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Area of hyperboloid

  1. Apr 22, 2004 #1
    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 [tex]\sqrt{\frac{3}{2}}[/tex] and the intersecting curve is a circle with the radius [tex]\sqrt{\frac{5}{2}}[/tex]. The hyperboloid is defined as
    [tex]
    \left\lbrace\begin{array}{lcl}
    x &=& \sqrt{1+t^2}\cos\varphi \\
    y &=& \sqrt{1+t^2}\sin\varphi \\
    z &=& t
    \end{array}\right.
    [/tex]
    where [tex]-\sqrt{\frac{3}{2}} \le t \le \sqrt{\frac{3}{2}}[/tex] and [tex]0 \le \varphi < 2\pi[/tex]. The area is given by the integral of the crossproduct of the derivates of t and [tex]\varphi[/tex]over an area:
    [tex]
    A = \iint_{\Omega}\left|\frac{\partial f}{\partial t}\times\frac{\partial f}{\partial\varphi}\right|\mathrm{d}t\mathrm{d}\varphi
    [/tex]

    Which yields
    [tex]
    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
    [/tex]
    where a and b are the limits of t.

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

    Thanks in advance!
     
    Last edited: Apr 22, 2004
  2. jcsd
  3. Apr 22, 2004 #2

    arildno

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    I believe your formulae are correct. Check up hyperbolic identities..
     
  4. Apr 25, 2004 #3
    Ok... I was thinking about something like this as well:

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

    This set can be approximated by cylinders as
    [tex]
    \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
    [/tex]

    The area of each cylinder is [tex]2\pi\sqrt{1+z_i^2}\Delta z[/tex], so
    [tex]
    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
    [/tex]

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

    Thanks in advance!
     
  5. Apr 25, 2004 #4

    arildno

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    It's wrong, because the side of your "cylinder" is slightly curved, that's why you get a different area element.
     
  6. Apr 25, 2004 #5
    Are you sure about this? The height of the cylinder [tex]\Delta z \rightarrow 0[/tex], 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
    [tex]
    \int_a^b \sqrt{2t^2+1}\mathrm{d}t
    [/tex]

    I tried to set [tex]t = \sinh s \Rightarrow \mathrm{d}t = \cosh s\mathrm{d}s[/tex]:
    [tex]
    \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
    [/tex]

    I honestly can't get any further! Please help!

    Thanks in advance!
     
  7. Apr 26, 2004 #6

    arildno

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    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).
     
  8. Apr 26, 2004 #7

    arildno

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    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)
     
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