Confusion about geometrical meaning of line integrals of a scalar field

Question

So, I've started to learn about line integrals in a scalar field some days ago; I learned about it first on Khan Academy's video (https://www.youtube.com/watch?v=_60sKaoRmhU&t=5s - if you want to see it).

It starts with a curve in $xy$ plane and the geometrical meaning of the line integral is presented as the area in the vertical from the curve to the surface above.

I understood that; However, now I'm doing some exercises and I have this:

$\int_C f(x,y,z)dS$ Where $f(x,y,z) = x + \sqrt{x} - z^2$

and $C = r(t) = (0,0,t)$, with $0\leq t\leq 1$

So, I thought: "Well, if it is the area in the vertical, I have $0$ area, so the line integral must be equal to $0$". But, when I calculated it, to my amaze it is not equal to 0. Why??

I must assume that my comprehension about the geometrical meaning of it is wrong. Can you explain me what am I missing here??

PS: the exercise doesn't specifically ask for geometrical interpretation, but I'm trying to find one, to get more intuition about these mathematical questions.

Answer 1

The “area below the curve” interpretation of an integral only makes intuitive sense for functions of one or two variables.

If you have a function of a single variable $f(x)$ then you can plot the curve $y=f(x)$ and think of the integral as the area between the curve and the $x$ axis. If you have a function of two variables $f(x,y)$ then you can plot the curve $z=f(x,y)$ and think of the integral as the area between the curve and the $x,y$ plane.

But if you have a function of three variables $f(x,y,z)$ then to plot a curve you would have to introduce a fourth variable $w=f(x,y,z)$ and plot the curve in four dimensional space. And then the interpretation of the integral as the area “below” the curve becomes much less intuitive and much harder (or impossible) to visualise.

Answer 2

As gandalf61 correctly says You must think of a fourth dimension and the integral could be interpreted as the (signed) area of the "curtain" ( as Khan says) along the curve $C$ with the height given by values of $w=f(x,y,z)$ "above" this curve. In Your case this should give $$\int_Cf(x,y,z)dS=\int_0^1f(0,0,t)\sqrt{0^2 +0^2+1^2}dt=\int_0^1 -t^2dt=-\frac{1}{3}.$$