"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?
\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}