Jacobian determinants and area integrals.

**Confused77** · August 23rd 2008, 05:20 AM

I have the following past exam question.

Write down the Cartesian coordinates x and y in terms of the plane polar coordinates r and $\theta$

Which is easy.

Next, evaluate the jacobian determiant

By considering the matrix equation relating the vectors
$\begin{pmatrix}{dx}\\{dy}\end{pmatrix}$
and
$\begin{pmatrix}{dr}\\{d\theta}\end{pmatrix}$
use this result to obtain an expression for the area element dxdy in plane polar coordinates.

b)Work out the numerical value of the same area integral

$\int(r^2+1)dA$

over the interior of the circle of radius 2 centred at the origin.

**wingless** · August 23rd 2008, 05:25 AM

Transformation from cartesian coordinates to polar:
$x = r \cos \theta$
$y = r \sin \theta$

Can you do it now? If you can't, where are you stuck?

**Confused77** · August 23rd 2008, 05:32 AM

That's the part I did get, I seem unable to evaluate that determinant correctly. and then do the part just after.

**wingless** · August 23rd 2008, 05:52 AM

OK.

I'll only work for two dimensions. Let us have a transform from (x,y) to (a,b) :
$x = f(a,b)$ and $y=g(a,b)$ .

The Jacobian matrix is defined as

$J = \frac{\partial (x,y)}{\partial (a,b)} = \begin{bmatrix} \dfrac{\partial x}{\partial a} & & \dfrac{\partial x}{\partial b} \\ \\ \dfrac{\partial y}{\partial a} & & \dfrac{\partial y}{\partial b}\end{bmatrix}$

$|J| = \left | \begin{array}{ccc} \dfrac{\partial x}{\partial a} & & \dfrac{\partial x}{\partial b} \\ \\ \dfrac{\partial y}{\partial a} & & \dfrac{\partial y}{\partial b}\end{array} \right |$

In our example, this is,

$|J| = \left | \begin{array}{ccc} \dfrac{\partial r \cos \theta }{\partial r} & & \dfrac{\partial r \cos \theta}{\partial \theta} \\ \\ \dfrac{\partial r \sin \theta}{\partial r} & & \dfrac{\partial r \sin \theta}{\partial \theta}\end{array} \right |$

And remember that the determinant of a 2x2 matrix is $\left | \begin{array}{cc} a & b \\ c & d \end{array} \right | = ad - bc$

**Confused77** · August 25th 2008, 03:42 AM

Thank you very much, have done that bit and part b now.

I'm still stuck on this part however.

By considering the matrix equation relating the vectors
$\begin{pmatrix}{dx}\\{dy}\end{pmatrix}$
and
$\begin{pmatrix}{dr}\\{d\theta}\end{pmatrix}$
use this result to obtain an expression for the area element dxdy in plane polar coordinates.

**wingless** · August 25th 2008, 09:28 AM

Let R be our region that we integrate f on.

$\int\int\limits_R f(x,y)~dy~dx = \int\int\limits_R f(x,y)~dx~dy = \int\int\limits_R f(r\cos\theta,r\sin\theta) |J|~dr~d\theta$

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