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Thread: Increments of area dxdy

  1. #1
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    Increments of area dxdy

    Hello everyone!

    I've lately been thinking about this: why isn't $\displaystyle dxdy$ equal to $\displaystyle rdrd\theta$.

    Okay now, $\displaystyle x=r\cos\theta$ and $\displaystyle y=r\sin\theta$.
    So using the derivative rule, we say that $\displaystyle dx=\cos\theta dr-r\sin\theta d\theta$ and that $\displaystyle dy=\sin\theta dr+r\cos\theta d\theta$...
    But when we carry out $\displaystyle dxdy$ we get something very long...
    How is that?

    Thanks!
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  2. #2
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    Area and volume elements in polar coordinate systems

    By the way, $\displaystyle dx\,dy = r\,dr\,d\theta$ IS true.
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  3. #3
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    You cannot just multiply differentials like that: If x= u(s, t) and y= v(s, t) then $\displaystyle dxdy= J(x,y;u,v)dsdt$ where J(x,y;u,v) is the Jacobian determinant: $\displaystyle \left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}\end{array}\right|$.

    In this case, u and v are r and $\displaystyle \theta$, respectively, $\displaystyle x= r cos(\theta)$ and $\displaystyle y= r sin(\theta)$ so $\displaystyle J(x, y, r, \theta)= \left|\begin{array}{cc}cos(\theta) & sin(\theta) \\ -r sin(\theta) & r cos(\theta)\end{array}\right|= (cos(\theta)(r cos(\theta))- (sin(\theta))(-r sin(\theta))$$\displaystyle = r cos^2(\theta)+ r sin^2(\theta)= r$.

    (In more advanced mathematics, differential geometry, we use that to define the "algebra of differentials" in such a way that it is anti-commutative- that is, that ab= -ba. From that $\displaystyle a^2= a(a)= -(a)a= -a^2$ so all squares are 0.

    If $\displaystyle x= r cos(\theta)$ then $\displaystyle dx= cos(\theta) dr- r sin(\theta)d\theta$ and if $\displaystyle y= r sin(\theta)$ then $\displaystyle dy= sin(\theta) dr+ r cos(\theta)d\theta$.

    Then $\displaystyle dxdy= (cos(\theta) dr- r sin(\theta)d\theta)(sin(\theta) dr+ r cos(\theta)d\theta)$. The terms we would get by multiplying dr and dr together or $\displaystyle d\theta$ and $\displaystyle d\theta$ together are 0 leaving $\displaystyle r cos^2(\theta)drd\theta- r sin^2(\theta)d\theta dr$ which, since multiplication is anti-commutative, is the same as $\displaystyle r cos^2(\theta)drd\theta+ r sin^2(\theta)drd\theta= r drd\theta$.)
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  4. #4
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    Dear HallsofIvy,

    Thank you for commenting, I am aware of the Jacobian (stretching factor)... But never considered this case.
    Regarding differential Geometry, can you recommend me a book (an intro) from calculus III to differential geometry? I would appreciate it...

    Best,
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