How to calculate this area?

Question

Calculate the area of $$S:=\lbrace (x,y,z)\in\mathbb{R}^3: y^2+(z-x^2)^2=x^2-1\rbrace$$

Anyone have an idea? I tried using cylindrical coordinates, but nothing.

Well, I have something, but I'm not really sure... Ok:

We have the surface $$y^2+(z-x^2)^2=x^2-1 \Rightarrow (z-x^2)^2=x^2-1-y^2$$ $$\Rightarrow z-x^2=\sqrt{x^2-1-y^2} \Rightarrow z(x,y) = \sqrt{x^2-1-y^2}+x^2$$

Now we put some reestriction on z. We have two cases:

$$S^{-} = \lbrace C \cap \lbrace z<x^2\rbrace \rbrace$$ $$S^{+} = \lbrace C \cap \lbrace z>x^2\rbrace \rbrace$$

Let's work on $S^{+}$. Using the polar parametrization $x=rcos(\theta)$ e $y=rsin(\theta)$ we have:

$$z(\theta, r) = \sqrt{r^2cos(2\theta)-1}+r^2cos^2 (\theta)$$ Where we have 2 restrictions: $r>1$ and $r^2cos(\theta)>1$ (looking the root). So we can calculate the integration limits.

And we have the vector: $$\gamma(\theta, r) = (rcos(\theta), rsin(\theta), z(\theta, t))$$ So we only need use the area equation $$\int \int \left| \frac{\partial\gamma}{\partial \theta}\times \frac{\partial\gamma}{\partial r}\right|d\theta dr$$

Answer 1

Reviewing the definition of an smooth surface again; we can form a Riemann Sum for a bounded function $f$ defined at all points of an smooth surface of finite area in $\mathbb R^3$. The area, as you noted above, seems to be infinite, so as @Greg said, I don't think the surface becomes finite.

Answer 2

The given equation simplifies to $x = (z^4+y^2+1)/2z^2$. So for every $(y,z)$ with $z\ne0$, there is a unique value of $x$ such that $(x,y,z)$ is on the surface. Furthermore, this set of $x$ is unbounded. So I'm pretty sure the surface area will be infinite.