The question in Stewart's Calculus book is the following:
Evaluate $\int_{C}y^2dx+xdy$, where C is the line segment from $(-5,-3)$ to $(0,2)$.
The parametrization of C is $x = 5t-5$, $y=5t-3$, where $t \in [0,1]$. Then, using the formulas
$\int_{C}f(x,y)dx$ = $\int_{C}f(x(t),y(t))\frac{dx}{dt}dt$
$\int_{C}f(x,y)dy$ = $\int_{C}f(x(t),y(t))\frac{dy}{dt}dt$, we have
Calculations give $-5/6$.
Then, I considered $\int_{C}y^2 + x ds$, where this integral gives the line integral with respect to the arc length. Calculations give $-5/6 \sqrt2$ which is equal to $\sqrt{(-5/6)^2+(-5/6)^2}$. Thus, it seems to me that the following formula might be true
$\int_{C}f(x,y)ds = \sqrt{(\int_{C}f(x,y)dx)^2 + (\int_{C}f(x,y)dx)^2}$
I think the problem is that either I did not understand the geometric concept completely and these calculations are just lucky or there is indeed a connection and I cannot prove/see precisely. Can you explain to me the connection between these line integrals?
Thanks for any help in advance.
Your equation is not rigot. Note that you are not considering all terms in the original integral.
The idea you are looking for is something related to the Pythagoras theorem and, in this sense, you are in the right direction. When considering line integrals with respect to arc lenght, say $ds$, the right formula reads $$\int_{C}f(x,y)ds = \int_{a}^{b}f(x(t),y(y))\sqrt{\bigg{(}\frac{dx}{dt}\bigg{)}^{2}+\bigg{(}\frac{dy}{dt}\bigg{)}^{2}}dt$$ where $\gamma(t) = (x(t),y(t))$ is a parametrization of the given curve. Note that this has a connection to the Pythagoras theorem, when you look closely to the square root term in the integral. A nice way to interpret this is as follows: take $f(x,y) \equiv 1$ for simplicity. If $dx/dt$ is interpreted as the speed of a particle along the $x$-axis on the curve and $dy/dt$ its speed along the $y$-axis (here, I'm interpreting $t$ and time), then $\sqrt{\bigg{(}\frac{dx}{dt}\bigg{)}^{2}+\bigg{(}\frac{dy}{dt}\bigg{)}^{2}} \equiv v$ is the total speed (in two dimensions) of the particle. Thus, we would expect that $$\int_{C}\sqrt{\bigg{(}\frac{dx}{dt}\bigg{)}^{2}+\bigg{(}\frac{dy}{dt}\bigg{)}^{2}}dt = \int_{a}^{b}v dt = \Delta S(a,b)$$ where $\Delta S(a,b)$ is the lenght of the curve travelled by the particle from $t=a$ to $t=b$.
Another approach that may convince yourself that this is the right formula is this: take $f(x,y)\equiv 1$ again and consider the curve $C$ along the function $y=x$, when $0\le x \le 1$. This is the hypothenuse of a right triangle of sides $l_{1}=l_{2}=1$ and $l_{3}=\sqrt{2}$, so we would expect that $\int_{C}ds = \sqrt{2}$ in this case. Note that a parametrization of such a curve is easy to get: take $\gamma(t) = (x(t),y(t))$ where $x(t) =y(y) =t$ with $0\le t \le 1$. Thus $$\int_{C}ds = \int_{0}^{1}\sqrt{\bigg{(}\frac{dx}{dt}\bigg{)}^{2}+\bigg{(}\frac{dy}{dt}\bigg{)}^{2}}dt = \int_{0}^{1}\sqrt{2}dt = \sqrt{2}\int_{0}^{1}dt = \sqrt{2}$$ as we wish.