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Dx in polar?

  1. Oct 21, 2013 #1
    Ok, so we know that [itex]x=rcos(\theta)[/itex]

    So what is dx?

    ***

    Furthermore, can I get dS in polar by finding dx and dy in polar and then substituting them into dS for rectangular? Is there an easier way to solve for dS in polar?
     
    Last edited: Oct 21, 2013
  2. jcsd
  3. Oct 21, 2013 #2
    You can figure out the differentials [itex]dx[/itex] and [itex]dy[/itex] from the general formula for a multivariable differential: $$ x = x(r, \theta) = r \cos \theta \Leftrightarrow dx = \frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta = \cos \theta\,dr - r \sin \theta\,d\theta $$
    $$ y = y(r, \theta) = r \sin \theta \Leftrightarrow dy = \frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta = \sin \theta\,dr + r \cos \theta\,d\theta $$
    For the relation between the area elements [itex]dA[/itex], you have to use the Jacobian:
    $$ dx dy = \frac{\partial(x,y)}{\partial(r,\theta)}dr\,d\theta = (r \cos^2 \theta + r \sin^2 \theta)\,dr\,d\theta = r\,dr\,d\theta
    $$
    This is also easily seen if you sketch out what is happening. If you change [itex]\theta[/itex] by a small amount, you change one side of your area element by [itex]r\,d\theta[/itex]. If you change [itex]r[/itex] by a small amount, you change the other side of your area element by [itex]dr[/itex].
     
  4. Oct 22, 2013 #3

    arildno

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    As you can see, JPaquim's formula for the Jacobian is simply the determinant of the transformation matrix between the differentials.
    And if you remember your linear algebra, it is precisely the determinant of a square matrix that tells you of how the area is transformed when going from one set of basis vectors to another by means of matrix transformation.
     
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