Length of a Curve and Area of Torus Consider a parametrised surface (the torus of revolution),
$$
M(u, v) = ((a + r \cos u) \cos v, (a + r \cos u) \sin v, r \sin u)   $$   where $a > r > 0$.
a) Calculate the arc length of the curve $u =\pi/2$ between the endpoints corresponding
to $v = 0$ and $v = 2\pi$.
b) Calculate the area of the region specified by $0 < u < 2\pi,\ 0 < v < 2\pi$
I have calculated the first and second fundamental forms but not too sure how to proceed from here. 

a) If we fix $u=\pi/2$ then we have a curve $c(v)=M(\pi/2,v)$ so that
$$ c'(v) = (-a\sin\ v,a\cos\ v, 0),\ {\rm length} =
\int_0^{2\pi} |c'(v)| dv = 2\pi a $$
(Intuitive Way : Draw a torus $T$ in ${\bf R}^3$. Then described curve is
the set $\{ (x,y,z)\in T\subseteq {\bf R}^3 |\ z=r \}$. So its
length is $2\pi a$.)
b) Specified area is area of whole torus. Not to use second we can
calculate :
$$ M_u=( (-r\sin\ u)\cos\ v,(-r\sin\ u)\sin\ v,\ r\cos \ u ) $$
$$ M_v = (-(a+r\cos\ u)\sin\ v
,(a+r\cos\ u)\cos\ v,0) $$
Then the normal vector $N$ to torus (not unit) is $$ N=M_u\times M_v
= (-r(a+r\cos\ u)\cos\ u\cos\ v, -r(a+r\cos\ u)\cos\ u\sin\ v,
(-r\sin\ u)(a+r\cos\ u))
$$
 Here $$ E=M_u\cdot M_u,\ F=M_u\cdot M_v,\ G=M_v\cdot M_v $$
So $$ |N| ={\rm area} (M_u,M_v)=|M_u||M_v| \sin\
 \angle (M_u,M_v) $$ $$= \sqrt{ EG-F^2}  = r
  (a+r\cos\ u) $$
So $$ {\rm area} (T) = \int_{[0,2\pi]^2} |N| dudv = 2\pi
\int_0^{2\pi } r
  (a+r\cos\ u) du =4\pi^2 ar  $$
