Possible Duplicate:
surface area of torus of revolution
Let $R>r>0$ fixed. I want to compute the Area of $S=\operatorname{Im} \phi$ given by $\phi(s,t):= \begin{pmatrix}(R+r\cos s )\cos t \\ (R+r\cos s) \sin t \\ r \sin s \end{pmatrix}$
First I compute: $D\phi = \begin{pmatrix}-r\sin(s)\cos t & -R\sin(t) \\ -r\sin(s)\sin(t) & R\cos(t) \\ r\cos(s) & 0 \end{pmatrix}$
then the cross product of the first vertical row with the second vertical row is: $\partial_{1}\phi \times \partial_{2} \phi = \begin{pmatrix} Rr\cos(s)\cos(t) \\ -Rr\sin(t)\cos(s)\\ -Rr\sin(s)\cos^{2}(t)-Rr\sin(s)\sin^{2}(t) \end{pmatrix},$
so $\left\|\partial_{1} \phi \times \partial_{2} \phi \right\| =\sqrt{R^{2}r^{2}(\cos^2 (s)(\cos^2(t)+\sin^2 (t))+\sin^2(s))} = Rr,$
and that gives:$\int_0^{2\pi} \int_0^{2\pi} Rr \;ds \;dt = 4\pi^2 Rr$
The results seems to be fine, but are my calculations fine, too? Thanks for any help.