line integral... Calculate $$\int_Γ f \, d\ell$$ for $f(x,y) = y,  \; y=x^{1/2}$, $ x $ is in $[2,6]$.
I know (now) that it means that: $$\int_\Gamma f \, d\ell=\int_a^b f(\Gamma(t)) \cdot \|\dot\Gamma(t)\| \, dt$$
But why will the line be $(x,x^{1/2})$ and not $(0,x^{1/2})$?
I just don't understand that thing. Hope for help. Thanks.
EDIT:
UNDERSTOOD. I'D LOVE IF ONE COULD SOLVE IT IN ORDER TO COMPARE MY RESULT.
 A: Your result should be $\frac{49}{6}$ which is approximately 8.16667.
Full solution:
Let $\Gamma(t) = \begin{pmatrix} t \\ \sqrt{t} \end{pmatrix}$ and $t \in [2,6]$ (this is the path parametrization) then we have
$$
 \dot{\Gamma}(t) = \begin{pmatrix} 1 \\ \frac{1}{2\sqrt{t}} \end{pmatrix}
$$
as well as
$$
 \left\| \dot{\Gamma}(t) \right\| = \sqrt{1+\frac{1}{4t}}
$$
and $$f(\Gamma(t)) = \sqrt{t}$$
which yields
$$
\int_\Gamma f \mathrm{d}\ell = \int_{2}^6 f(\Gamma(t)) \left\| \dot{\Gamma}(t) \right\| \mathrm{d}t =  \int_{2}^6 \sqrt{t} \sqrt{1+\frac{1}{4t}} \mathrm{d} t = \int_{2}^6 \sqrt{t+\frac{1}{4}} \mathrm{d} t = \frac{2}{3}{\left(t+\frac{1}{4}\right)}^\frac{3}{2}\Biggr|_{t=2}^{6} = \frac{1}{12}{\left(4t+1\right)}^\frac{3}{2}\Biggr|_{t=2}^{6} = \frac{98}{12} = \frac{49}{6} \;.
$$
P.S.: You might get more answers if you would edit your post to show the steps you have already done. Moreover this looks pretty much like a homework problem. Which is why I hesitated in providing you with a full answer right away...
