# Area of hyperboloid

1. Apr 22, 2004

### LostInSpace

Hi! I'm supposed to find the area of a hyperboloid within a sphere with the radius 2. The hyperboloid and the sphere intersect at $$\sqrt{\frac{3}{2}}$$ and the intersecting curve is a circle with the radius $$\sqrt{\frac{5}{2}}$$. The hyperboloid is defined as
$$\left\lbrace\begin{array}{lcl} x &=& \sqrt{1+t^2}\cos\varphi \\ y &=& \sqrt{1+t^2}\sin\varphi \\ z &=& t \end{array}\right.$$
where $$-\sqrt{\frac{3}{2}} \le t \le \sqrt{\frac{3}{2}}$$ and $$0 \le \varphi < 2\pi$$. The area is given by the integral of the crossproduct of the derivates of t and $$\varphi$$over an area:
$$A = \iint_{\Omega}\left|\frac{\partial f}{\partial t}\times\frac{\partial f}{\partial\varphi}\right|\mathrm{d}t\mathrm{d}\varphi$$

Which yields
$$A = \iint_{\Omega}\sqrt{2t^2 + 1}\mathrm{d}u\mathrm{d}t = 2\pi\int_a^b\sqrt{2t^2 + 1}\mathrm{d}t$$
where a and b are the limits of t.

Is this correct? Then what? How can I calculate the final integral?

Last edited: Apr 22, 2004
2. Apr 22, 2004

### arildno

I believe your formulae are correct. Check up hyperbolic identities..

3. Apr 25, 2004

### LostInSpace

Ok... I was thinking about something like this as well:

The hyperboloid can be defined as
$$\lbrace (x,y,z) \mid x^2 + y^2 - z^2 = 1 \rbrace$$

This set can be approximated by cylinders as
$$\lim_{n\rightarrow\infty}\bigcup_{i=0}^n \lbrace (x,y) \mid x^2 + y^2 = 1 + z_i^2, z_i \le z \le z_i + \Delta z \rbrace$$

The area of each cylinder is $$2\pi\sqrt{1+z_i^2}\Delta z$$, so
$$A = \lim_{n\rightarrow\infty}2\pi\sum_{i=0}^n\sqrt{1+z_i^2}\Delta z = 2\pi\int_a^b\sqrt{z^2 + 1}\mathrm{d}t$$

This is roughly the same, but roughly isn't very good... is this wrong?

4. Apr 25, 2004

### arildno

It's wrong, because the side of your "cylinder" is slightly curved, that's why you get a different area element.

5. Apr 25, 2004

### LostInSpace

Are you sure about this? The height of the cylinder $$\Delta z \rightarrow 0$$, so the curve shouldn't be any problem. Or should it?

Anyway, I assume you are right, but I just wanted to test this. But how do you integrate
$$\int_a^b \sqrt{2t^2+1}\mathrm{d}t$$

I tried to set $$t = \sinh s \Rightarrow \mathrm{d}t = \cosh s\mathrm{d}s$$:
$$\int_a^b 2\pi\sqrt{\mathrm{2\sinh^2 s + 1}}\cosh s\mathrm{d}s = 2\pi\int_a^b \sqrt{\cosh 2s}\cosh s\mathrm{d}s$$

6. Apr 26, 2004

### arildno

Hmm, I thought about that myself today, so I went back to your first post, but at the moment, I can't figure out why your answers differ.
As to the integration, simply set u=sqrt(2)*t, and later on substitute u=sinh(s).

7. Apr 26, 2004

### arildno

I've got it: In your cylindrical approach, you basically use the tangent vectors:
T1=sqrt(1+z^(2))(-sin(w),cos(w),0), T2=(0,0,1).
But T2 is NOT a tangent vector to the surface area!
(T1 is, though)
The true tangent vector makes an angle cos(a)=sqrt((1+z^(2))/(1+2z^(2))) to the vertical.
We then have that the proper area element dA fulfill the relation:
dAcos(a)=dS,
where dS is the cylinder area element (having the vertical as a tangent)