# Thread: Increments of area dxdy

1. ## 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!

2. Area and volume elements in polar coordinate systems

By the way, $\displaystyle dx\,dy = r\,dr\,d\theta$ IS true.

3. 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 $\displaystyle r cos^2(\theta)drd\theta+ r sin^2(\theta)drd\theta= r drd\theta$.)

4. 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,