Torus, manifolds I have some trouble with the following questions: 
$\mathbb{R}^3$ has standard coördinates $(x, y, z)$. Regard in the plane $x=0$ the circle with centre $(x,y,z) = (0,0,b)$ and radius $a$, $0<a<b$. The area that arise when you turn the circle around the y-axis is called T. 
1A. Give the equation of T and prove that it's a manifold of dimension 2.
I thought the following: 
$$T= \int_{C} \pi (f(y))^2 dy $$ where C is the circle described above and $f(y)= \sqrt(a-y^2+2zb-b^2)$ But now I don't know how to continue, cause I don't really have any boundaries.
B. Regard now $\mathbb{S}^1 \subseteq \mathbb{R}^2$. Write for the standard 2-Torus $\mathbb{T}^2= \mathbb{S}^1 \times \mathbb{S}^1$, then $\mathbb{T}^2 \subseteq \mathbb{R}^4$ is a two dimensional manifold. Prove that $\mathbb{T}^2$ and $T$ are diffeomorph.  
 A: a)
Try $T=\lbrace (\sqrt{y^2+z^2} - b)^2 + z^2 = a,  (x,y,z) \in \mathbb{R}^3 \rbrace$, i think it gives you the required parametrization of your torus, embedded in the 3 dimensional euclidian space.
It's a 2 manifold because it's the zeros of the submersion 
$f:\mathbb{R}^3 \to \mathbb{R}, (x,y,z) \mapsto (\sqrt{y^2+z^2} - b)^2 + z^2 - a$
(you can check it easily).
b) For showing it's diffeomorphic to the product of two circles, you need to give first a two dimensional parametrization of T:
$\phi : \mathbb{R}^2 \to \mathbb{R}^3, (\theta, \psi) \mapsto \left(
\matrix
{
\cos 2\pi \theta&0&-\sin 2\pi \theta \\
0&1&0 \\
\sin 2\pi \theta&0&\cos 2\pi \theta    
}
\right)\left(
\matrix
{
0 \\
\sqrt{a} \cos 2\pi\psi \\
b+\sqrt{a} \sin 2\pi \psi   
}
\right) $
Note that the vectr of the right defines just your circle C, when the matrix gives you the rotation around the y axis.
Then you have to show that
_this map is smooth
_it's $\mathbb{Z}^2$ periodic, so induces a map $(\mathbb{R/Z})^2 \to \mathbb{R}^3$
_this new map is the diffeomorphism you need!!
