myininaya one year ago Show $r dr d \theta =dx dy$ .

1. myininaya

Also I know one way is to use the Jacobian. I wonder if there is any other way.

2. iPwnBunnies

Idk bout Jacobian...But I gotta do alot of drawing.

3. iPwnBunnies

|dw:1396312983170:dw|

4. iPwnBunnies

arclength = r*theta dA = r*dtheta*dr dy dx, dx dy = r dr dtheta

5. Kainui

You can do it geometrically. But you can definitely do it with the jacobian. The Jacobian is a lot simpler than some people make it out to be. So if you change the integral to another coordinate set then just right away we can see: |dw:1396312970588:dw| So obviously when we scale it the squares get larger as we move out and we need to account for that. The determinant is just the area of a matrix right? Well similarly, the determinant of a matrix of partial derivatives is just saying how much a little bit of area changes at each point. In fact you were doing this with u-substitution you just didn't know it because the derivative you took was only in 1 dimension.

6. iPwnBunnies

What I drew was a polar rectangle, R... $R = [(r,\Theta) | a \le r \le b, \alpha \le \Theta \le \beta]$

7. myininaya

@ipwnbunnies I like your geometric picture to show this. You found the area of that one piece using the fact that looked sorta rectangular-ish (use the area of a rectangle) Yep jacobian is pretty easy. $J(r, \theta) dr d \theta =dx dy$ $(x_r \cdot y_\theta -x_\theta \cdot y_ r) dr d \theta = dx dy$ $(\cos(\theta) \cdot r \cos(\theta)--rsin(\theta) \cdot \sin(\theta)) d r d \theta=dx dy$ $r(\cos^2(\theta)+\sin^2(\theta))dr d \theta = dx dy$ $r \cdot 1 d r d \theta= dx dy$ $r dr d \theta=dx dy$ I think it is great you guys like math enough to discuss things like this. Kainui, I like that you posting problems to get us advanced math people thinking a little more.

8. iPwnBunnies

Not familiar with Jacobian anything...But it looks interesting.

9. iPwnBunnies

Yeah, my proof is a rectangular-like, but as the polar rectangle gets smaller, it "looks" like a rectangular rectangle. :)

10. Kainui

@iPwnBunnies Have you taken linear algebra? Basically it's just the area of a parallelogram made by two vectors in a matrix. |dw:1396314943369:dw| So if you have a matrix $\left[\begin{matrix}u_x & u_y \\ v_x & v_y\end{matrix}\right]$ the determinant of this is the area. The jacobian just has how the coordinates change in each of the directions, so it's the area of the transformation really. I hope that sort of makes it a little more understandable. Of course if you have 3 vectors or higher the determinant is the volume, or higher dimensional analogue.

11. iPwnBunnies

Oh yeah, I know the area is the cross product of the two vectors. I guess I'm not familiar with the term 'matrix'. Just vectors and determinants lol.