Scalar Line Integrals

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Objectives:

1.: Know what a scalar line integral is and what it represents geometrically.
2.: Understand the scaling factor when computing a scalar line integral.
3.: Be able to compute scalar line integrals.

Recap and Understanding

First, take a look at this recap video going over the basics of scalar line integrals.

Important things out of the video:

We have a procedure for computing scalar line integrals:
To compute the scalar line integral $\displaystyle \int _C f(x,y)ds$ :
- Parametrize the curve $C$ as $\textbf {r}(t)$
- Compute $\int _C fds = \int _a^b f(\textbf {r}(t)) \| \textbf {r}'(t)\| dt.$
Geometrically, the scalar line integral of any function $f(x,y)$ over a curve $C$ represents what?
The arc length of the curve $C$ The signed area under the function $z = f(x,y)$ above the curve $C$
We can get the arc length of a finite curve $C$ by taking the scalar line integral over $C$ of the function $f(x,y) = \answer {1}$ .

Example Video

Take a look at the following video working through the following problem:

If $C$ is the portion of $x=y^2$ from $(1,-1)$ to $(9,-3)$ , and $f(x,y) = \sqrt {x}$ , evaluate $\displaystyle \int _C f(x,y)ds$ .

Something to think about: what purpose does the $\| \textbf {r}'(t)\|$ serve in the formula for a scalar line integral?

There are two ways to think about this. We learned the formula for arc length of a curve already, and that formula essentially says that if $\Delta s$ is a small change along the curve, then $\Delta s \approx \| \textbf {r}'(t)\| \Delta t$ , and it becomes exact when taking a limit.

The other way to look at it is to say that $r$ bends and stretches a time interval $[a,b]$ into the curve $C$ . The $\|\textbf {r}'(t)\|$ is giving you how much you are stretching the interval, which will consequently scale your area by the same amount.

Problems

If $f(x,y) = \sqrt {1+9xy}$ and $C$ is the portion of $y = x^3$ from $(0,0)$ to $(1,1)$ , evaluate $\displaystyle \int _C f(x,y)ds$ .

We follow the steps described in the video.

We first parametrize $C$ . For this curve, we could let $x = t$ , in which case $y = \answer {t^3}$ and our parametrization is: $\textbf {r}(t) = \left \langle t,\answer {t^3}\right \rangle .$
The bounds for $t$ when we do this: $t$ must be in the interval $[\answer {0},\answer {1}]$ .
In this case, $f(\textbf {r}(t)) = \answer {\sqrt {1+9t^4}}$
We also compute $\| \textbf {r}'(t) \|$ : we get $\textbf {r}'(t) = \left \langle \answer {1},\answer {3t^2} \right \rangle$ , and so $\|\textbf {r}'(t)\| = \answer {\sqrt {1+9t^4}}$
The integral can therefore be written as: $\int _C f(x,y)ds = \int _a^b f(\textbf {r}(t)) \| \textbf {r}'(t) \| dt = \int _{\answer {0}}^{\answer {1}} \answer {1+9t^4}dt.$
Evaluating this integral gives an answer of $\answer {\frac {14}{5}}$

As we will see in this next problem, the process is exactly the same for functions of three variables. Just apply the formula in the same way, but now your parametrization will have three components instead of two (but still only one parameter!).

Integrate the function $f(x,y,z) = x+y+z$ over $C$ , which is the portion of the helix given by $\textbf {r}(t) = \left \langle \cos (t),\sin (t),t \right \rangle$ for $0 \leq t \leq \pi$ .

We already have the parametrization and the bounds for $t$ , so we have fewer steps:

We compute $\textbf {r}'(t) = \left \langle \answer {-\sin (t)},\answer {\cos (t)},\answer {1} \right \rangle .$
The magnitude is $\| \textbf {r}'(t) \| = \answer {\sqrt {2}}$
$f(\textbf {r}(t)) = \answer {\cos (t)+\sin (t)+t}$
The integral can be written as $\int _C f(x,y,z) ds = \int _{\answer {0}}^{\answer {\pi }} \answer {\sqrt {2}(\cos (t)+\sin (t)+t)}dt.$
The integral evaluates to: $\answer {2\sqrt {2}+\frac {\sqrt {2}}{2}\pi ^2}$

To find the arc length of the curve $C$ in the previous problem, we would have to integrate $f(x,y,z) = \answer {1}$

If $f(x,y,z) = z^2$ and $C$ is the curve $\textbf {r}(t) = \left \langle 2t,3t,4t \right \rangle$ for $0 \leq t \leq 2$ , then $\int _C f(x,y,z)ds = \answer {\frac {128\sqrt {29}}{3}}$

If $f(x,y,z) = xe^{z^2}$ , and $C$ is the piecewise linear path from $(0,0,1)$ to $(0,2,0)$ to $(1,1,1)$ (so two line segments), then $\int _C f(x,y,z) ds = \answer {\frac {\sqrt {3}}{2}(e-1)}$

True/False

Here are some true/false questions to get you thinking a bit more about scalar line integrals.

If $f(x,y)$ is a continuous function with $f(x,y) \geq 0$ at all points, and $C$ is a finite curve, then $\displaystyle \int _C f(x,y)ds \geq 0$ .

True False

If $f(x,y)=3$ , and $C$ is a curve of length $5$ , then $\displaystyle \int _C f(x,y)ds=15$ .

True False

If $f(x,y)$ is a continuous function with $\displaystyle \int _C f(x,y)ds \leq 0$ , then $f(x,y) \leq 0$ for every point on the curve $C$ .

True False