# Change of variable, why can I not multiply the differentials directly

1. Feb 7, 2010

### phucnguyen

Hi, I'm learning to do double integration by changing variables and wondering about this.

Suppose we have f(x, y) and want to find the volume under the surface over some bounded area in the xy plane.

Say, I want to change the variables into u and v by:
u = 3x - 2y
v = x + y

I need to find the relations between dxdy and dudv.

Now I have:
du = 3dx - 2dy
dv = dx + dy

So
dudv = (3dx - 2dy)(dx + dy) = 3(dx)^2 + dxdy - 2(dy)^2

Dividing both sides by dxdy, we obtain:
(dudv)/(dxdy) = 3(dx/dy) + 1 - 2(dy/dx)

Since x and y are independent, dx/dy and dy/dx are 0.

Hence I conclude dudv = dxdy.

It's easily to find a counter example to this. The ratio is actually a constant of 5.

Where have I been wrong here? Thank you very much.

2. Feb 7, 2010

### Landau

From what text or course are you learning this stuff? You shouldn't perform magical tricks with differentials. You just need to compute the absolute value of the determinant of the Jacobian, see here.

In your case, $$D\phi=\begin{bmatrix} 3 & -2\\ 1 & 1 \end{bmatrix}.$$
so that |det(D\phi)|=3*1-(-2*1)=5.

3. Feb 7, 2010

### phucnguyen

Hi Landau,

Thanks for showing me the right way to do it. I mean I was learning the Jacobian. And before continuing with the lesson, I attempted to do it this way. I knew it was very likely that I would be wrong, but I tried it anyway, so that I could later find out where my thinking was wrong and gain more insights.

So I'm currently confused about why this is wrong. Exactly at what line?

If I think of these as approximations of deltaX and deltaY, the approximation will hold until the last line, where we can then take the limits of both sides?

I'm sorry if this is annoying but it's just my habit of looking straight into my errors and try to be clear why I'm wrong :D

Many thanks.

4. Feb 8, 2010

### torquil

If you want your calculation to give the correct result, think of it geometrically:

Assume that dxdy represents the area of a small square in the coordinate system (x,y).

Then you use a different coordinate system (u,v) on the same space. The quantity dudv represents the area of a small square in this coordinate system. A square in one coordinate system may not look much like a square in another. But the areas of those two squares are related, since you can express (u,v) as functions of (x,y). You need to find the area of dudv in terms of the area dxdy. This is what the Jacobi determinant measures, the relative change of such a small area, under a coordinate transformation.

Notice that it is not a coincidence that the number 5 appears in this vector product calculation (coinsidering dx and dy as orthogonal vectors):

du x dv = 3 dx x dx + 3 dx x dy - 2 dy x dx + 2 dy x dy
= 0 + 3 dx x dy + 2 dx x dy + 0
= 5 dx x dy

The absolute value of the cross product is of course related to areas...

Torquil