Alex Holyoake
Posted on Wednesday, 09 July, 2003 - 06:43 pm:
I am getting rather annoyed with the fact that I can't prove that r dr dq = dx dy

When I do the partial differentations I get dx dx/dr dq-dx dy/dr dq = r

Which is obviously a load of rubbish!

Alex Miller Alex
Posted on Wednesday, 09 July, 2003 - 08:01 pm:
x=rcos(q) and y=rsin(q) so
x
r
 =cos(q),

x
q
 =-rsin( q),

y
r
 =sin(q),
and
y
q
 =rcos( q).
On the top row of our det matrix is
x
r

and
x
q

. On the bottom row of our det matrix is
y
r

and
y
q

just sub the things in and find the det. It comes out to be cos(q)×rcos(q)-sin(q) ×(-rsin(q)) so r(cos2(q)+ sin2(q))=r.
Alex Miller Alex
Posted on Wednesday, 09 July, 2003 - 08:06 pm:
However, I am not too clear on your notation. So, if my work was not pertinent to your post, please post back.
Arun Iyer
Posted on Wednesday, 09 July, 2003 - 08:17 pm:
Thinking Geometrically,
dxdy = infinitesimal area

in polar coordinates,
infinitesimal area = (length of arc)*(small change in radius)
 =(r×dq)×(dr)

=r×dr×dq

love arun