I am trying the compute the surface of a given hyperboloid: $x^2+y^2-z^2=1$ between the planes $z=-4$ and $z=2$.

  1. I have found that the surface element is $$d\vec{S}=(\sqrt{1+z^2}cos\theta,\sqrt{1+z^2}sin\theta,-z),$$ so in order to compute the surface I should use: $$dS=||d\vec{S}||=\sqrt{1+z^2+z^2}=\mathbf{\sqrt{1+2z^2}}d\theta dz.$$ So my doubt it is now:
  2. Why if I use the cilindrical surface element $dS=\rho d\theta dz$, where $\rho=\sqrt{1+z^2}$, giving $$dS=\mathbf{\sqrt{1+z^2}}d\theta dz,$$ I don't get the same answer? It is that $2$ in front of the $z$ that bothers me.

Thanks in advance.


1 Answer 1


wiritnig our surface under the form $f(x,y) = z$ if $z \ge 0$ then $z = \sqrt {{x^2} + {y^2} - 1} $ and if $z \le 0$ then $z = - \sqrt {{x^2} + {y^2} - 1} $. now we pass to polar cordinates to getting V=$2\pi (\int\limits_1^{\sqrt {17} } {\sqrt {{r^2} - 1} rdr + } \int\limits_1^{\sqrt 5 } {\sqrt {{r^2} - 1} rdr)} $

  • $\begingroup$ Thanks but this doesn't answer my question. $\endgroup$
    – mcejalvo
    Jan 16, 2017 at 9:09

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