This is the problem:

Find intersection of this two surfaces and then calculate its length. $$x^2+4y^2=4$$ and
$$y\sqrt3 +z=1$$ Method 1 :

Common way is to take:
$$x=2\cos t ,\: y=\sin t ,\: z=1-\sqrt3 \sin t$$
$$r(t) =(2\cos t,\:\sin t,\:1-\sqrt3 \sin t)$$ and
$$r'(t) = (-2\sin t,\:\cos t,\:-\sqrt3\cos t) ,\: |r'(t)|=2$$

Finally the length is:
$$L = \int_{0}^{2\pi} 2\:dt=4\pi$$

But in another parameterization I find something strange

Method 2:

$$x=2\sqrt{1-t^2} ,\: y=t ,\: z=1-t\sqrt3$$ So
$$r(t) = (2\sqrt{1-t^2} ,\: t ,\: 1-t\sqrt3)$$ and
$$r'(t) = \left(\frac {-2t}{\sqrt{1-t^2}},1,-\sqrt3\right) ,\: |r'(t)|=\frac {2}{\sqrt{1-t^2}}$$ (and that's different from method 1 $|r'(t)|$)

Finally the length
$$L=\int_{0}^{2\pi} \frac {2}{\sqrt{1-t^2}}\:dt\\ =2\int_{0}^{\arcsin{2\pi}}\frac {1}{\cos{\theta}}\cos{\theta}\:d\theta\\ =2\int_{0}^{\arcsin{2\pi}}1\:d\theta$$

and because of $\arcsin(2\pi)$ its undefined it cant be solved.

Now my first question is that:
If we know from past that we can create many parameterizations for a single curve, so why the result of they length is different?

and the second one:

How to be sure that a parameterization is the correct one?


As a complement to @Axel's answer

Let $t_j$ denote the $j$th method's $t$ so $t_2=\sin t_1$ and $\tfrac{dt_2}{\sqrt{1-t_2^2}}=d\arcsin t_2=\pm dt_1$. If for example $t_1$ increases from $0$ to $2\pi$, $t_2$ is monotonic in three parts, cut at $t=\pi/2$ and $t=3\pi/2$. The monotonic pieces have alternating $\pm$, in this case with the pattern $+-+$, so the length is$$\begin{align}\int_0^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}+\int_{-1}^0\frac{2dt_2}{\sqrt{1-t_2^2}}&\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\\&\color{red}{=4\pi}\\&\color{blue}{=\left(\int_0^{\pi/2}+\int_{\pi/2}^{3\pi/2}+\int_{3\pi/2}^{2\pi}\right)2dt_1}\\&=\color{blue}{4\pi},\end{align}$$where the $t_2$-space calculation can continue from the black expression with either the red or the blue treatment. If $t_1$ varies over another length-$2\pi$ interval the monotinicity division details are different, but any choice requires at least one such cut, succumbs to the above technique, and gets the same answer. Let's check another example from $t_1=-\pi/2$ to $t_1=3\pi/2$, giving two pieces (meeting at $t=\pi/2$) where the $\pm$ are $+-$:$$\int_{-1}^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\color{blue}{=\left(\int_{-\pi/2}^{\pi/2}+\int_{\pi/2}^{3\pi/2}\right)2dt_1}.$$

  • $\begingroup$ Thank you both.I understand it now. $\endgroup$ – program_craft Jan 22 at 4:58

The problem lies in the way you parametrize the curve in the second method.

The reason why you integrate from $0$ to $2\pi$ in the first method is because you have implicitly define you parametrization to go from $0$ to $2\pi$ because you are looking for a mapping $[0;2\pi) \longrightarrow \textrm{your intersection curve}$ such that that mapping is biyective with the entire curve.

If you take a look to what you limits of integrations are in the second method you see they are the same from the first one. That again implicitly define you parametrization to go from $0$ to $2\pi$ but this time (because of the equation $y=t$) you are saying that $y$ runs in the range $[0;2\pi)$ when in reallity in ranges $[-1;1)$ so you actually end up measuring the length of a totally different curve (one with that parametrization but with $y \in [0;2\pi)$ insted of $[-1;1)$).

Now given that $t$ should be in $[-1;1)$ the correct way to measure the lenght in the second method would be: $$ \int\limits_{-1}^1 \frac{2}{\sqrt{1-t²}} \ dt + \int\limits_{1}^{-1} \frac{2}{\sqrt{1-t²}} \ dt $$

That is beacuse you can't map the entire curve with just the $[-1;1)$ range (in fact you only map one half of it). Thus the second integral.

So finally you have:

$$ 2\int\limits_{-1}^1 \frac{2}{\sqrt{1-t²}} \ dt = 4\pi $$

  • $\begingroup$ Can you explain more why t is in range[-1,1] $\endgroup$ – program_craft Jan 21 at 17:36

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