Describe the intersection of the sphere $$x^2+y^2+z^2=1$$ and the cylinder $$x^2+2z^2=1$$ Find the total lenth of this intersection curve.
I parametized curve as $$x=cos(t)$$ $$z=\frac{1}{\sqrt{2}} \sin(t)$$ $$y^2=\frac{sin^2(t)}{2}$$ Applied formula and get $$\int_0^{2\pi} 1dt$$ I found $2\pi$, but my teacher said it's $4\pi$. I suspect that because of y, since has 2 possible values positive and negative. My teacher solved as follows:
Since the path is an elliptic curve in the space we can divide curve in 8 octanes. Thus, $S =8\int_0^{\pi/2}1dt$
$S=4\pi$
How can we say that the curve passes all octanes and it's length is exactly same as others? And what to do when it wasn't symetric about axes?
Your suspicion is correct. The integral $\int_0^{2\pi}{1}{dt}$ only calculate the incoming intersection of the "squashed" cylinder on one side of the circle, but does not account for when it leaves on the other side.
Note that as long the cylinder intersects the circle with, its centre common to the circle's centre, then the arc length will always $\int_0^{4\pi}{1}{dt}$.
However, if the intersection is not perfect to a common centre, I would suggest creating $2$, $2$D side profiles to understand how your intersection will look.
Ie. for your equations, on the XZ plane:
and the YZ plane: