The Jacobian

**demode** · April 30th 2010, 10:53 PM

Here's my question:

(a)

$\int^{2 \pi}_0 \int^1_0 [r^2(cos^2\theta + sin^2\theta)]rdr d\theta$

$\int^{2 \pi}_0 \int^1_0 r^3 dr d \theta$ $= \int^{2 \pi}_0 \frac{r^4}{a} |^1_0 d \theta$

$=\frac{1}{4} \theta |^{2 \pi}_0 = \frac{\pi}{2}$

Is this correct?

(b) So, is the Jacobian for this problem given by the following?

$ J(x,y)= \frac{\partial(r,\theta)}{\partial (x,y)}=\begin{bmatrix} {\partial r\over \partial x} & {\partial r\over \partial y} \\ {\partial \theta\over \partial x} & {\partial \theta\over \partial y} \end{bmatrix} $

If so, how can I obtain the four partials in the matrix?

**Redding1234** · April 30th 2010, 11:25 PM

The answer for (a) looks correct to me.

When you're changing to polar coordinates, you need to use the Jacobian. It is r, as you have used in (a).

To calculate the Jacobian, you use the same idea as you have written, but the numerator and denominator should be switched. So

$ \frac{\partial(x,y)}{\partial (r,\theta)}=\begin{bmatrix} {\partial x\over \partial r} & {\partial x\over \partial \theta} \\ {\partial y\over \partial r} & {\partial y\over \partial \theta} \end{bmatrix} $

Using $x = r\cos\theta$ and $y = r\sin\theta$ , the rest will follow.

**slovakiamaths** · April 30th 2010, 11:52 PM

first part of your problem is correct

for secondone see attachment

**HallsofIvy** · May 1st 2010, 02:38 AM

Originally Posted by demode

Here's my question:

(a)

$\int^{2 \pi}_0 \int^1_0 [r^2(cos^2\theta + sin^2\theta)]rdr d\theta$

$\int^{2 \pi}_0 \int^1_0 r^3 dr d \theta$ $= \int^{2 \pi}_0 \frac{r^4}{a} |^1_0 d \theta$

$=\frac{1}{4} \theta |^{2 \pi}_0 = \frac{\pi}{2}$

Is this correct?

(b) So, is the Jacobian for this problem given by the following?

$ J(x,y)= \frac{\partial(r,\theta)}{\partial (x,y)}=\begin{bmatrix} {\partial r\over \partial x} & {\partial r\over \partial y} \\ {\partial \theta\over \partial x} & {\partial \theta\over \partial y} \end{bmatrix} $

If so, how can I obtain the four partials in the matrix?

The other responders have shown that. I want to point out that your use of " $r drd\theta$ " already has the Jacobian in it! If u(x,y) and v(x,y) are a change of variable then $dx dy= \frac{\partial (u, v)}{\partial (x, y)} dudv$ . Saying that $dx dy= r dr d\theta$ is the same as saying $\frac{\partial (r, \theta)}{\partial (x, y)}= r$

**demode** · May 1st 2010, 04:56 PM

I followed Slovakiamaths' working and the Jacobian is $J=r$ . So to answer part (b), the only point interior to R at which the Jacobian equals zero is (0,0), right?

Furthermore, the question asks: "Prove that the mapping used is 1 to 1 on region R excluding the boundary points, and excluding the origin. Note: one way to do this is to produce the inverse mapping." I know that a mapping is one to one if distinct points in the $r, \theta$ plane have distinct images in the xy-plane. But in this case how do I need to prove it?

**demode** · May 2nd 2010, 01:23 PM

Okay, my mistake. Here's the mapping

$x = rcos\theta$
$y = rsin\theta$

I don't know how to show it's one to one. I have to show that every point on the unit circle (except (1,0) ) is the mapping of exactly one angle in the interval 0 to 2π. But I don't know how to check that for all the points. Can anyone help?

**maddas** · May 2nd 2010, 01:40 PM

Are you aware that the sine and cosine are 1-1 on the interval [0,2pi)?

**demode** · May 3rd 2010, 12:42 AM

Originally Posted by maddas

Are you aware that the sine and cosine are 1-1 on the interval [0,2pi)?

Yes, I think they are. But how do we show/prove it, in order to answer the question?

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