Question

Prove the polar differential rdrdΘ

Answer 1

@amistre64

Answer 2

it requires a jacobian i believe

Answer 3

http://ocw.mit.edu/resources/res-18-007-calculus-revisited-multivariable-calculus-fall-2011/part-v/lecture-3-multiple-integration-and-the-jacobian/

Answer 4

i just LOVE herbert gross :)

Answer 5

Your change of variables is:\[x=r\cos \theta\]\[y=r\sin \theta\]and you should calculate:\[\det \left[\begin{array}{cc} \frac{\partial x}{\partial r}&\frac{\partial r}{\partial r} \\ \frac{\partial x}{\partial \theta}& \frac{\partial y}{\partial \theta}\end{array}\right].\]Thats what goes in front of the dr dtheta.

Answer 6

eh component 1,2 should read \(\partial y/\partial r\) and anyways what you wrote there is actually the transpose of the Jacobian... the determinant will be the same but be careful