# Surface integral of hyperboloid using polar coordinates fails?

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?

• 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 $n = \left<\frac{-x}{z}, \frac{-y}{z}, 1 \right>$. – Nick May 19 '19 at 3:40
• That is interesting, why would that be? – dlp May 19 '19 at 3:43
• 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. – Nick May 19 '19 at 3:44
• 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? – dlp May 19 '19 at 20:25

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 $$\sqrt{1+f_x^2+f_y^2}$$.