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Cartesian and polar coordinates - integrals

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data
    When dealing with an integral integrated with respect to dxdy, I can convert this to polar coordinates, and then integrate with respect to dr d\theta. But I have to multiply with a "r" before integrating.

    If I am dealing with an integral with respect to dydz, I can substitute this with d\theta dz, but not have to multiply with "r". Why is that? (And is what I wrote even correct?)
  2. jcsd
  3. Mar 10, 2008 #2
    - or do I have to multiply with "r" (being the radius, btw) every time, but when it's d\theta dz, I just insert the value for r, but when it's drd\theta, I have to integrate it?
  4. Mar 10, 2008 #3


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    Hi Niles!

    I'm not sure what your y and z are …

    But anyway, if you think about the dimensions, x y z and r (and dx dy and dz) are all lengths, but theta (and dtheta) is only a number.

    So, to keep everything the right dimensions, you'll always need an extra r for every theta or dtheta, to make it a length.

    (That's not a proof, of course … just a useful check when you're not sure you have the right formula!)
  5. Mar 10, 2008 #4


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    The area of the rectangle between x,y and x+dx,y+dy is dx*dy. The area of a the region from r,theta to r+dr,theta+dtheta is r*dr*dtheta. So if you are integrating a function over an area, that's where the extra r comes from.
  6. Mar 10, 2008 #5


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    I seriously doubt that yo can "substitute d[itex]\theta[/itex]dz" without multiplying by r but I don't understand what coordinates you are using. In converting from Cartesian, x, y, z, coordinates to cylindrical r, [itex]\theta[/itex], z the "differential of volume is r dr d[itex]\theta[/itex] dz. The only kind of integral I can think of that would involve "d[itex]\theta[/itex] would be on the surface of a cylinder. And even then, you would multiply by the radius of the cylinder. That is because lines of constant [itex]\theta[/itex] grow farther apart as r increases.
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