Line integral of a Vector Field around a Closed path

This is just a little question.

Suppose you want to evaluate an integral around a closed path formed by a curve $C(t) $(only one curve), I suspect that the result would be $0$, because you will do an integral from the point $P$ to the same point.

so for example if $P=C(a)$, then your integral is $$\int_CF=\int_a ^a F(C(t))\cdot C'(t)\,dt = 0$$

Is that true?

1 Answer

You will do an integral from the point $P$ to the same point.

This is true in certain situations. The Fundamental Theorem For Line Integrals says that if your vector function $\mathbf F$ is the gradient of a scalar function $f$, then you can replace these sorts of line integrals (like $\int_C \mathbf F(\mathbf C(t))\cdot \mathbf C'(t)\,\mathrm dt$) with a difference in the values of $f$ at the endpoints. In a situation like that, you are correct that the integral over a closed loop $C$ must be zero since the two endpoints would be the same. (It doesn't matter whether if the loop $C$ is given by single formula for $\mathbf{C}(t)$ or if there is a piecewise definition.)

However, if $\mathbf F$ is not the gradient of a scalar function, then we don't have an analog of that theorem, and the integrals around closed loops are often nonzero. The line integral of $\mathbf F$ measures how much $\mathbf F$ goes along with the direction of the curve $C$; see Lucas Vieira's animation featured in the line integral of a vector field section of the English Wikipedia article for "Line Integral", for one illustration. So the integral around a loop can be nonzero as long as the direction of $\mathbf F$ is aligned with the loop.

A famous example would be $\mathbf F(x,y)=\left\langle\dfrac{-y}{\sqrt{x^2+y^2}},\dfrac{x}{\sqrt{x^2+y^2}}\right\rangle$ (inspired by the polar $\left\langle-\sin\theta,\cos\theta\right\rangle$). Wolfram|Alpha plots this vector field like this:

Since the vectors of $\mathbf F$ are all going counterclockwise around the origin, we should expect the line integral of $\mathbf F$ around a counterclockwise loop around the origin to be positive. Indeed, if we integrate over the counterclockwise loop $C$ of radius $r$ parametrized by $\mathbf C(t)=\left\langle r\cos t,r\sin t\right\rangle$ for $t$ from $0$ to $2\pi$, then \begin{align*}&\phantom{=}\oint_C\mathbf F\cdot\mathrm d\mathbf C\\&=\int_0^{2\pi}\mathbf F(\mathbf C(t))\cdot\mathbf C'(t)\,\mathrm dt\\&=\int_0^{2\pi}\mathbf F(r\cos t,r\sin t)\cdot\left\langle -r\sin t,r\cos t\right\rangle\,\mathrm dt\\&=\int_0^{2\pi}\left\langle \dfrac{-r\sin t}{\sqrt{r^2\cos^2t+r^2\sin^2t}},\dfrac{r\cos t}{\sqrt{r^2\cos^2t+r^2\sin^2t}}\right\rangle\cdot\left\langle -r\sin t,r\cos t\right\rangle\,\mathrm dt\\&=\int_0^{2\pi}\left\langle -\sin t,\cos t\right\rangle\cdot\left\langle -r\sin t,r\cos t\right\rangle\,\mathrm dt\\&=r\int_0^{2\pi}\left\langle -\sin t,\cos t\right\rangle\cdot\left\langle -\sin t,\cos t\right\rangle\,\mathrm dt\\&=r\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt\\&=r\int_0^{2\pi}1\,\mathrm dt\\&=2\pi r>0\end{align*}