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orientable surface- revolution hyperboloid of a paper

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0

$begingroup$

I have to prove that the equation $$ fracx^2a^2+fracy^2a^2-fracz^2b^2=1$$ $$ a,b>0$$

makes a orientable surface. For do it first of all I prove that the equation $f(x,y,z)$ is differentiable. And it is because is a polynomial. But I also have to prove that the surface is a regular one. And I don´t know how to prove it. I did:$$x=a cosh(u) cos(v)$$$$y=a cosh(u)sin v$$$$z=b sinh(u)$$ and I prove that the gradient is different from $0$ but is not a injective because I can take $v'=v+2*pi$. So I don't know if I have to take another way or I have to prove that it is not orientable.

Thanks in advance!

differential-geometry

share|cite|improve this question

edited Mar 14 at 15:55

Max

9071318

asked Mar 14 at 15:37

MeliodasMeliodas

216

$endgroup$

add a comment | 

0

$begingroup$

I have to prove that the equation $$ fracx^2a^2+fracy^2a^2-fracz^2b^2=1$$ $$ a,b>0$$

makes a orientable surface. For do it first of all I prove that the equation $f(x,y,z)$ is differentiable. And it is because is a polynomial. But I also have to prove that the surface is a regular one. And I don´t know how to prove it. I did:$$x=a cosh(u) cos(v)$$$$y=a cosh(u)sin v$$$$z=b sinh(u)$$ and I prove that the gradient is different from $0$ but is not a injective because I can take $v'=v+2*pi$. So I don't know if I have to take another way or I have to prove that it is not orientable.

Thanks in advance!

differential-geometry

share|cite|improve this question

edited Mar 14 at 15:55

Max

9071318

asked Mar 14 at 15:37

MeliodasMeliodas

216

$endgroup$

add a comment | 

$begingroup$

I have to prove that the equation $$ fracx^2a^2+fracy^2a^2-fracz^2b^2=1$$ $$ a,b>0$$

makes a orientable surface. For do it first of all I prove that the equation $f(x,y,z)$ is differentiable. And it is because is a polynomial. But I also have to prove that the surface is a regular one. And I don´t know how to prove it. I did:$$x=a cosh(u) cos(v)$$$$y=a cosh(u)sin v$$$$z=b sinh(u)$$ and I prove that the gradient is different from $0$ but is not a injective because I can take $v'=v+2*pi$. So I don't know if I have to take another way or I have to prove that it is not orientable.

Thanks in advance!

differential-geometry

share|cite|improve this question

edited Mar 14 at 15:55

Max

9071318

asked Mar 14 at 15:37

MeliodasMeliodas

216

$endgroup$

share|cite|improve this question

edited Mar 14 at 15:55

Max

9071318

asked Mar 14 at 15:37

MeliodasMeliodas

216

share|cite|improve this question

edited Mar 14 at 15:55

Max

9071318

asked Mar 14 at 15:37

MeliodasMeliodas

216

edited Mar 14 at 15:55

Max

9071318

edited Mar 14 at 15:55

Max

9071318

asked Mar 14 at 15:37

MeliodasMeliodas

216

asked Mar 14 at 15:37

MeliodasMeliodas

216