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- How do we get rdrdt from dxdy?

The area differential ##dA## in Cartesian coordinates is ##dxdy##.

The area differential ##dA## in polar coordinates is ##r dr d\theta##.

How do we get from one to the other and prove that ##dxdy## is indeed equal to ##r dr d\theta##?

$$

dx d\theta = \cos \theta dr d\theta -r \sin \theta\; , \;dy d\theta = \sin \theta dr d\theta +r \cos \theta

$$

The differentials then become

$$

dx= \cos \theta dr -r \sin \theta d\theta \; , \;dy = \sin \theta dr + r \cos \theta dr \theta

$$

which to me does not make any sense.

It seems as though finding ##dr d\theta ## is what we must do, but how? What am I missing?

How do we prove that indeed ...

$$

dx dy = r dr d\theta

$$

Is it in Griffiths? On what page?

The area differential ##dA## in polar coordinates is ##r dr d\theta##.

How do we get from one to the other and prove that ##dxdy## is indeed equal to ##r dr d\theta##?

##dxdy=r dr d\theta##

The trigonometric functions are used to obtain the conversions ##x=r \cos \theta ## and ##y=r \sin \theta ## . If we differentiate both with respect to the polar angle ##\theta##,

$$

dx d\theta = \cos \theta dr d\theta -r \sin \theta\; , \;dy d\theta = \sin \theta dr d\theta +r \cos \theta

$$

The differentials then become

$$

dx= \cos \theta dr -r \sin \theta d\theta \; , \;dy = \sin \theta dr + r \cos \theta dr \theta

$$

which to me does not make any sense.

It seems as though finding ##dr d\theta ## is what we must do, but how? What am I missing?

How do we prove that indeed ...

$$

dx dy = r dr d\theta

$$

Is it in Griffiths? On what page?

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