LostInSpace
Apr22-04, 06:44 AM
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 \varphiover an area:
A = \iint_{\Omega}\left|\frac{\partial f}{\partial t}\times\frac{\partial f}{\partial\varphi}\right|\mathrm{d}t\mathrm{d}\va rphi
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?
Thanks in advance!
\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 \varphiover an area:
A = \iint_{\Omega}\left|\frac{\partial f}{\partial t}\times\frac{\partial f}{\partial\varphi}\right|\mathrm{d}t\mathrm{d}\va rphi
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?
Thanks in advance!