So I was watching some videos on Line-Integrals on Khan Acadamy and when Sal was discussing about Line Integrals of Scalar Functions, Sal mentioned that one can think of a line integral as the area of a curtain drawn from your curve to your function.
To elaborate:
$$ f(x,y) = xy \\ C: x = \cos{t}, y=\sin{t}\\ 0 \le t \le \frac{\pi}{2} $$
Now we can think of $$ \int_C f(x,y) \ ds $$ where $ds = \sqrt{dx^2 + dy^2} $ as the area of a curtain as shown below:
Now my question is:
If we have a vector field $$ \vec{f}(x,y) = P(x,y) \hat{i} + Q(x,y) \hat{j}$$
and we are taking the line integral to calculate Work Done as:
$$ W = \int_C \vec{f} . d\vec{r} $$
Sal said line integrals can be thought of as the area of a curtain along some curve between the xy-plane and some surface z = f(x,y). This new use of the line integral in a vector field seems to have no resemblance to the area of a curtain.
How are the two concepts connected?