# What is the geometrical meaning of the integral of a vector valued function?

If $f:\mathbb{R}\rightarrow\mathbb{R}$ is an integrable function.

then $\int_a^b f(x)dx$ can be considered as the area between the graph and the x-axis.

But what if $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$?

Let $\gamma:[0,1]\rightarrow \mathbb{R}^n$ be a smooth curve.

what is the geometrical meaning of $\int_\gamma f\cdot dl$? (or in case $n=m$, $= \int_0^1 f(\gamma(t))\gamma '(t)\cdot dt$?)

Thanks :)

(for simplicity, you can assume $m,n$ are small numbers... i.e $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ )

• Well, it's a vector. Isn't a vector a geometrical object? – John Wayland Bales Jul 27 '16 at 20:32
• Hope I made it clearer :) I mean the line/contour integral of a vector valued function – Daniel Jul 27 '16 at 20:38
• Wikipedia has a neat animation that you should look at. – user137731 Jul 27 '16 at 20:38
• @Daniel What does $f(\gamma(t))\gamma'(t)$ mean? Do you mean $f(\gamma(t))\cdot \gamma'(t)$? If so, then $n=m$. – user137731 Jul 27 '16 at 20:42
• I would like to interject that if $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ where $n\neq m$ then $\int_0^1f(\gamma(t))\gamma'(t)\ dt$ does not make sense, as $f$ is $\mathbb{R}^m$ valued while $\gamma'$ is $\mathbb{R}^n$ valued. Also OP's wording makes me think he wants to do a vector valued integral, not a usual line integral. – Bence Racskó Jul 27 '16 at 20:43

If you imagine drawing the curve $y=f(x)$, the area under the curve can be roughly interpreted as the sum of all the infinitesimal-width vertical 'strips' which connect the point $(x,0)$ to the point $(x,f(x))$. Of course, this doesn't make much sense, because there's no such thing as a rectangle with 'infinitely small width'. One way of turning this intuitive picture into a sensible idea is the Riemann sums, which are roughly of the form:
$$\sum_{i=1}^{n}f(x_{i}^{*})(x_{i+1}-x_{i})$$ where $x_{0},\ldots,x_{n}$ are points that partition your interval $[a,b]$ into the intervals $[x_{i-1},x_{i}]$, and $x_{i}^{*}$ is some point in that interval. But this can also be viewed as a weighted sum of the values $f$ takes in the interval $[a,b]$. If we take a suitable limit of the partition which makes the intervals $[x_{i-1},x_{i}]$ uniformly small, we get your area interpretation of the integral.
So, the line integral $$\int_{\gamma}f(x)dx:=\int_{a}^{b}f(\gamma(t))|\gamma'(t)|dt$$ can be interpreted as a continuous sum of vector contributions of $f$, as we travel along small portions of the curve $\gamma$. The $|\gamma'(t)|$ factor, reminiscent of the substitution rule for 1-D integrals, can be thought of as a scale factor relating how 'fast' we go along the curve $\gamma$ - this ensures that if we reparametrise the curve so that we go faster or slower in places, we don't change the value of the integral.
For a physical example, let's suppose $\gamma$ is a uniformly charged wire, and you want to calculate the electric field at a point $x$ due the wire. Well, there is a charge contribution from each point $\gamma(t)$, which is given approximately by $\lambda |\gamma'(t)|dt|$, where we have written $\lambda$ for the charge per unit length of the wire. The electric field due to this point-like charge is simply $$E(\gamma(t))=\int_{a}^{b}\frac{x-\gamma(t)}{4\pi\epsilon_{0}|x-\gamma(t)|^{3}}\lambda|\gamma'(t)|dt = \int_{\gamma}\frac{x-x'}{4\pi\epsilon_{0}|x-x'|^{3}}\lambda dx'$$