Area under a parametric curve in $\mathbb{R}^3$

Say we are given a space curve described by some parametrization $f(t)=(f_1(t),f_2(t),f_3(t))$. Is there a formula that computes the "area under the curve"?

For example, if $f(t)=(\cos(t), \sin(t),t)$, $0 \leq t \leq 2\pi$, the area under the curve should be half the surface area of the ("uncapped") unit cylinder about the $z$-axis: $2\pi^2$. It turns out this is what I get if compute $\| \int_0^2 f(t)dt \|$, which was what my naive guess at such a formula would be. But this would always return a non-negative number; what if I want an unsigned area?

Additionally, such an integral would have to take into account some form of "orientation," right? I.e., that I'm looking at the "area" between my curve and the $xy$-axis in $\mathbb{R}^3$, as opposed to some other plane. Should I be looking at one of the kinds of integrals that show up in vector calculus (line/path/etc. integrals)?

Thanks in advance for any help!

• Your curve $f(t)=(\cos t, \sin t, t)$ is a helix. How do you get a cylinder from this? – A. Goodier Mar 5 '18 at 17:36
• Maybe my wording was poor. I'm imagining that for each point on the helix corresponding to times $0\leq t \leq 2$, I'm dropping down a line from that point to the $xy$-plane; the resulting surface is some portion of the cylinder and takes up half of that truncated cylinder. – Bachmaninoff Mar 5 '18 at 18:26
• You want $0\le t \le 2\pi$ right? – zhw. Mar 5 '18 at 19:12

Here's a way to see $|\int_0^{2\pi} f(t)\,dt|$ can't be right in general. Take $f_1(t) = R\cos t,$ $f_2(t) = R\sin t,$ $f_3(t)=1.$ That's a cylinder of radius $R$ and height $1.$ The area of this surface is $2\pi R\cdot 1 =2\pi R .$ But the formula on the table would give $|(0,0,2\pi)|=2\pi.$ No good. I think you just lucked out with you helix example.
$$\int_a^{b} \sqrt {(f_1'(t))^2 + (f_2'(t))^2}\, f_3(t)\,dt$$
should work for positive geometric area in most cases. Here we are adding up increments of arclength of the planar curve $(f_1,f_2)$ times height $f_3.$ You can visualize the approximation being done. We probably want $(f_1,f_2)$ to be smooth and injective, while we only need to assume $f_3$ is continuous and nonnegative.
Let be $\gamma = (f_1,f_2)$. Using reparametrization if required, we can suppose that $\gamma$ is parametrized by arc length. The area "under" $f$ is the (real, scalar) line integral $$\int_\gamma f_3.$$