- #1
phucnguyen
- 6
- 0
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.
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.