\| &= \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}, \end{align*} the line integral giving the mass is \begin{align*} \slint{\dlc}{f} &= \pslint{0}{2\pi}{f}{\dllp}\\ &= \int_0^{2\pi}(1+\cos t + t)\sqrt{2}dt\\ &= \sqrt{2}(t + \sin t + t^2/2)\Big|_0^{2\pi} = 2\sqrt{2}\pi + 2\sqrt{2}\pi^2. \end{align*}