Originally Posted by

**chopet** I am trying to convert dxdy in cartesian into polar plane.

and have to prove: dxdy = rd$\displaystyle \theta$dr

Here is my conversion proof:

x = rcos$\displaystyle \theta$

y = rsin$\displaystyle \theta$

dx = -rsin$\displaystyle \theta$d$\displaystyle \theta$ + cos$\displaystyle \theta$dr

dy = rcos$\displaystyle \theta$d$\displaystyle \theta$ + sin$\displaystyle \theta$dr

ignoring 2nd order differentials, and keeping only drd$\displaystyle \theta$,

we have:

dxdy = -rsin$\displaystyle \theta$d$\displaystyle \theta$sin$\displaystyle \theta$dr

+ rcos$\displaystyle \theta$drcos$\displaystyle \theta$d$\displaystyle \theta$

= rd$\displaystyle \theta$dr(-sin $\displaystyle ^2 \theta$ + cos$\displaystyle ^2 \theta$)

I need to equate (-sin $\displaystyle ^2 \theta$ + cos$\displaystyle ^2 \theta$) = 1,

but i can't because of the -ve sign. Where did I go wrong?

Thanks for your time.