Vtyjkyuk

I have some trouble with the following questions:

$mathbbR^3$ 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 $mathbbS^1 subseteq mathbbR^2$. Write for the standard 2-Torus $mathbbT^2= mathbbS^1 times mathbbS^1$, then $mathbbT^2 subseteq mathbbR^4$ is a two dimensional manifold. Prove that $mathbbT^2$ and $T$ are diffeomorph.

a)
Try $T=lbrace (sqrty^2+z^2 - b)^2 + z^2 = a, (x,y,z) in mathbbR^3 rbrace$, 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
$f:mathbbR^3 to mathbbR, (x,y,z) mapsto (sqrty^2+z^2 - b)^2 + z^2 - a$
(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 : mathbbR^2 to mathbbR^3, (theta, psi) mapsto left(
matrix

cos 2pi theta&0&-sin 2pi theta \
0&1&0 \
sin 2pi theta&0&cos 2pi theta

right)left(
matrix

0 \
sqrta cos 2pipsi \
b+sqrta sin 2pi 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 $mathbbZ^2$ periodic, so induces a map $(mathbbR/Z)^2 to mathbbR^3$

_this new map is the diffeomorphism you need!!