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I have to calculate a line integral with respect to the arc length in the form $\int_C \varphi\ ds$ where there curve $C$ is the intersection between $x^2 + y^2 = z^2 $ and $y^2 = x$ and goes from $(0,0,0)$ to $(1,1,\sqrt{2})$. I have found that I can write that as $x^2 + x = z^2$ and if I "complete the square" I have $x^2 + x + \frac{1}{4} - \frac{1}{4} = z^2$, which is $(x+\frac{1}{2})^2 - z^2 = \frac{1}{4}$ or $$4(x+\frac{1}{2})^2 - 4z^2 = 1$$, which is the equation of a hyperbola centered at $(\frac{1}{2}, 0)$.

I read on Wikipedia that one can parametrize the hyperbola using $x = h + a \sinh t$ and $z = h + b \cosh t$ but if I do this type of parametrization it's not consistent with the points that are given originally, that is, $(0,0,0)$ to $(1,1,\sqrt{2})$.

Is there another parametrization I could look into? Is the above derivation correct so it can be parametrized?

PS: I know this question has been answered before in this site, but unfortunately the answers given were not very helpful, even though I tried them all.

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  • $\begingroup$ Why is not consistent? Just pick right bounds on $t$. $\endgroup$ – Ahmed S. Attaalla Feb 6 '17 at 1:50
  • $\begingroup$ The bound for $t$ is given by the points $(0,0,0)$ and $(1,1,\sqrt{2})$, is it not? $\endgroup$ – lorenzattractor Feb 6 '17 at 1:55
  • $\begingroup$ Yes, it is given by initial and point @lorenzattractor But you have to solve for $t$ when you set $x=f(t)$ ... $\endgroup$ – Ahmed S. Attaalla Feb 6 '17 at 2:04
  • $\begingroup$ So that is where the parametrization using $\sinh$ and $\cosh$ falls apart, because if I use the point $(0,0,0)$ I don't get the point $(1,1,\sqrt{2})$. Both points should be on the same curve, right? $\endgroup$ – lorenzattractor Feb 6 '17 at 2:06
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Let,

$$x=-\frac{1}{2}+\frac{1}{2}\cosh (t)=\sinh^2(\frac{t}{2})$$

$$z=\frac{1}{2}\sinh (t)$$

With $t \in [0,\sinh^{-1} (2\sqrt{2})]$. Can you see why this works?

As for $y$ it follows from $y^2=x$ and the fact that we are on the part of $C$ where $y \geq 0$.

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  • $\begingroup$ I actually had done this before, maybe it slipped away because I've been trying to solve this for almost 3 hours. Thanks for the answer! $\endgroup$ – lorenzattractor Feb 6 '17 at 3:10
  • $\begingroup$ Please see edit @lorenzattractor $\endgroup$ – Ahmed S. Attaalla Feb 8 '17 at 4:22
  • $\begingroup$ Thank you, I was reading more about it and it actually made more sense what you wrote before. $\endgroup$ – lorenzattractor Feb 10 '17 at 0:00

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