# Change to polar coordinates

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• Nov 7th 2010, 01:50 PM
courteous
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 $\frac{dv}{d\theta}=r\cos(\theta) \Rightarrow dudv = r dr d\theta\cos^2(\theta)$(Wondering)

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 $\tan^{-1}(\infty)=\frac{\pi}{2}$.)
• Nov 7th 2010, 02:21 PM
TheEmptySet
Quote:

Originally Posted by courteous
$\displaystyle{ \hdots=\int_0^\infty\int_0^\infty e^{-(u^2+v^2)}du dv }$

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

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

Attachment 19622

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
• Nov 7th 2010, 10:58 PM
courteous
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?
• Nov 7th 2010, 11:03 PM
Prove It
Quote:

Originally Posted by courteous
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.
• Nov 7th 2010, 11:36 PM
courteous
I meant differentiate $\frac{du}{dr}$ and $\frac{dv}{d\theta}$. What is wrong with my algebraic "reasoning" above?
• Nov 7th 2010, 11:54 PM
mr fantastic
Quote:

Originally Posted by courteous
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.
• Nov 8th 2010, 12:54 PM
courteous
I've just been to library today. (Nod)
Quote:

The lyf so short, the craft so longe to lerne.
Thank you for a nice pdf!