- #1
geoduck
- 258
- 2
I just wanted to verify which order you put the rows of the Jacobian. If your initial variables are (x,y) and change to new variables u(x,y)=f(x,y) and v(x,y)=g(x,y), then you'll get a Jacobian. If this Jacobian is negative, would you change your definition to u(x,y)=g(x,y) and v(x,y)=f(x,y), since dxdy is assumed positive, so dudv needs to be positive?
So can you always get away with putting absolute values on the Jacobian?
My worry is that region of integration might be negative, but the measure positive, when making a change of variables.
An example is [itex]\int_{-5}^{6} f(x)dx [/itex] with [itex]u(x)=-x [/itex].
Then [itex]\int_{-5}^{6} f(x)dx=\int_{5}^{-6}f(-u) (-du) [/itex].
So it would have been a mistake to make the Jacobian positive:
[itex]\int_{-5}^{6} f(x)dx \neq \int_{5}^{-6}f(-u) (|-1|du) [/itex]
Can I say that in all integrals, we automatically put the lowest limit on the bottom, and the highest limit on the top, and make sure all Jacobians are positive? Would that work every time?
So can you always get away with putting absolute values on the Jacobian?
My worry is that region of integration might be negative, but the measure positive, when making a change of variables.
An example is [itex]\int_{-5}^{6} f(x)dx [/itex] with [itex]u(x)=-x [/itex].
Then [itex]\int_{-5}^{6} f(x)dx=\int_{5}^{-6}f(-u) (-du) [/itex].
So it would have been a mistake to make the Jacobian positive:
[itex]\int_{-5}^{6} f(x)dx \neq \int_{5}^{-6}f(-u) (|-1|du) [/itex]
Can I say that in all integrals, we automatically put the lowest limit on the bottom, and the highest limit on the top, and make sure all Jacobians are positive? Would that work every time?