# Thread: surface area of the torus

1. ## surface area of the torus

Hello everybody...

I have a problem, can help me with this??

A torus of radius 3 (and cross-sectional radius 1) can be represented parametrically by the function

by:
where D is the rectangle given by .

2. You have to integrate the length of the normal vector to the surface over the region given in the $\theta \phi$ plane.

3. First find the tangent vectors $T_{\theta}$ and $T_{\phi}$

In this problem, for example, $T_{\theta} = \frac {\partial \Phi}{\partial \theta} = (-(3+cos\phi)sin \theta, (3+cos \phi)cos \theta,0)$

Then find the cross product of $T_{\theta}$ and $T_{\phi}$

Now you have a vector normal to the surface of the torus. Find the magnitude of this vector.

then surface area of the torus = $\int^{2 \pi}_{0} \int^{2 \pi}_{0} ||T_{\theta} \times T_{\phi}|| d \phi d \theta$

4. ThanKsssss You Saved My Life... !!!!!!