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I am trying to find the surface area of the hyperboloid $x^2 + y^2 − z^2 = 1$ where $0\le z \le 1 $. My book goes ahead making hyperbolic substitutions, however I don't understand why the simple approach fails.

$$\mathbf n= \langle 2x, 2y, -2z \rangle$$ $$=\langle x, y, -z \rangle$$ $$=\langle x, y, -\sqrt{x^2+y^2-1} \rangle$$ $$||\mathbf n||=\sqrt{2x^2+2y^2-1}$$ $$=\sqrt{2r^2-1}$$

Now, since $z=\sqrt{x^2+y^2-1}$ and $0\le z \le 1 $, so if $x^2+y^2=r^2$, then $1 \le r \le \sqrt{2}$. So I continue

$$\int_0^{2\pi}\int_1^\sqrt{2}r\sqrt{2r^2-1}\ dr \ d\theta$$ $$\approx 4.3942$$

My book comes out with $7.9665$, what did I do wrong?

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  • $\begingroup$ I think if you use the normal vector in $x,y$-coordinates, then you need to re-scale so that the $z$-coordinate is 1 before taking the norm. So your normal vector should be $n = \left<\frac{-x}{z}, \frac{-y}{z}, 1 \right>$. $\endgroup$ – Nick May 19 '19 at 3:40
  • $\begingroup$ That is interesting, why would that be? $\endgroup$ – dlp May 19 '19 at 3:43
  • $\begingroup$ If you write your surface as a graph (meaning solve for $z$ to get $z = \sqrt{x^2+y^2-1}$), then the area differential is $dA = \sqrt{z_x^2+z_y^2+1}$, which corresponds to using the normal vector I mentioned. $\endgroup$ – Nick May 19 '19 at 3:44
  • $\begingroup$ Maybe you could explain a little more, I originally thought that all we need is the magnitude of the normal. I hadn't thought about the fact that there are an infinite number of normals and a corresponding number of magnitudes. So how does one decide precisely which normal to use when doing a surface integral? $\endgroup$ – dlp May 19 '19 at 20:25
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When doing a surface integral, as is mentioned above in the comments, it is necessary to choose the "correct" normal vector. You start by choosing a parameterization of your surface, of the form $$r(u,v) = (x(u,v), \, y(u,v), \, z(u,v)) $$ In other words, express the $x,y,$ and $z$ coordinates as functions of two parameters. Next, you compute the tangent vectors to the surface given by the partial derivative vectors:

$$ r_u = (x_u,y_u,z_u) $$ $$ r_v = (x_v,y_v,z_v) $$

The "correct" normal vector to use for the surface integral is the cross product $r_u \times r_v$. The area differential is then the magnitude $|r_u \times r_v|$.

One particular common situation is when the surface is a graph (i.e. it is given by $z=f(x,y)$). In this case, you can choose your parameterization to simply be $u=x$ and $v=y$, with $z=f(x,y)$. So it looks like

$$ r(x,y) = (x,y,f(x,y)) $$

If you take the cross product, you get

$$ r_x \times r_y = (-f_x,-f_y,1) $$

Therefore the area differential is $\sqrt{1+f_x^2+f_y^2}$.

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