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(adsbygoogle = window.adsbygoogle || []).push({});

[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!

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# Area of hyperboloid

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