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"Find the line integral of F over the path composed of the connected lines $(0,0,0)$ to $(1,1,0),$ $(1,1,0)$ to $(1,1,1). $ F = $⟨\sqrt z, -2x, \sqrt y ⟩$ "

Alright, so the formula for one of these line segments is:

$\int_a^b$ F $\cdot \frac {dr} {dt} dt $

And I need to add both line segments together. This gives me two integrals to sum up:

$\int_0^1 ⟨0,-2t,\sqrt t⟩ \cdot⟨1,1,0 ⟩ dt$

$\int_0^1 ⟨\sqrt t,0,0⟩ \cdot⟨0,0,1 ⟩ dt$

The problem is, the top integral solves to -1. But the bottom integral's dot product is $0$, so I end up with $\int_0^1 0dt = C.$

a) This very strange, because I'm fairly sure I'm not supposed to have the integral of zero, and

b) The answer to this question is 0, meaning I'd expect the bottom integral to solve as 1. Unless the $C$ that's tacked onto the end of every integral is supposed to be 1 in this case, but that doesn't seem right. Can someone point out where I'm going off here?

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    Although $\int 0\,dt=C$, $\int_0^10\,dt=C-C=0$. – 2017-01-31
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    Okay, that still doesn't explain the overall incorrect solution though. In that case, I end up with $ -1+0 = -1$, but the answer is supposed to be $0$. – 2017-01-31
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    Shouldn't the first vector in the second integral be $\langle \sqrt{t},-2,1\rangle$ since at $t=1$, $-2t=-2$ and $\sqrt{t}=1$? – 2017-01-31
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    I think you're right, actually. I started $t$ all over again but I should have accounted for the fact that we're already partway through the path. Will try again and confirm. – 2017-01-31
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    It's ok to start $t$ over again for the second segment, just don't forget that you are already at $(1,1,0)$ so the vector is already at $\langle 0,-2,1\rangle$ and on its way to $\langle 1,-2,1\rangle$ as $t$ goes from $0$ to $1$. – 2017-01-31
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    Yep, that works, and the integral solves to exactly 1, just what I was looking for. Marking this as solved, thank you! – 2017-01-31
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    You're welcome. – 2017-01-31
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    Oh, I actually can't because you left this as a comment. If you change it to an answer I'll give you the points. – 2017-01-31

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\begin{eqnarray} &&\int_0^1 ⟨0,-2t,\sqrt t⟩ \cdot⟨1,1,0 ⟩ dt+\int_0^1 ⟨\sqrt t,-2,1⟩ \cdot⟨0,0,1 ⟩ dt\\ &=&\int_0^1-2t\,dt+\int_0^11\,dt\\ &=&\left[-t^2\right]_0^1+\left[t\right]_0^1\\ &=&-1+1\\ &=&0 \end{eqnarray}