13
$\begingroup$

I ran into a problem that, initially, I thought was a typo.

$$\int_C\ e^xdx $$

where C is the arc of the curve $x=y^3$ from $(-1,-1)$ to $(1,1)$.

I have only encountered line integrals with $ds$ before, not $dx$ (or $dy$, for that matter). At first, I thought the $dx$ was supposed to be a $ds$, but that led to an unsolvable integral.

Unfortunately, my book does not cover this topic very well, and the online answers I have found are rather vague. I tried plugging in $x=y^3$ to get $$ \int_{-1}^{1}\ e^{y^3}3y^2dy $$

which evaluates to $e - \frac{1}{e}$. This is also what I get when I do $ \int_{-1}^{1}\ e^{x}dx $, so I am inclined to believe it is correct. However, I'm not sure, and I'd like to know for certain if my intuition is valid.

$\endgroup$
1
  • 1
    $\begingroup$ I think you mean “Line integral not with respect to arc length” in your title, right? $\endgroup$ Dec 2, 2016 at 15:01

5 Answers 5

11
$\begingroup$

Let $C$ be any curve in $\mathbb{R}^d$. A parametrization of $C$ is a map $\gamma: [a,b] \to \mathbb{R}^d$ which trace along the points on $C$ in a specific order. For any two functions $f$, $g$ defined on the set of points belong to $C$, we define the line integral over $C$ by

$$\int_C f dg \stackrel{def}{=} \int_\gamma f dg \stackrel{def}{=}\int_a^b f(\gamma(t))\,(g \circ \gamma)'(t) dt$$

i.e. the line integral is defined through a integral over a specific parametrization of the curve. The key is the value of the integral on the right is independent of the choice of parametrization. For clarity, one can drop the explicit parameter $t$ from the expression.

For your case, $\int e^{x} dx$ really means $\int e^{x(\gamma(t))}\,(x\circ\gamma)'(t) dt$ for whatever parametrization you choose to evaluate the integral. The $s$ you usually see stands for the arc length parametrization, it is only one possible choice of parametrization. You don't need to use it if it make your life harder.

$\endgroup$
2
  • $\begingroup$ Just to be clear, was I correct in using $\int_{-1}^{1}\ e^{x}dx$? I think I understand what you're saying, but I'd like to be sure. $\endgroup$
    – Somatic
    Dec 2, 2016 at 6:14
  • 1
    $\begingroup$ Yes. if the integrand can be expressed as a deriviative of a single value function of the integration variable, then the usual rule of evaluate $1d$-integral by taking difference of the antiderivative at the end points continue to work. $\endgroup$ Dec 2, 2016 at 6:17
5
$\begingroup$

Define $\vec{r}(t) = <t^3,t>$. Define $\vec{F}(t)= <e^{t^3}, 0>$. Then you want $\int_C\vec{F}\cdot d\vec{r}$, which can now be integrated along the $t$ line from -1 to 1. But this is looking at it as a vector field not a scalar field. But it is consistent with the notation.

$\endgroup$
2
$\begingroup$

It is fairly standard to write $\int_C P (x,y)dx+Q (x,y)dy $ for the line integral $\int_C\vec F\cdot d\vec r $, where $\vec F (x,y)=(P (x,y),Q (x,y))$. So in this case $\vec F (x,y)=(e^x ,0)$.

Here one can solve as you did; or you can notice that $\vec F $ is conservative, with potential function $f (x,y)=e^x $. Thus $$\int_C e^x\,dx=f (1,1)-f (-1,-1)=e-\frac1e. $$

It is important to note all this works because $P $ depends only on $x $.

$\endgroup$
1
$\begingroup$

You need to find $s.$

Do you know how to find the length of an arc?

$S(t) = \int_1^t \sqrt {\frac {dx}{dt}^2+ \frac {dy}{dt}^2} dt$

$ds = \sqrt {\frac {dx}{dt}^2+ \frac {dy}{dt}^2} dt$

In this case let $y=t, x=t^3$

$\int_{-1}^{1} e^{t^3} \sqrt {9t^4+ 1} dt$

Not sure how to calculate that integral outright. I would need to do a numerical aproximation.

$\endgroup$
1
  • $\begingroup$ There is no $ds$ in the problem. It's over $dx$. I did what you did at first, until I realized I was misreading it. I know there is an exact answer. $\endgroup$
    – Somatic
    Dec 2, 2016 at 5:37
1
$\begingroup$

This is an example of line integral with respect to $x$. Line integrals with respect to coordinate functions are not entirely intuitive (to me). One possible motivation for such a thing would be computing work. If you roll a snowball up a hill, you do work. The work depends on the mass of the snowball, which depends on how long you've been rolling it, and the elevation gain of the hill. But any horizontal component to the rolling takes no work, since the only force is gravity. Thus, if the hill has a profile in the shape of a curve $C$, and $f(x,y)$ is the mass of the snowball at position $(x,y)$ on the hill, ($y$ is the vertical coordinate), then the work is proportional to $\int_C f(x,y)\,dy$.

There is a simple, formal procedure to compute integrals like these. For yours, parametrize $C$ by $x=t^3$, $y=t$ over $-1 \leq t\leq 1$. Then the integrand is $e^{t^3}$, and the differential is $dx= 3t^2 \,dt$. The integral becomes $$ \int_C e^{x}\,dx=\int_{-1}^t e^{t^3} 3t^2\,dt = \left. e^{t^3}\right|^{1}_{-1} = e - e^{-1} $$

Actually, I think the formality is the major advantage to these integrals. If $\mathbf{F} = P\mathbf{i} + Q \mathbf{j} + R\mathbf{k}$ is a vector field on $C$, then $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_C P\,dx + \int_C Q\,dy + \int_C R\,dz $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.