Calculate the line integral with respect to arc length of $\int_Czds$, where $C$ is parametrized by $γ(t) = (tcost,tsint,t), 0 ≤ t ≤ t_0.$
I calculate the norm of $γ'(t)$ and I get $\sqrt {2+t^2}$. But how can I calculate $\int_0^{t_0}z \sqrt {2+t^2}dt$? I thought I might did something wrong.
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$\begingroup$Parametrized curve $C$ is given by $γ(t) = (t \cos t, t \sin t, t), 0 ≤ t ≤ t_0$. What this means is that points on the curve are given by $x = t \cos t, y = t \sin t, z = t$.
So the integral of scalar function $f(x, y, z) = z$ with respect to arc length,
$\displaystyle \int_C z \ ds = \int_0^{t_0} t \ \gamma'(t) \ dt = \int_0^{t_0} t \ \sqrt{2+t^2} \ dt$
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