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If $f(x,y)=y^2$ and we consider the path $(5t-5, 5t-3)$ with $0\le t\le 1$, then the line integral with respect to $x$ is $\int_C y^2 dx=\int_0^1 5(5t-3)^2 dt$; and with respect to $y$ it is the same $\int_C y^2 dy=\int_0^1 5(5t-3)^2 dt$.

However, the line integral of $f$ with respect to arc length is $\int_C y^2 ds=\int_0^1 (5t-3)^2\sqrt{50} dt$.

So, we see that $\int_0^1 5(5t-3)^2 dt+\int_0^1 5(5t-3)^2 dt \ne \int_0^1 (5t-3)^2\sqrt{50} dt$.

It seems to me like if I take a line integral with respect to each variable independently, it would equal the whole line integral.

What is going on here?

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Recall that for a planar curve line integral of a function $f$ is $ \displaystyle \int_C f ds$, where $s$ is the arclength, so that $\displaystyle ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 }dt$, not the simple sum $\displaystyle \frac{dx}{dt} + \frac{dy}{dt} \neq ds$.

Therefore you cannot simply add $\displaystyle \int_C f dx $ and $\displaystyle \int_C f dy $ and expect the sum to be equal $\displaystyle \int_C f ds $.

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