# Legal move?

1. Mar 8, 2013

### eyesontheball1

Suppose we have the double integral of a function f(x,y) with domain of integration being some rectangular region in the 1st quadrant: 0≤a≤x≤b, 0≤c≤y≤d. Would the following transformation generally be acceptable? (I've quickly tried it out several times with arbitrary integrands and domains, with some instances in which I arrive at the correct answer, and others in which I do not; this leads me to believe that either it is a permissible change of variables but one has to be careful in certain cases, or it's generally not a permissible change of variables, hence the inconsistency in my answers.)

x=rtan(θ), y = rcot(θ), with 0≤$\sqrt{ac}$≤r≤$\sqrt{bd}$, and 0≤arctan($\sqrt{b/c}$)≤θ≤arctan($\sqrt{a/d}$)≤π/2, s.t. we have dxdy -> |J|drdθ = 2r/cos(θ)sin(θ) drdθ.

I'm unsure of this transformation because although it fails to be one to one on the boundary of the region in the r-theta plane, it's one to one everywhere else, and the determinant only fails to be nonzero at r=0, which lies on the boundary of the said region in the r-theta plane as well.

2. Mar 19, 2013

### Vargo

No, it is not quite valid.

The transformation is fine. The problem is that the rectangle in the xy plane does not transform into a rectangle in your new coordinates.

r and theta usually denote polar coordinates, which these are not, so it is clearer to change the notation a bit.

x = uv, y = u/v

I am setting u=r and v = tan(theta). Now a rectangle in the xy plane has boundaries with either x = const. or y = const.

x = const means uv=const, so you get a hyperbola in the uv plane. Changing v to tan(theta) will give you a different curve, but u and theta will still be dependent on one another.

y=const means u/v=const so you get a straight line through the origin in the u,v plane. Again u depends on v, so changing v=tan(theta) will just mean u depends on theta so you will not get the horizontal or vertical sides of a rectangle.

3. Mar 19, 2013

### eyesontheball1

Thanks Vargo!