# Line Integral. Why parametrization is necessary?

I have a doubt regarding integration of line integrals, in the books that I refer the integrand is usually parameterized to bring it in terms of a single variable.

But I don't know why parametrization is necessary, can we just integrate the differential form?

Example : If we need to calculate $$\int \vec F\cdot\mathrm d\vec r$$ (Work) going from $$(0,0)$$ to $$(2,1)$$ over the straight line path from $$(0,0)$$ to $$(2,1)$$ i.e $$x=2y$$ and the force equation having $$x$$-component = $$xy$$ and $$y$$-component = $$y^2$$ then $$\int \vec F\cdot\mathrm d\vec r$$ will become $$\int\left((xy)\mathrm dx - (y^2)\mathrm dy\right)$$.

Can't we just integrate this by integrating the two parts of the integrand separately? Why do we need to bring both in terms of $$\mathrm dx$$ or $$\mathrm dy$$ or parameterize?

I know this might sound absurd because maybe I have not understood things properly but any help in clarifying the doubt will be appreciated.

• What do you get instead, when parametrising? Commented May 6, 2016 at 15:50
• @KennyLau Well not exactly parametrising in case of the path being a line in my example , I would bring everything in a single variable if I had to follow the book , as in y=x/2 and dy = dx/2 , but I would parametrise in case the path would be y=x^2 as x=t and y=t^2 Commented May 6, 2016 at 15:54
• Well, do you get the same result when parametrising? Commented May 6, 2016 at 15:57
• @KennyLau Okay yeah i am getting the same result (=1) when I am integrating by bringing everything in one variable and when I am not too , so does that mean parametrization is not necessary ? Commented May 6, 2016 at 16:01
• I don't really think that it is necessary to bring everything to one variable. Commented May 6, 2016 at 16:04

If you integrate $\int xy\,dx = \frac{x^2}{2}y$, plug-in $(2,1)$ and $(0,0)$, you would get $2-0$. For the other half, $\int -y^2\,dy = -\dfrac{y^3}{3}$, plug in end points, and get $-\dfrac{1}{3}-0$. This would yield a line integral value $2-\dfrac{1}{3}=\dfrac{5}{3}$. While that would make this an easy computation, it's totally wrong.

First, let's get the correct answer. Then we can discuss why the above technique doesn't work.

We have $x=2y$ so use $x=2t$ and $y=t$ where $0\leq t \leq 1$ ($t=0$ corresponds with $(0,0)$ and $t=1$ corresponds with $(2,1)$). Then $dx=2\,dt$ and $dy=dt$. So $$\int_C xy\,dx-y^2\,dy = \int_0^1 (2t)(t)\cdot 2\,dt-(t^2)\cdot dt = \int_0^1 (4t^2-t^2)\,dt = \int_0^1 3t^2\,dt=t^3 \Bigg|_0^1=1$$

Why didn't the first method give the same answer? You can see it with this integral: $\int xy\,dx = \dfrac{1}{2}x^2y + C$. Doing this "partial integration" (undoing a partial derivative) assumes that we can treat $y$ like a constant. But this is only ok if $x$ and $y$ are independent variables. However, we have declared that we're integrating along the line $x=2y$. This means that $x$ and $y$ depend on each other!

Also, if the parameterization doesn't play a role in the computation of your line integral, your line integral must only depend on the end points of your curve. This would mean that your line integrals are path independent.

Notice that your example is path dependent: Instead of traveling along the line $x=2y$, go along the $x$-axis first: $y=0$ for $0 \leq x \leq 2$ (call this $C_1$). Then go along the vertical line: $x=2$ for $0 \leq y \leq 1$ (call this $C_2$). So $C_1+C_2$ goes from $(0,0)$ to $(2,0)$ then on to $(2,1)$.

If we do this, $C_1$ gives us $\int_{C_1} xy\,dx-y^2\,dy = \int_0^2 x(0)\,dx-0^2(0) = 0$. Along $C_2$ gives, $\int_{C_2} xy\,dx-y^2\,dy = \int_0^1 2y(0)-y^2\,dy = -\dfrac{y^3}{3}\Bigg|_0^1=-\dfrac{1}{3}$. Thus this path, $C_1+C_2$, yields the answer $0-1/3=-1/3$ (not the same as $1$).

This did give us an answer similar answer to our "incorrect method". Why? Well, we broke down our curve and treated $x$ and $y$ independently. :)

However, there are cases when we can "integrate the form". But this assumes that there is an integral at all. If your vector field $F$ is of the form $\nabla f$ (i.e. $\mathrm{grad}(f)$ = gradient of $f$), then $F$ can be "integrated" and you can apply the Fundamental Theorem of Line Integrals. In this case, you don't need to parameterize your curve to compute a line integral.

But not all vector fields can be integrated. To see this, suppose that you could integrate $F=P{\bf i}+Q{\bf j}.$ Then $F=\nabla f$ and so $P=f_x$ and $Q=f_y$. If this were the case, $P_y=f_{xy}=f_{yx}=Q_x$ (assuming continuous 2nd partial derivatives). So if $P_y \not= Q_x$, we cannot integrate/un-gradient $F$. This is true in your example: $P=xy$ and $Q=-y^2$. Notice $P_y=x \not= 0=Q_x$.

In fact, under some mild assumptions one can shows that $F=P{\bf i}+Q{\bf j}=\nabla f$ for some $f$ if and only if $F$'s line integrals are path independent. Also, (again assuming some technical conditions) $P{\bf i}+Q{\bf j} = \nabla f$ for some $f$ if and only if $P_y=Q_x$.

• But by putting $y=x/2$ in the first part of the integral it becomes $\int x^2/2 dx$ and that gives 8/6 and the second part $\int -y^2 dy$ gives -1/3 and so the answer becomes 8/6 - 1/3 which gives the correct answer for the given path . I now understand there would be a different answer for a different path but in the case i mentioned above am i actually removing independency of a variable by putting $y=x/2$ and is that why i am getting the correct answer (i.e 1). Learnt a bit of mathjax :) Commented May 6, 2016 at 16:48
• Ok. I see. You wanted to use the parameterization: x=x and y=x/2. Yes, that's perfectly fine. Any parameterization of your curve will yield the correct answer. Line integrals are independent of (orientation preserving) reparameterization...just not independent of path. :) Commented May 6, 2016 at 16:52
• okay so just to summarise for my understanding , parameterization is necessary in case the line integral is not path independent since otherwise we are letting the curve break down . Also we can use any parameterization (will read up on that ) . Also we can directly integrate if the vector field is the gradient of some function . Is that correct ? . Commented May 6, 2016 at 17:00
• Yes. That's essentially it. Parameterization is necessary to capture what your curve actually is but the particular parameterization used doesn't matter. If you setup the same line integral with two different parameterizations, those integrals will just differ by a u-substitution. Commented May 6, 2016 at 17:05
• @BillCook Your answer solved, like, 10 misunderstandings I had with vector calculus. Thank you! Commented Jun 20, 2023 at 12:18