Re: proof needed

hagman <google@xxxxxxxxxxxxx> wrote in message
82b13c6d-7655-47f4-8f7c-2d4c333d81ae@xxxxxxxxxxxxxxxxxxxxxxxxxxx
On 2 Jul., 17:22, Kiuhnm <kiuhnm03.4t.yahoo.it> wrote:
Let f:R->R^2 be defined as
f(t) = ( r(t)cos(w(t)), r(t)sin(w(t)) ),
where r and w are continuous R->R functions.
Let t1 and t2 be real numbers such that 0 <= t1 < t2 < 2pi, and let
S_{t1,t2} = {(x,y) | there exist k,t in R . ( k in [0,1], t in
[t1,t2] and (x,y) = kf(t) )}.
How can we express the area of S_{t1,t2}?

In this case I would try something like the following.
Let A(t1,t2) be the area of S_{t1,t2}, and h(x) = t1 + (t2-t1)x/n (it
should be h(x,n,t1,t2) but it would be too cumbersome).
A(t1,t2) = lim_{n->inf}
Sum_{x=1}^n |f(h(x-1))| |f(h(x))| sin[h(x)-h(x-1)].
If n->inf then h(x-1)->h(x), f(h(x-1))->f(h(x)),
sin[h(x)-h(x-1)]->h(x)-h(x-1).
Therefore we have
A(t1,t2) = Int_{t1}^{t2} |f(x)|^2 dx.

But how can I prove it?

Kiuhnm

I'd rather guess (at least if w is differentiable; if r' makes
problems, it can surely be eliminated in the argument):

A = 1/2 Int det(x, y; dx/dt, dy/dt) dt
= 1/2 Int (x*dy/dt - y*dx/dt)dt
= 1/2 Int (r cos(w) (r' sin(w) + r w' cos(w)) - r sin(w) (r' cos(w) -
r w' sin(w)) dt
= 1/2 Int r^2 w' dt

That make more sense :-)

Dirk Vdm
.