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my question is: Could we prove the this conversion of variable work by my formula on the bottom?

$$\iint_R f(r,\theta) \ dxdy = \int_a^b \int_0^{r(\theta)} f(r,\theta) r (dr)\ d\theta$$

as $d r$ and $d \theta$ approach $0$. Prove or disprove that:

$$((r+\Delta r) \cos(a +\Delta \theta) -r \cos a) \cdot ((r+\Delta r) \sin(a + \Delta \theta) -r \sin a) / (r \;\Delta \theta \; \Delta r)=1 .$$ where the variable represent as in this graph :

sketch of the situation

as $\Delta r $ and $\Delta \theta$ approach $0$

This question is inspired from $dx\;dy=r \;dr \;d \theta$.

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This question is meaningless in that form. The equation doesn't hold for small finite $\Delta r$ and $\Delta\theta$, and you haven't specified what you mean by "$\Delta r$ and $\Delta\theta$ approach $0$". If you mean whether the two sides have the same limit when $\Delta r$ and $\Delta\theta$ approach $0$, the answer is obviously yes, since the limit on both sides is $0$ (independent of the order in which you take the two limits, which you didn't specify). A meaningful question that you may have had in mind is whether the quotient of the two sides tends to $1$ in an appropriate limit. –  joriki Dec 10 '11 at 20:32
    
@joriki - i know my question is still not perfect after the edition, please help to correct it! –  Victor Dec 10 '11 at 20:42
    
Unfortunately the question doesn't make much more sense now than it did before. $\Delta r$ and $\Delta\theta$ don't occur in the first equation, so it doesn't make sense to add "as $\Delta r$ and $\Delta\theta$ approach $0$". Also, the part that I already commented on is unchanged and thus still meaningless. In case the first line is intended to mean "Can we prove that this substitution of variables works", the answer is yes, we don't need any special manipulations for that, we can apply substitution. –  joriki Dec 10 '11 at 20:50
2  
No, it doesn't look much better, unfortunately. I suggest that you take an elementary course in analysis before you spend more time trying to understand double integrals. It makes no sense to write "as $\mathrm dr$ and $\mathrm d\theta$ approach $0$" because these aren't free variables in the equation; in fact they're not even variables, they're just symbols used to denote an integral. Also, the second displayed equation ($\ldots=1$) is evidently false; you'd have to add some limit like $\lim_{\Delta r\to0}\lim_{\Delta\theta\to0}$ to let it make sense. –  joriki Dec 10 '11 at 21:15
4  
You can read about the Jacobian determinant on Wikipedia. However, as I said, since almost everything you write is full of elementary errors, I believe you should start at a more basic level and familiarize yourself with the elementary concepts and notations of analysis before trying to tackle multiple integrals and Jacobian determinants. Just my subjective bit of advice, of course you're free to ignore it, but I'm also free to ignore further questions ;-) –  joriki Dec 10 '11 at 21:18
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2 Answers

up vote 2 down vote accepted

I've understood your question as the question whether $$\lim\limits_{\substack{\Delta r\to 0\\\Delta a\to 0}} \frac{((r+\Delta r) \cos(a +\Delta \theta) -r \cos a) \cdot ((r+\Delta r) \sin(a + \Delta \theta) -r \sin a)}{r \;\Delta \theta \; \Delta r}=1.$$ I hope this was correct.


The expression you're asking about has the form $[f(a+\Delta a,r+\Delta r)-f(a,r)]\cdot[g(a+\Delta a,r+\Delta r)-g(a,r)]$ for the functions $$f(a,r)=r\cos a$$ $$g(a,r)=r\sin a$$

The following picture should show what this expression represents. (With $(x_1,y_1)=(r\cos a,r\sin a)$ and $(x_2,y_2)$ the same for $r+\Delta r$ and $a+\Delta a$, the given expression is just the are of rectangle in the picture.)

Generic position

The partial derivatives of these functions are $f_r(a,r)=\cos a$, $f_a(a,r)=-r\sin a$, $g_r(a,r)=\sin a$, $g_a(a,r)=r\cos a$.

Using the first order Taylor polynomial we can approximate this expression as $$(\cos a \Delta r-r\sin a\Delta a)(\sin a\Delta r+r\cos a\Delta a)=$$ $$\cos a\sin a (\Delta r)^2 + r(\cos^2a-\sin^2a) \Delta a \Delta r - r^2 \sin a \cos a (\Delta a)^2.$$


Let us check at lest in some special cases whether we get what we would expect.

If $a\approx 0$, then the above expression is close to $r\Delta a\Delta r$. From the following figures we see, that we would really expect it to approximate quite well the area between the two arcs.

r approx 0

r approx 0

But for $\cos a=\sin a = \frac{\sqrt 2}2$ this expression is approximately $\frac{(\Delta r)^2}2-r^2\sin a\cos a(\Delta a)^2$. The following picture show that it can really attain positive values, values close to zero, and negative values.

close to 45 deg

close to 45 deg

close to 45 deg


I am sorry for the large size of the pictures - obviously I am not too good in converting graphics. If someone would like to see the metapost source, it is available here.

Adding an explanation in order to capture intuition behind the above computations.

First let us work with only one variable and with a differentiable functions $h(x)$. The definition of the derivative says that $$\lim\limits_{\Delta x\to 0} \frac{h(x_0+\Delta x)-h(x_0)}{\Delta x}=h'(x_0),$$ i.e. derivative is the number such that $\frac{h(x_0+\Delta x)-h(x_0)}{\Delta x_0} \approx h'(x_0)$ for small $\Delta x$. In other words, if we replace $h(x)$ by a linear function $h(x) \approx h(x_0)+h'(x_0)(x-x_0) =h(x_0)+h'(x_0)\Delta x$, then we do not make too big mistake. For more precise introduction what it means that the mistake is not "too big" see e.g. wikipedia article or your favorite analysis textbook.

Note that the above can be rewritten as $h(x)-h(x_0) \approx h'(x_0)\Delta x$.

Taylor's theorem in two variables is slightly more complicated, but the basic idea is the same. We approximate the function by a linear function. In two dimension this is not a line but a plane. For nice functions (with enough continuous partial derivatives) it can be again shown that mistake will not be "too big".

Partial derivatives are precisely derivatives in the direction parallel to axes. So in two dimension the tangent plane is given by the equation $$h(x,y)-h(x_0,y_0) \approx \frac{\partial h}{\partial x}(x_0,y_0) \Delta x + \frac{\partial h}{\partial y}(x_0,y_0) \Delta y,$$ where $\Delta x=x-x_0$, $\Delta y=y-y_0$.


Perhaps some other questions on this site can be useful in getting intuition about Taylor's polynomial in several variables, e.g.:

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May you explain how to substitude the value into the taylor theorem to get your equation? –  Victor Dec 10 '11 at 21:49
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The diagram that you are giving is not what is going on at all. $\mathrm{d}x\;\mathrm{d}y$ is an element of area intended to represent the plane broken up into small rectangles.:

rectangular grid

$\mathrm{d}r\;\mathrm{d}\theta$ is an element of area in a space whose small squares get mapped to small annular wedges by $x=r\cos(\theta)$, $y=r\sin(\theta)$:

polar grid

The Jacobian is the matrix that locally maps between two coordinate systems. $$ \frac{\partial(x,y)}{\partial(u,v)}=\begin{bmatrix}\frac{\partial x}{\partial u}&\frac{\partial y}{\partial u}\\\frac{\partial x}{\partial v}&\frac{\partial y}{\partial v}\end{bmatrix}\tag{1} $$ From polar to rectangular coordinates, the Jacobian is $$ \frac{\partial(x,y)}{\partial(r,\theta)}=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta)\end{bmatrix}\tag{2} $$ Note that $\begin{vmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta)\end{vmatrix}=r$.

A small area in $\mathrm{d}r\;\mathrm{d}\theta$, the green square, is mapped by the polar coordinate map to the blue annular wedge, which has approximately the same area as the red rectangle. The Jacobian matrix maps the green square to the red rectangle. The ratio of the area of the red rectangle to the green square is the determinant of the Jacobian (this is just linear algebra).

Jacobian

Therefore, since a small square in $\mathrm{d}r\;\mathrm{d}\theta$ is mapped by the coordinate transform so that its area is $\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|\;\mathrm{d}r\;\mathrm{d}\theta$ $$ \begin{align} \iint f(x,y)\;\mathrm{d}x\;\mathrm{d}y &=\iint f(r,\theta)\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|\;\mathrm{d}r\;\mathrm{d}\theta\\ &=\iint f(r,\theta)\;r\;\mathrm{d}r\;\mathrm{d}\theta\tag{3} \end{align} $$

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robjohn: What software did you use for these pictures? // Is there a specific reason why you prefer \int \int to \iint? –  Srivatsan Dec 29 '11 at 14:37
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@Srivatsan: I used Intaglio for these diagrams. As far as \int\int vs \iint, it is simply a lack of sophistication on my part. I will edit my answer since the combined symbol looks better. –  robjohn Dec 29 '11 at 15:58
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