$$x=e^t , y = \sqrt{2} t , z= \ e^{-t} ; ~~ ~~ 0\leq t \leq 1$$ Find the centroid of this curve.
I know one formula for centroid which is $x = \frac{\int {x dx dy}}{\int dx dy}$ and for y as well, which is for a surface. I don't know any formula for centroid of curve. I've also drawn the curve in mathematica and it is not a closed curve in given boundary. How do we calculate this?
The question is from "Calculus, Adams - Chapter 15.3"
To calculate centroid of a curve, first we compute the $\mathrm{d}s$: $$\mathrm{d}s = \sqrt{x^\prime(t)^2+y^\prime(t)^2+z^\prime(t)^2}=\sqrt{e^{2t}+2+ e^{-2t}}$$
Now note that $$\int_0^1 \mathrm{d}s = \int_0^1 \sqrt{e^{2t}+2+ e^{-2t}} \mathrm{d}t = \int_0^1 \sqrt{e^{-2t}(e^{4t}+2e^{2t}+1)} \mathrm{d}t = \int_0^1 e^{-t}(e^{2t}+1)\mathrm{d}t$$
$$=\int_0^1 e^t+e^{-t}\mathrm{d}t = e-\frac1e\approx 2.35$$
Then you have
$$\bar{x}=\frac{\int_0^1 e^t(e^t+e^{-t})\mathrm{d}t}{2.35}\approx 1.78$$
$$\bar{y}=\frac{\int_0^1 t(e^t+e^{-t})\mathrm{d}t}{2.35}\approx 0.761$$
$$\bar{z}=\frac{\int_0^1 e^{-t}(e^t+e^{-t})\mathrm{d}t}{2.35}\approx 0.609$$