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Math Help - Change to polar coordinates

  1. #1
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    Question Change to polar coordinates

    \displaystyle{  \hdots=\int_0^\infty\int_0^\infty e^{-(u^2+v^2)}du dv  }

    We write u=r\cos(\theta) and v=r\sin(\theta) which simplifies \displaystyle{  e^{-(u^2+v^2)}  } into \displaystyle{  e^{-r^2}  }, but why is differential area now \mathbf{r} dr d\theta?

    \frac{du}{dr}=\cos(\theta) and \frac{dv}{d\theta}=r\cos(\theta) \Rightarrow dudv = r dr d\theta\cos^2(\theta)


    Also on a sidenote, if I rewrite  r as r=\frac{u}{\cos(\theta)}=\frac{v}{\sin(\theta)}\Ri  ghtarrow v =u\tan(\theta) \Rightarrow \theta=\tan^{-1}(\frac{v}{u})\stackrel{u,v\to\infty}{=}\tan^{-1}(\frac{\infty}{\infty})\stackrel{?}{=}\frac{\pi}  {2}.
    (I know that \tan^{-1}(\infty)=\frac{\pi}{2}.)
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  2. #2
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    Quote Originally Posted by courteous View Post
    \displaystyle{  \hdots=\int_0^\infty\int_0^\infty e^{-(u^2+v^2)}du dv  }

    We write u=r\cos(\theta) and v=r\sin(\theta) which simplifies \displaystyle{  e^{-(u^2+v^2)}  } into \displaystyle{  e^{-r^2}  }, but why is differential area now \mathbf{r} dr d\theta?

    \frac{du}{dr}=\cos(\theta) and \frac{dv}{d\theta}=r\cos(\theta) \Rightarrow dudv = r dr d\theta\cos^2(\theta)


    Also on a sidenote, if I rewrite  r as r=\frac{u}{\cos(\theta)}=\frac{v}{\sin(\theta)}\Ri  ghtarrow v =u\tan(\theta) \Rightarrow \theta=\tan^{-1}(\frac{v}{u})\stackrel{u,v\to\infty}{=}\tan^{-1}(\frac{\infty}{\infty})\stackrel{?}{=}\frac{\pi}  {2}.
    (I know that \tan^{-1}(\infty)=\frac{\pi}{2}.)
    Here is why
    Polar coordinate system - Wikipedia, the free encyclopedia

    You can also draw a picture of a polar rectangle and calculate is area directly this is a very informal justification

    Change to polar coordinates-capture.jpg

    Finally you need to calculate the new limits of integration in Polar coordinates.

    Sketch the first quadrant and now decided what values r and \theta need to vary between to sweep out the first quadrant.

    I hope this helps
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    I am not questioning the correctness of rdrd\theta. So, you have to consider the Jacobian determinant (of the coordinate conversion formula)? And if you just differentiate as usual (the way I did it) it is just wrong?
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    Quote Originally Posted by courteous View Post
    I am not questioning the correctness of rdrd\theta. So, you have to consider the Jacobian determinant (of the coordinate conversion formula)? And if you just differentiate as usual (the way I did it) it is just wrong?
    You don't differentiate it, you integrate it.
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    I meant differentiate \frac{du}{dr} and \frac{dv}{d\theta}. What is wrong with my algebraic "reasoning" above?
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    Quote Originally Posted by courteous View Post
    I meant differentiate \frac{du}{dr} and \frac{dv}{d\theta}. What is wrong with my algebraic "reasoning" above?
    Read pp6-7 here: http://mathstat.carleton.ca/~amingar...oordinates.pdf

    Most textbooks that cover vector calculus have a derivation of the jacobian. I suggest you visit your library.
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  7. #7
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    I've just been to library today.
    The lyf so short, the craft so longe to lerne.
    Thank you for a nice pdf!
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