I am told:
$C$ is the part of the circle $\frac{x^2}{6^2}+\frac{y^2}{6^2}=1$ in the first quadrant, find the following line integral with respect to arc length:
$\int_C(8x-5y)ds$
So I understood that we are simply computing:
$\int_C f(r(t))|c'(t)|dt$
we can express the curve $C$ as $y=\sqrt{36-x^2}$ or equivalently as $r(t)=\langle t,\sqrt{36-t^2} \rangle$
The range in this case would be $0\leq t \leq 6$ and thus I get:
$r'(t)=\langle1,\frac{-t}{\sqrt{36-t^2}}\rangle$
$|r'(t)|=\sqrt{1+\frac{t^2}{36-t^2}}=\sqrt{\frac{36}{36-t^2}}$
And thus the integral is:
$\int_6^0 (8t-5\sqrt{36-t^2})\frac{6}{\sqrt{36-t^2}})dt$
$=\int_6^0\frac{48t}{\sqrt{36-t^2}}-30$
The indefinite integral being:
$F(t)=-48\sqrt{36-t^2}-30t+C$
Thus the final answer should be: $-108$
(I skipped a couple of steps for shortness)
This answer is apparently wrong and I am not sure why exactly. I would appreciate help mostly with the reading comprehension of the question. What exactly does it mean by "with respect to arc length" is that equivalent with "along the curve $C$"? and if not why not?