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 d θ = dx dy

When I do the partial differentations I get $dx dx / dr d θ - dx dy / dr d θ = r$

Which is obviously a load of rubbish!

Alex Miller Alex

Posted on Wednesday, 09 July, 2003 - 08:01 pm:

x = r \cos (θ)

and

y = r \sin (θ)

\frac{\partial x}{\partial r} = \cos (θ),

\frac{\partial x}{\partial θ} = - r \sin (θ),

\frac{\partial y}{\partial r} = \sin (θ),

and

\frac{\partial y}{\partial θ} = r \cos (θ) .

On the top row of our det matrix is

\frac{\partial x}{\partial r}

and

\frac{\partial x}{\partial θ}

. On the bottom row of our det matrix is

\frac{\partial y}{\partial r}

and

\frac{\partial y}{\partial θ}

just sub the things in and find the det. It comes out to be

\cos (θ) \times r \cos (θ) - \sin (θ) \times (- r \sin (θ))

r (\cos^{2} (θ) + \sin^{2} (θ)) = 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 \times d θ) \times (dr)

$= r \times dr \times d θ$

love arun