What's the meaning and how to integrate a function over a curve Okay, I have the following problem; "Integrate the function f over the curve $\gamma$"
$$f(x,y,z)=x+y+z, \gamma (t)=(\sin t,\cos t,t), t\in[0,2\pi]$$
I don't understand what's the meaning of integrating a function over a curve, I guess it's related with the parameterization of the curve, but I don't understand the concept. 
Also, I have the formula $\int_a^bf(\alpha(t))||\alpha'(t)||$, but I don't know if i can apply directly the values like this: $$\int_a^b(\sin t+ \cos t+t)\sqrt{(\cos t)^2+(-\sin t)^2+(2t)^2}dt$$
 A: I suppose you were told that by evaluating
$\int_a^bf(\alpha(t)) \lVert \alpha'(t) \rVert\, dt$
you could integrate the function $f$ over the curve $\alpha(t)$.
That is correct.
Integrating a function over an interval on the real number line
(something you presumably have done a lot already),
such as $\int_a^b f(x) \,dx,$
means that for each small increment $\Delta x$ of "distance" along the segment
from $x=a$ to $x=b$, you add $f(x) \cdot \Delta x$ to the value of the integral.
The idea of integrating a function over a curve is similar, except that the
"distance" you move isn't from $x$ to $x+\Delta x$, it's a distance along the curve. Parameterized over $t$, you might take a small step from 
$\alpha(t)$ to $\alpha(t+\Delta t)$; the length of that step is
a distance between three-dimensional points,
$\lVert \alpha(t+\Delta t) - \alpha(t) \rVert \approx 
 \lVert \alpha'(t) \rVert \Delta t$.
So it's the same idea as $\int_a^b f(x) \,dx,$
but using $\lVert \alpha'(t) \rVert \,dt$ instead of $dx$
to keep track of the "distance" over which we integrate.
But I wouldn't say the meaning of integration over a curve has anything
to do with the parameterization of the curve; the meaning has to do
with the distance we travel along the curve to get from one point to 
another, and the values of the function between those points.
The parameterization of the curve merely helps us compute the integral,
because it gives us a relatively simple way to measure distances along the curve.
A: This is call line integral. 
You have the correct formula but the calculation of $|\gamma'(t)|$ is incorrect.
Note that
$$
\sin^2t+\cos^2t=1
$$
and the derivative of $t$ is $1$, not $2t$.
What you want to calculate is 
$$\int^{2\pi}_0(\sin t+ \cos t+t)\sqrt{(\cos t)^2+(-\sin t)^2+(1)^2}dt\\
$$
But note that for any $t$, 
$$
\sqrt{(\cos t)^2+(-\sin t)^2+(1)^2}=\sqrt{2}.
$$
