show that the parameterization of a torus is ____________ I need help on a homework problem. 
a) A torus of revolution (doughnut) is obtained by rotating a circle C in the xz-plane about the z-axis in space.
See accompanying figure. If C has a radius r>0 and center (R,0,0), show that a parametirization of the torus is
r(u,v) = ((R + rcos(u)*cos(v))i + ((R + cos(u))*sin(v))j + (rsin(u))k

where 
0 </= u </= 2*pi  and 0 </= v </= 2*pi

are the angles in the figure. 
b) Show that the surface area of the torus is 
A = 4*pi^2*R*r

I am not good a "proof" sort of problems, but this kind of looks like the resulting cross product of the parameterization. How do I do this? Anything helps, thanks!
 A: The circle centered at $(R,0,0)$ of radius $r$ in the $xz$-plane can be parametrized as
$$ u \mapsto \begin{pmatrix}R+r\cos u \\ 0 \\ r\sin u\end{pmatrix}, \quad u \in [0,2\pi). $$
The rotation about the $z$-axis by an angle $v\in [0, 2\pi)$ is given by
$$ \begin{pmatrix} \cos v & -\sin v & 0 \\ \sin v & \cos v & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$
Hence, a parametrization of the torus is obtained by composing the latter two, i.e.
the parametrization is given by
$$ \Psi: (u,v) \mapsto \begin{pmatrix} \cos v & -\sin v & 0 \\ \sin v & \cos v & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}R+r\cos u \\ 0 \\ r\sin u\end{pmatrix}
= \begin{pmatrix}(R+r\cos u)\cos v \\ (R+r\cos u)\sin v \\ r\sin u\end{pmatrix}. $$
The surface area of a surface $S$ parametrized by $\Psi:U \to \mathbb{R}^3$ defined as
$$ A(S) = \int_S \,\mathrm dA = \int_U |\partial_u\Psi \times \partial_v \Psi| \, \mathrm du \, \mathrm dv, $$
so you need to compute the tangent vectors $\partial_u\Psi$ and $\partial_v \Psi$, as well as the modulus of their cross product and insert it back into the integral. Can you do that on your own?
